Mathematical Morsels : IV

Posted on July 4, 2020July 4, 2020 by Nalin Pithwa

The words set and space are often used in loose context to one another. A set is merely an amorphous collection of elements, without coherence or form. When some kind of algebraic or geometric structure is imposed on a set, so that its elements are organized into a systematic whole, then it becomes a space.

Symmetric Difference is associative

Posted on July 1, 2020July 1, 2020 by Nalin Pithwa

Symmetric difference is associative

Problem Set based on VII. Complete Metric Spaces

Posted on May 31, 2020May 31, 2020 by Nalin Pithwa

Problem 1.

Prove that the limit $f(t)$ of a uniformly convergent sequence of functions $\{ f_{n}(t)\}$ continuous on $[a,b]$ is itself a function continuous on $[a,b]$ .

Hint: Clearly,

$|f(t)-f(t_{0})| \leq |f(t) - f_{n}(t)|+|f_{n}(t) - f_{n}(t_{0})|+|f_{n}(t_{0}) - f(t_{0})|$ where $t, t_{0} \in [a,b]$ .

Use the uniform convergence to make the sum of the first and third terms on the right small for sufficiently large n. Then, use the continuity of $f_{n}(t)$ to make the second term small for t sufficiently close to $t_{0}$ .

Problem 2.

Prove that if R is complete, then the intersection $\bigcap_{n=1}^{\infty}S_{n}$ figuring in the theorem 2 consists of a single point. Theorem 2 is recalled here: Nested sphere theorem: A metric space R is complete if and only if nested sequence $S_{n} = \{ S[x_{n}, r_{n}]\}$ of closed spheres in R such that $t_{n} \rightarrow 0$ as $n \rightarrow \infty$ has a non empty intersection $\bigcap_{n=1}^{\infty}S_{n}$ .

Problem 3.

Prove that the space m is complete. Recall: Consider the set of all bounded infinite sequences of real numbers $x = (x_{1}, x_{2}, \ldots, x_{k}, \ldots)$ and let $\rho(x,y) = \sup_{k}|x_{k}-y_{k}|$ . This is a metric space which we denote by m.

Problem 4:

By the diameter of a subset A of a metric space R is meant the number $d(A) = \sup_{x, y \in A} \rho(x,y)$

Suppose R is complete and let $\{ A_{n}\}$ be a sequence of closed sets of R nested in the sense that

$A_{1} \supset A_{2} \supset \ldots \supset A_{n} \supset \ldots$

Suppose further that

$\lim_{n \rightarrow \infty} d(A_{n})=0$ .

Prove that the intersection $\bigcap_{n=1}^{\infty}A_{n}$ is non empty.

Problem 5.

A subset A of a metric space R is said to be bounded if its diameter $d(A)$ is finite. Prove that the union of a finite number of bounded sets is bounded.

Problem 6.

Give an example of a complete metric space R and a nested sequence $\{ A_{n}\}$ of closed subsets of R such that

$\bigcap_{n=1}^{\infty}A_{n} = \phi$ .

Reconcile this example with Problem 4 here.

Problem 7.

Prove that a subspace of a complete metric space R is complete if and only if it is closed.

Problem 8.

Prove that the real line equipped with the distance metric function

$\rho(x,y) = |\arctan{x} - \arctan{y}|$ is an incomplete metric space.

Problem 9.

Give an example of a complete metric space homeomorphic to an incomplete metric space.

Hint: Consider the following example we encountered earlier: The function $y = f(x) = \frac{2}{\pi}\arctan{x}$ establishes a homeomorphism between the whole real line $(-\infty, \infty)$ and the open interval $(-1,1)$ .

Comment: Thus, homeomorphic metric spaces can have different metric properties.

Problem 10:

Carry out the program discussed in the last sentence of the following example:

Ir R is the space of all rational numbers, then $R^{*}$ is the space of all real numbers, both equipped with the distance $\rho(x,y) = |x-y|$ . In this way, we can “construct the real number system.” However, there still remains the problem of suitably defining sums and products of real numbers and verifying that the usual axioms of arithmetic are satisfied.

Hint: If $\{ x_{n}\}$ and $\{ y_{n}\}$ are Cauchy sequences of rational numbers serving as “representatives” of real numbers $x^{*}$ and $y^{*}$ respectively, define $x^{*} = y^{*}$ as the real number with representative $\{ x_{n} + y_{n}\}$ .

I will post the solutions in about a week’s time.

Regards,

Nalin Pithwa.

mathhothouse01@gmail.com

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Cauchy’s Mean Value Theorem and the Stronger Form of l’Hopital’s Rule

Posted on May 1, 2020May 20, 2020 by Nalin Pithwa

Reference: G B Thomas, Calculus and Analytic Geometry, 9th Indian Edition.

The stronger form of l’Hopital’s rule is as follows:

Suppose that $f(x_{0})=g(x_{0})=0$ and the functions f and g are both differentiable on an open interval $(a,b)$ that contains the point $x_{0}$ . Suppose also that $g^{'} \neq 0$ at every point in $(a,b)$ except possibly $x_{0}$ . Then,

$\lim_{x \rightarrow x_{0}} \frac{f(x)}{g(x)} = \lim_{x \rightarrow x_{0}}\frac{f^{'}(x)}{g^{'}(x)}$ …call this I, provided the limit on the right exists.

Remarks:

The proof of the stronger from of l’Hopital’s rule in based on Cauchy’s mean value theorem, a mean value theorem that involves two functions instead of one. We prove Cauchy’s theorem first and then show how it leads to l’Hopital’s rule.

Cauchy’s Mean Value Theorem:

Suppose the functions f and g are continuous on $[a,b]$ and differentiable through out $(a,b)$ and suppose also that $g^{'} \neq 0$ through out $(a,b)$ . Then, there exists a number c in $(a,b)$ at which

$\frac{f^{'}(c)}{g^{'}(c)} = \frac{f(b)-f(a)}{g(b)-g(a)}$ .

(Note this becomes the ordinary mean value theorem when $g(x)=x$ ).

Proof of Cauchy’s Mean Value theorem:

We apply the ordinary mean value theorem twice. First, we use it to show that $g(b) \neq g(a)$ . Because if $g(b)=g(a)$ , then the ordinary Mean Value theorem says that

$g^{'}(c) = \frac{g(b)-g(a0}{b-a}=0$ for some c between a and b. This cannot happen because $g^{'}(x) \neq 0$ in $(a,b)$ .

We next apply the Mean Value Theorem to the function

$F(x) = f(x)-f(a) - \frac{f(b)-f(a)}{g(b)-g(a)}(g(x)-g(a))$

This function is continuous and differentiable where f and g are, and note that $F(b)=F(a)=0$ . Therefore, by the ordinary mean value theorem, there is a number c between a and b for which $F^{'}(c)=0$ . In terms of f and g, this says

$F^{'}(c) = f^{'}(c) - \frac{f(b)-f(a)}{g(b)-g(a)}(g^{'}(c)) = 0$

or $\frac{f^{'}(c)}{g^{'}(c)} = \frac{f(b)-f(a)}{g(b)-g(a)}$ which is equation II above.

Proof of the stronger form of L’Hopital’s Rule:
We first establish equation I for the case $\lim x \rightarrow x_{0}^{+}$ . The method needs almost no change to apply to the case $\lim x \rightarrow x_{0}^{-}$ , and the combination of these two cases establishes the result.

Suppose that x lies to the right of $x_{0}$ . Then, $g^{'}(x) \neq 0$ and we can apply Cauchy’s Mean Value theorem to the closed interval from $x_{0}$ to x. This produces a number c between x and $x_{0}$ such that

$\frac{f^{'}(c)}{g^{'}(c)} = \frac{f(x)-f(x_{0})}{g(x)-g(x_{0})}$

But, $f(x_{0})=g(x_{0})=0$ so

That $\frac{f^{'}(c)}{g^{'}(c)}= \frac{f(x)}{g(x)}$

As x approaches $x_{0}$ , c approaches $x_{0}$ as it lies between x and $x_{0}$ . Therefore,

$\lim_{x \rightarrow x_{0}^{+}} \frac{f(x)}{g(x)} = \lim_{x \rightarrow x_{0}^{+}} \frac{f^{'}(c)}{g^{'}(c)} = \lim_{x \rightarrow x_{0}^{+}} \frac{f^{'}(x)}{g^{'}(x)}$ .

This establishes l’Hopital’s Rule for the case where approaches $x_{0}$ from right. The case where x approaches $x_{0}$ from the left is proved by applying Cauchy’s Mean Value Theorem to the closed interval $[x,x_{0}]$ when $x < x_{0}$ .

QED.

Regards,

Nalin Pithwa

VII. Complete Metric Spaces

Posted on April 21, 2020May 31, 2020 by Nalin Pithwa

Reference: Introductory Real Analysis by Kolmogorov and Fomin. Translated by Richard A. Silverman. Dover Publications.

Available on Amazon India and Amazon USA. This text book can be studied in parallel with Analysis of Walter Rudin.

7.1. Definition and examples:

The reader is presumably already familiar with the notion of completeness of the real line. (One good simple reference for this could be: Calculus and analytic geometry by G B Thomas. You can also use, alternatively, Advanced Calculus by Buck and Buck.)The real line is, of course, a simple example of a metric space. We now make the natural generalisation of the notion of completeness to the case of an arbitrary metric space.

DEFINITION 1:

A sequence $\{ x_{n}\}$ of points in a metric space R with metric $\rho$ is said to satisfy the Cauchy criterion if given any $\epsilon >0$ , there is an integer $N_{\epsilon}$ such that $\rho(x_{n}, x_{n^{'}})<\epsilon$ for all $n,n^{'}> N_{\epsilon}$ .

DEFINITION 2:

A subsequence $\{ x_{n}\}$ of points in a metric space R is called a Cauchy sequence (or a fundamental sequence ) if it satisfies the Cauchy criterion.

THEOREM 1:

Every convergent sequence $\{ x_{n}\}$ is fundamental.

Proof 1:

If $\{ x_{n} \}$ converges to a limit x, then, given any $\epsilon>0$ , there is an integer $N_{\epsilon}$ such that

$\rho(x_{n}, x) \leq \frac{\epsilon}{2}$ for all $n > N_{\epsilon}$ .

But, then

$\rho(x_{n}, x_{n^{'}}) \leq \rho(x_{n},x)+\rho(x_{n^{'}},x)<\epsilon$

for all $n, n^{'} >N_{\epsilon}$ . QED.

DEFINITION 3:

A metric space R is said to be complete if every Cauchy sequence in R converges to an element of R. Otherwise, R is said to be incomplete.

Example 1:

Let R be the “space of isolated points” (discrete metric space) defined as follows: Define $\rho(x,y)=0$ , if $x=y$ ; let $\rho(x,y)=1$ , when $x \neq y$ . Then, the Cauchy sequence in R are just the “stationary sequences,” that is, the sequences $\{ x_{n}\}$ all of whose terms are the same starting from some index n. Every such sequence is obviously convergent to an element of R. Hence, R is complete.

Example 2:

The completeness of the real line R is familiar from elementary analysis:

Example 3:

The completeness of the Euclidean n-space $\Re^{n}$ follows from that of $\Re^{1}$ . In fact, let

$x^{(p)} = (x_{1}^{(p)}, x_{2}^{(p)}, \ldots, x_{n}^{(p)})$ where $p = 1, 2, \ldots$

be fundamental sequence of points in $\Re^{n}$ . Then, given $\epsilon >0$ , there exists an $N_{\epsilon}$ such that

$\Sigma_{n=1}^{\infty} (x_{k}^{(p)}-x_{k}^{(q)})^{2} < \epsilon^{2}$

for all $p, q > N_{\epsilon}$ . It follows that

$|x_{k}^{(p)}-x_{k}^{(q)}|<\epsilon$ for $k=1,2,\ldots, n$ for all $p,q > N_{\epsilon}$ , that is, each $\{x_{k}^{(p)} \}$ is a fundamental sequence in $\Re^{1}$ .

Let $x = (x_{1}, \ldots, x_{n})$ where $x_{k} = \lim_{p \rightarrow \infty} x_{k}^{(p)}$

Then, obviously $\lim_{p \rightarrow \infty} x^{(p)} = x$ .

This proves the completeness of $\Re^{n}$ . The completeness of the spaces $R_{0}^{n}$ and $R_{1}^{n}$ introduced in earlier examples/blogs is proved in almost the same way. (HW: supply the details). QED.

Example 4:

Let $\{ x_{n}(t)\}$ be a Cauchy sequence in the function space $C_{[a,b]}$ introduced earlier. Then, given any $\epsilon >0$ , there is an $N_{\epsilon}$ such that

$|x_{n}(t) - x_{n^{'}}(t)|< \epsilon$ ….I

for all $n, n^{'} > N_{\epsilon}$ and all $t \in [a,b]$ . It follows that the sequence $\{ x_{n}(t)\}$ is uniformly convergent. But the limit of a uniformly convergent sequence of continuous functions is itself a continuous function (see Problem 1 following this Section). Taking the limit as $n^{'} \rightarrow \infty$ in I, we find that

$|x_{n}(t) - x(t)|\leq \epsilon$ for all $n > N_{\epsilon}$ and all $t \in [a,b]$ , that is, $\{ x_{n}(t)\}$ converges in the metric space $C_{[a,b]}$ to a function $x(t) \in C_{[a,b]}$ . Hence, $C_{[a,b]}$ is a complete metric space.

Example 5:

Next, let $x^{(n)}$ be a sequence in the space $l_{2}$ so that

$x^ {(n)} = (x_{1}^{(n)}, x_{2}^{(n)}, \ldots, x_{k}^{(n)}, \ldots)$

$\Sigma_{k=1}^{\infty}(x_{k}^{(n)})^{2} < \infty$ , where $n = 1, 2, , \ldots$

Suppose further that $\{ x^{(n)}\}$ is a Cauchy sequence. Then, given any $\epsilon > 0$ there is a $N_{\epsilon}$ such that

$\rho^{2}(x^{(n)},x^{(n^{'})}) = \Sigma_{k=1}^{\infty}(x_{k}^{(n)}-x_{k}^{(n^{'})})^{2}< \epsilon$ …let us call this II.

if $n, n^{'} > N_{\epsilon}$ .

It follows that $(x_{k}^{(n)}-x_{k}^{(n^{'})})^{2} < \epsilon$ (for $k =1, 2, \ldots$ )

That is, for every k the sequence $\{ x_{k}^{(n)}\}$ is fundamental and hence, convergent. Let

$x_{k} = \lim_{n \rightarrow \infty} x_{k}^{(n)}$ , $x = (x_{1}, x_{2}, \ldots, x_{k}, \ldots)$

Then, as we now show, x is itself a point of $l_{2}$ and moreover, $\{ x^{(n)}\}$ converges to x in the $l_{2}$ metric space, so that $l_{2}$ is a complete metric space.

In fact, the Cauchy criterion here implies that $\Sigma_{k=1}^{M}(x_{k}^{(n)}-x_{k}^{(n^{'})})^{2}<\epsilon$ for any fixed M. …let us call this III.

Holding n fixed in III, and taking the limit as $n^{'} \rightarrow \infty$ , we get

$\Sigma_{k=1}^{M}(x_{k}^{(n)}-x_{k})^{2} \leq \epsilon$ ….call this IV.

Since IV holds for arbitrary M, we can in turn take the limit of IV as $M \rightarrow \infty$ , obtaining

$\Sigma_{k=1}^{\infty}(x_{k}^{(n)}-x_{k})^{2} \leq \epsilon$ .

But, as we have learnt earlier in this series of blogs, the convergence of the two series $\Sigma_{k=1}^{\infty}(x_{k}^{(n)})^{2}$ and $\Sigma_{k=1}^{\infty}(x_{k}^{(n)}-x_{k})^{2}$ implies that of the series $\Sigma_{k=1}^{\infty}x_{k}^{2}$ .

This proves that $x \in l_{2}$ . Moreover, since $\epsilon$ is arbitrarily small, III implies that

$\lim_{n \rightarrow \infty}\rho( x^{(n)} ,x ) = \lim_{n \rightarrow \infty} \sqrt{\Sigma_{k=1}^{\infty} (x_{k}^{(n)}-x_{k})^{2}} = 0$

That is, $\{ x^{(n)}\}$ converges to x in the $l_{2}$ metric space, as asserted.

QED.

Example 6.

Consider the space $C_{[a,b]}^{2}$ . To recap: consider the set of all functions continuous on the closed interval $[a,b]$ with the distance metric defined by: $\rho(x,y) = (\int_{a}^{b}|x(t)-y(t)|^{2}dt)^{\frac{1}{2}}$ .

It is easy to show that the space $C_{[a,b]}^{2}$ is incomplete. If

$\phi_{n}(t) = -1$ , if $-1 \leq t \leq -\frac{1}{n}$

$\phi_{n}(t) = nt$ , if $-\frac{1}{n} \leq t \leq \frac{1}{n}$

$\phi_{n}(t) = 1$ , if $\frac{1}{n} \leq t \leq 1$

then $\{ \phi_{n}(t)\}$ is a fundamental sequence in $C_{[a,b]}^{2}$ since

$\int_{-1}^{1}(\phi_{n}(t)-\phi_{n^{'}}(t))^{2} dt \leq \frac{2}{\min{ \{ n,n^{'}}\}}$

However, $\{ \phi_{n}(t)\}$ cannot converge to a function in $C_{[-1,1]}^{2}$ . In fact, consider the discontinuous function

$\psi(t) = -1$ , when $t \leq 0$

$\psi(t) = 1$ , when $t \geq 0$ .

Then, given any function $f \epsilon C_{[-1,1]}^{2}$ , it follows from Schwarz’s inequality (obviously still valid for piecewise continuous functions) that

$(\int_{-1}^{1}(f(t)-\psi(t))^{2}dt)^{\frac{1}{2}} \leq (\int_{-1}^{1}(f(t)-\phi_{n}(t))^{2}dt)^{\frac{1}{2}} + (\int_{-1}^{1}(\phi_{n}(t) - \psi(t))^{2}dt)^{\frac{1}{2}}$

But the integral on the left is non-zero, by the continuity of f, and moreover, it is clear that

$\lim_{n \rightarrow \infty}\int_{-1}^{1}(\phi_{n}(t)-\psi(t))^{2}dt = 0$

Therefore, $\int_{-1}^{1}(f(t)-\phi_{n}(t))^{2}dt$ cannot converge to zero as $n \ rightarrow \infty$ .

QED.

7.2 The nested sphere theorem.

A sequence of closed spheres $S[x_{1}, r_{1}], S[x_{2}, r_{2}], \ldots, S[x_{n}, r_{n}], \ldots$

in a metric space R is said to be nested (or decreasing) if

$S[x_{1}, r_{1}] \supset S[x_{2}, r_{2}] \supset \ldots \supset S[x_{n}, r_{n}] \supset \ldots$

Using this concept, we can prove a simple criterion for the completeness of R:

THEOREM 2: The Nested Sphere Theorem:

A metric space R is complete if and only if every nested sequence $\{ S_{n}\} = \{ S_{[x_{n}, r_{n}]}\}$ of closed spheres in R such that $r_{n} \rightarrow 0$ as $n \rightarrow \infty$ has a non empty intersection

$\bigcap_{n=1}^{\infty}S_{n}$ .

Proof of the nested theorem:

Part I: Assume that R is complete and that if $\{ S_{n}\} = \{ S[x_{n}, r_{n}] \}$ is any nested sequence of closed spheres in R such that $r_{n} \rightarrow 0$ as $n \rightarrow \infty$ , then the sequence $\{ x_{n}\}$ of centres of the spheres is fundamental because

$\rho(x_{n}, x_{n^{'}}) < r_{n}$ for $n^{'}>n$ and $r_{n} \rightarrow 0$ as $n \rightarrow \infty$ . Therefore, $\{ x_{n} \}$ has a limit. Let

$x = \lim_{n \rightarrow \infty} x_{n}$ .

Then, $x \in \bigcap_{n=1}^{\infty}S_{n}$

Not only that, we can in fact say that $S_{n}$ contains every point of the sequence $\{ x_{n}\}$ except possibly the points $x_{1}, x_{2}, \ldots, x_{n-1}$ and hence, x is a limit point of every sphere $S_{n}$ . But, $S_{n}$ is closed, and hence, $x \in S_{n}$ for all n.

Conversely, suppose every nested sequence of closed spheres in R with radii converging to zero has a non empty intersection, and let $\{ x_{n}\}$ be any fundamental sequence in R. Then, x has a limit point in R. To see this, use the fact that $\{x_{n}\}$ is fundamental to choose a term $x_{n_{1}}$ of the sequence $\{ x_{n} \}$ such that

$\rho(x_{n}, x_{n_{1}}) < \frac{1}{2}$ for all $n \geq n_{1}$ , and let $S_{1}$ be the closed sphere of radius 1 with centre $x_{n_{1}}$ . Then, choose a term $x_{n_{1}}$ of $\{ x_{n}\}$ such that $n_{2} > n_{1}$ and $\rho(x_{n}, x_{n_{1}}) < \frac{1}{2^{2}}$ for all $n > n_{2}$ , and let $S_{2}$ be the closed sphere of radius $\frac{1}{2}$ with centre $x_{n_{2}}$ .

Continue this construction indefinitely, that is, once having chosen terms $x_{n_{1}}, x_{n_{2}}, \ldots, x_{n_{k}}$ (where $n_{1}<n_{2}< \ldots n_{k}$ ), choose a term $x_{n_{k+1}}$ such that $n_{k+1}>n_{k}$ and

$\rho(x_{n}, x_{n_{k+1}}) < \frac{1}{2^{k+1}}$ for all $n \geq n_{k+1}$ .

Let $S_{k+1}$ be the closed sphere of radius $\frac{1}{2^{n}}$ with centre $x_{n_{k+1}}$ , and so on. This gives a nested sequence $S_{n}$ of closed spheres with radii converging to zero. By hypothesis, these spheres have a non empty intersection, that is, there is a point x in all the spheres. This point is obviously the limit of the sequence $\{ x_{n}\}$ . But, if a fundamental sequence contains a subsequence converging to x, then the sequence itself must converge to x (HW quiz). That is,

$\lim_{n \rightarrow \infty}x_{n} =x$ .

QED.

7.3 Baire’s theorem:

We know that a subset A of a metric space R is said to be nowhere dense in R if it is dense in no (open) sphere at all, or equivalently, if every sphere $S \subset R$ contains another sphere $S^{'}$ such that $S^{'} \bigcap A = \phi$ . (Quiz: check the equivalence).

This concept plays an important role in the following:

THEOREM 3: Baire’s Theorem:

A complete metric space R cannot be represented as the union of a countable number of nowhere dense sets.

Proof of Theorem 3: Baire’s Theorem:

Suppose the contrary. Let $R = \bigcup_{n=1}^{\infty}A_{n}$ ….call this VI.

where every set $A_{n}$ is nowhere dense in R. Let $S_{0} \subset R$ be a closed sphere of radius 1. Since $A_{1}$ is nowhere dense in $S_{0}$ , being nowhere dense in R, there is a closed sphere $S_{1}$ of radius less than $\frac{1}{2}$ such that $S_{1} \subset S_{0}$ and $S_{1} \bigcap A_{1} =\phi$ . Since $A_{2}$ is nowhere dense in $S_{1}$ , being nowhere dense in $S_{0}$ , there is a closed sphere $S_{2}$ of radius less than $\frac{1}{3}$ such that $S_{2} \subset S_{1}$ and $S_{2} \bigcap A_{2} = \phi$ , and so on. In this way, we get a nested sequence of closed spheres $\{ S_{n}\}$ with radii converging to zero such that

$S_{n} \bigcap A_{n} = \phi$ , where $n=1,2,3,\ldots$

By the nested sphere theorem, the intersection $\bigcap_{n=1}^{\infty}S_{n}$ contains a point x. By construction, x cannot belong to any of the sets $A_{n}$ , that is,

$x \notin \bigcup_{n=1}^{\infty}A_{n}$

It follows that $R \neq \bigcup_{n=1}^{\infty} A_{n}$ contrary to VI.

Hence, the representation VI is impossible.

QED.

COROLLARY TO Baire’s theorem:

A complete metric space R without isolated points is uncountable.

Proof:

Every single element set $\{ x\}$ is nowhere dense in R.

QED.

7.4 Completion of a metric space:

As we now show, an incomplete metric space can always be enlarged (in an essentially unique way) to give a complete metric space.

DEFINITION 4: Completion of a metric space:

Given a metric space R with closure $[R]$ , a complete metric space $R^{*}$ is called a completion of R if $R \subset R^{*}$ and $[R]=R^{*}$ , that is, if R is a subset of $R^{*}$ everywhere dense in $R^{*}$ .

Example 1.

Clearly, $R^{*}=R$ if R is already complete. (Quiz: homework).

Example 2:

The space of all real numbers is the completion of the space of all rational numbers.

THEOREM 4:

Every metric space R has a completion. This completion is unique to within an isometric mapping carrying every point $x \in R$ into itself.

Proof of Theorem 4:

(The proof is somewhat lengthy but quite straight forward).

First , we prove the uniqueness showing that if $R^{*}$ and $R^{**}$ are two completions of R, then there is a one-to-one mapping $x^{**} = \phi(x^{*})$ onto $R^{**}$ such that $\phi(x)=x$ for all $x \in R$ and

$\rho_{1}(x^{*}, y^{*}) = \rho_{2}(x^{**}, y^{**})$ ….call this VII.

( $y^{**}=\phi(y^{*})$ ), where $\rho_{1}$ is the distance metric in $R^{*}$ and $\rho_{2}$ the distance metric in $R^{**}$ . The required mapping $\phi$ is constructed as follows: Let $x^{*}$ be an arbitrary point of $R^{*}$ . Then, by the definition of a completion, there is a sequence $\{ x_{n}\}$ of points of R converging to $x^{*}$ . The points of the sequence $\{ x_{n}\}$ also belong to $R^{**}$ , where they form a fundamental sequence (quiz: why?). Therefore, $\{ x_{n}\}$ converges to a point $x^{**} \in R^{**}$ since $R^{**}$ is complete. It is clear that $x^{**}$ is independent of the choice of the sequence $\{ x_{n}\}$ converging to the point $x^{*}$ (homework quiz: why?). If we set $\phi(x^{*})=x^{**}$ , then $\phi$ is the required mapping. In fact, $\phi(x)=x$ for all $x \in R$ , since if $x_{n} \rightarrow x \in R$ , then obviously $x = x^{*} \in R^{*}$ , $x^{**}=x$ . Moreover, suppose $x_{n} \rightarrow x^{*}$ , $y_{n} \rightarrow y^{*}$ in $R^{*}$ , while $x_{n} \rightarrow x^{**}$ , $y_{n} \rightarrow y^{**}$ . Then, if $\phi$ is the distance in R,

$\rho_{1}(x^{*},y^{*}) = \lim_{n \rightarrow \infty} \rho(x_{n}, y_{n}) = \lim_{n \rightarrow \infty}(x_{n}, y_{n})$ …call this VIII.

While at the same time, $\rho_{2}(x^{**}, y^{**})=\lim_{n \rightarrow \infty}\rho_{2}(x_{n}, y_{n})=\lim_{n \rightarrow \infty}\rho(x_{n}, y_{n})$ ….call this VIII-A. But VIII and VIII-A imply VII.

We must now prove the existence of a completion of R. Given an arbitrary metric space R, we say that two Cauchy sequences $\{x_{n} \}$ and $\{\overline{x}_{n} \}$ in R are equilvalent and write $\{ x_{n}\} \sim \{ \overline{x}_{n}\}$

if $\lim_{n \rightarrow \infty} \rho(x_{n}, \overline{x}_{n})=0$

As anticipated by the notation and terminology, $\sim$ is reflective, symmetric and transitive, that is, $\sim$ is an equivalence relation. Therefore, the set of all Cauchy sequences of points in the space R can be partitioned into classes of equivalent sequences. Let these classes be the points of a new space $R^{*}$ . Then, we define the distance between two arbitrary points $x^{*}, y^{*}$ by the formula

$\rho_{1}(x^{*}, y^{*}) = \lim_{n \rightarrow \infty} \rho(x_{n}, y_{n})$ ….call this IX.

where $\{ x_{n}\}$ is any “representative” of $x^{*}$ (namely, any Cauchy sequence in the class $x^{*}$ ) and $\{ y_{n}\}$ is any representative of $y^{*}$ .

The next step is to verify that IX is indeed a distance metric. That is, also to check that IX exists, independent of the choice of sequence $\{x_{n} \} \in x^{*}$ , $\{ y_{n} \} \in y^{*}$ , and satisfies the three properties of a distance metric function. Given any $\epsilon >0$ , it follows from the triangle inequality in R (this can be proved with a little effort: homework quiz) that

$|\rho(x_{n}, y_{n}) - \rho(x_{n^{'}},y_{n^{'}})| = |\rho(x_{n},y_{n}) -\rho(x_{n^{'}},y_{n}) + \rho(x_{n^{'}},y_{n}) - \rho(x_{n^{'}}, y_{n^{'}})|$

That is, $\leq |\rho(x_{n},y_{n}) -\rho(x_{n^{'}}, y_{n}) | + |\rho(x_{n^{'}}, y_{n}) - \rho(x_{n^{'}}, y_{n^{'}}) |$

that is, $\leq \rho(x_{n}, x_{n^{'}}) + \rho(y_{n}, y_{n^{'}}) \leq \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon$ ….call this X

for all sufficiently large $n$ and $n^{'}$ .

Therefore, the sequence of real numbers $\{ x_{n}\} = \{ \rho(x_{n},y_{n})\}$ is fundamental and hence, has a limit. This limit is independent of the choice of $\{x_{n} \} \in x^{*}$ , $\{ y_{n}\} \in y^{*}$ . In fact, suppose that

$\{ x_{n}\}, \{ \overline{x}_{n}\} \in x^{*}$ , $\{ y_{n}\}, \{ \overline{y}_{n} \in y^{*}$

Then,

$|\rho(x_{n}, y_{n}) - \rho(\overline(x)_{n}, \overline{y}_{n})| \leq \rho(x_{n}, \overline{x}_{n}) + \rho(y_{n}, \overline{y}_{n})$ by a calculation analogous to X. But,

$\lim_{n \rightarrow \infty} \rho(x_{n}, \overline{x}_{n}) = \lim_{n \rightarrow \infty}(y_{n}, \overline{y}_{n})=0$

since $\{ x_{n} \} \sim \{ \overline{x}_{n}\}$ and $\{ y_{n} \} \sim \{ \overline{y}_{n}\}$ , and hence,

$\lim_{n \rightarrow \infty} \rho(x_{n}, y_{n}) = \lim_{n \rightarrow \infty} \rho(\overline{x}_{n}, \overline{y}_{n})$ .

As for the three properties of a metric, it is obvious that

$\rho_{1}(x^{*}, y^{*}) = \rho_{1}(y^{*}, x^{*})$ , and the fact that

$\rho_{1}(x^{*}, y^{*}) =0$ if and only if $x^{*}=y^{*}$ is an immediate consequence of the definition of equivalent Cauchy sequences.

To verify the triangle inequality in $R^{*}$ , we start from the triangle inequality:

$\rho(x_{n}, z_{n}) \leq \rho(x_{n}, y_{n}) + \rho(y_{n}, z_{n})$

in the original space R, and then take the limit as $n \rightarrow \infty$ , obtaining

$\lim_{n \rightarrow \infty} rho(x_{n}, z_{n}) \leq \lim_{n \rightarrow \infty} \rho(x_{n}, y_{n}+ \lim_{n \rightarrow \infty} \rho(y_{n}, z_{n})$

That is, $\rho_{1}(x^{*}, z^{*}) \leq \rho_{1}(x^{*}, y^{*}) + \rho_{1}(y^{*}, z^{*})$

We now come to the crucial step of showing that $R^{*}$ is a completion of R. Suppose that with every point $x \in R$ , we associate the class $x^{*} \in R^{*}$ of all Cauchy sequences converging to x. Let

$x = \lim_{n \rightarrow \infty} x_{n}$ , $y = \lim_{n \rightarrow \infty} y_{n}$

Then, clearly $\rho(x,y) = \lim_{n \rightarrow \infty}(x_{n}, y_{n})$

(the above too can be proven with a slight effort: HW quiz); while, on the other hand,

$\rho_{1}(x^{*}, y^{*}) = \lim_{n \rightarrow \infty} \rho(x_{n}, y_{n})$ by definition. Therefore,

$\rho(x,y) = \rho_{1}(x^{*}, y^{*})$ and hence, the mapping of R into $R^{*}$ carrying x into $x^{*}$ is isometric. Accordingly, we need no longer distinguish between the original space R and its image in $R^{*}$ , in particular between the two metrics $\rho$ and $\rho_{1}$ . In other words, R can be regarded as a subset of $R^{*}$ . The theorem will be proved once we succeed in showing that

(i) R is everywhere dense in $R^{*}$ , that is, $[R] = R$ ;

(2) $R^{*}$ is complete.

Towards that end, given any point $x^{*} \in R^{*}$ and any $\epsilon >0$ , choose a representative of $x^{*}$ , namely a Cauchy sequence $\{ x_{n}\}$ in the class $x^{*}$ . Let N be such that $\rho(x_{n}, x_{n^{'}}) < \epsilon$ for all $n, n^{'} >N$ . Then,

$\rho(x_{n}, x^{*}) = \lim_{n \rightarrow \infty} \rho(x_{n}, x_{n^{'}}) \leq \epsilon$ if $n >N$ , that is, every neighbourhood of the point $x^{*}$ contains a point of R. It follows that $[R] = R$ .

Finally, to show that $R^{*}$ is complete, we first note that by the very definition of $R^{*}$ , any Cauchy sequence $\{ x_{n}\}$ consisting of points in R converges to some point in $R^{*}$ , namely to the point $x^{*} \in R^{*}$ defined by $\{ x_{n}\}$ . Moreover, since R is dense in $R^{*}$ , given any Cauchy sequence $x_{n}^{*}$ consisting of points in $R^{*}$ , we can find an equivalent sequence $\{ x_{n}\}$ consisting of points in R. In fact, we need only choose $x_{n}$ to be any point of R such that $\rho(x_{n}, x_{n}^{*}) < \frac{1}{n}$ . The resulting sequence $\{ x_{n}\}$ is fundamental, and, as just shown, converges to a point $x^{*} \in R^{*}$ . But, then the sequence $x_{n}^{*}$ also converges to $x^{*}$ .

QED.

Example.

If R is the space of all rational numbers, then $R^{*}$ is the space of all real numbers, both equipped with the distance $\rho(x,y) = |x-y|$ . In this way, we can “construct the real number system.” However, there still remains the problem of suitably defining sums and products of real numbers and verifying that the usual axioms of arithmetic are satisfied.

Regards,

Nalin Pithwa.

mathhothouse01@gmail.com

+91 97426 00025

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Problem Set based on VI. Convergence, open and closed sets.

Posted on April 5, 2020April 13, 2020 by Nalin Pithwa

Problem 1.

Give an example of a metric space R and two open spheres $S(x,r_{1})$ , $S(y, r_{2})$ in R such that $S(x, r_{1}) \subset S(y,r_{2})$ although $r_{1}> r_{2}$ .

Problem 2:

Prove that every contact point of a set M is either a limit point of M or is an isolated point of M.

Comment. In particular, $[M]$ can only contain points of the following three types:

a) Limit points of M belonging to M.

b) Limit points of M which do not belong to M.

c) Isolated points of M.

Thus, $[M]$ is the union of M and the set of all its limit points.

Problem 3:

Prove that if $x_{n}\rightarrow x$ and $y_{n} \rightarrow y$ as $n \rightarrow \infty$ then $\rho(x_{n},y_{n}) \rightarrow \rho(x,y)$ .

Hint : use the following problem: Given a metric space $(X,\rho)$ prove that $|\rho(x,z)-\rho(y,u)| \leq |\rho(x,y)|+|\rho(z,u)|$

Problem 4:

Let f be a mapping of one metric space X into another metric space Y. Prove that f is continuous at a point $x_{0}$ if and only if the sequence $\{ y_{n}\} = \{ f(x_{n})\}$ converges to $y=f(x_{0})$ whenever the sequence $x_{n}$ converges to $x_{0}$ .

Problem 5:

Prove that :

(a) the closure of any set M is a closed set.

(b) $[M]$ is the smallest closed set containing M.

Problem 6:

Is the union of infinitely many closed sets necessarily closed? How about the intersection of infinitely many open sets? Give examples.

Problem 7:

Prove directly that the point $\frac{1}{4}$ belongs to the Cantor set F, although it is not the end point of any of the open interval deleted in constructing F. Hint: The point $\frac{1}{4}$ divides the interval $[0,1]$ in the ratio $1:3$ and so on.

Problem 8:

Let F be the Cantor set. Prove that

(a) the points of the first kind form an everywhere dense subset of F.

(b) the numbers of the form $t_{1}+t_{2}$ where $t_{1}, t_{2} \in F$ fill the whole interval $[0,2]$ .

Problem 9:

Given a metric space R, let A be a subset of R, and $x \in R$ . Then, the number $\rho(A,x) = \inf_{a \in A}\rho(a,x)$ is called the distance between A and x. Prove that

(a) $x \in A$ implies $\rho(A,x)=0$ but not conversely

(b) $\rho(A,x)$ is a continuous function of x (for fixed A).

(d) $[A]=A \bigcup M$ , where M is the set of all points x such that $\rho(A,x)=0$ .

Problem 10:

Let A and B be two subsets of a metric space R. Then, the number $\rho(A,B)= \inf_{a \in A, b \in B}\rho(a,b)$ is called the distance between A and B. Show that $\rho(A,B)=0$ if $A \bigcap B \neq \phi$ , but not conversely.

Problem 11:

Let $M_{K}$ be the set of all functions f in $C_{[a,b]}$ satisfying a Lipschitz condition, that is, the set of all f such that $|f(t_{1}-f(t_{2})| \leq K|t_{1}-t_{2}|$ for all $t_{1}, t_{2} \in [a,b]$ , where K is a fixed positive number. Prove that:

a) $M_{K}$ is closed and in fact is the closure of the set of all differentiable functions on $[a,b]$ such that $|f^{'}(t)| \leq K$

(b) the set $M = \bigcup_{K}M_{K}$ of all functions satisfying a Lipschitz condition for some K is not closed;

Problem 12:

An open set G in n-dimensional Euclidean space $R^{n}$ is said to be connected if any points $x,y \in G$ can be joined by a polygonal line(by a polygonal line we mean a curve obtained by joining a finite number of straight line segments end to end.) lying entirely in G. For example, the open disk $x^{2}+y^{2}<1$ is connected, but not the union of the two disks $x^{2}+y^{2}<1$ , $(x-2)^{2}+y^{2}<1$ (even though they share a contact point). An open subset of an open set G is called a component of G if it is connected and is not contained in a larger connected subset of G. Use Zorn’s lemma to prove that every open set G in $R^{n}$ is the union of no more than countably many pairwise disjoint components.

Comment: In the case $n=1$ , that is, the case on the real line, every connected open set is an open interval, possibility one of the infinite intervals $(-\infty, \infty)$ , $(a, \infty)$ or $(-\infty, b)$ . Thus, theorem 6 (namely: Every open set G on the real line is the union of a finite or countable system of pairwise disjoint open intervals) on the structures of open sets on the line is tantamount to two assertions:

(i) Every open set on the line is the union of a finite or countable number of components.

(ii) Every open connected set on the line is an open interval.

The first assertion holds for open sets in $R^{n}$ (and, in fact, is susceptible to further generalizations), while the second assertion pertains specifically to the real line.

Cheers,

Happy analysis !!

Nalin Pithwa

VI Convergence. Open and Closed Sets.

Posted on January 16, 2020April 5, 2020 by Nalin Pithwa

Reference: Introductory Real Analysis by Kolmogorov and Fomin.

Reference: Introduction to Analysis by Rosenlicht.

Reference: Analysis by Walter Rudin.

Sec 6.1 Closure of a set. Limit points.

By the open sphere (or open ball) $S(x_{0},r)$ in a metric space R we mean the set of points $x \in R$ satisfying the inequality $\rho(x_{0},r) <r$

( $\rho$ is the metric of R.)

Note: Any confusion between “sphere” meant in the sense of spherical surface and “sphere” meant in the sense of a solid sphere (or ball) will always be avoided by judicious use of the adjectives “open” or “closed.”

The fixed point $x_{0}$ is called the center of the sphere and the number r is called its radius. By the closed sphere (or closed ball) $S[x_{0},r]$ with center $x_{0}$ and radius r we mean the set of points $x \in R$ satisfying the inequality

$\rho(x_{0},x) \leq r$

An open sphere of radius $\epsilon$ with center $x_{0}$ will also be called an $\epsilon$ –neighbourhood of $x_{0}$ denoted by $O_{\epsilon}(x_{0})$ .

A point $x \in R$ is called a contact point of a set $M \subset R$ if every neighbourhood of x contains at least one point of M. The set of all contact points of a set M is denoted by $[M]$ and is called the closure of M. Obviously, $M \subset [M]$ , since every point of M is a contact point of M. By the closure operator in a metric space R, we mean the mapping of R carrying each set $M \subset R$ into its closure $[M]$ .

Theorem 1:

The closure operator has the following properties:

If $M \subset N$ , then $[M] \subset [N]$
$[[M]] = [M]$
$[M \bigcup N] = [M] \bigcup [N]$
$[\phi] = \phi$ .

Proof I:

Property I is obvious.

Proof of property 2:

Let $x \in [[M]]$ . Then, any given neighbourhood $O_{\epsilon}(x)$ contains a point $x_{1} \in [M]$ . Consider the sphere $O_{\epsilon_{1}}(x_{1})$ of radius

$\epsilon_{1} = \epsilon - \rho(x,x_{1})$

PS: It helps to draw an illustrative diagram to understand this proof.

Clearly, $O_{\epsilon_{1}}(x_{1})$ is contained in $O_{\epsilon}(x)$ . In fact, if $z \in O_{\epsilon_{1}}(x_{1})$ then $\rho(z,x_{1}) < \epsilon_{1}$ and hence, since $\rho(x,x_{1}) = \epsilon - \epsilon_{1}$ , it follows from the triangle inequality that

$\rho(z,x) < \epsilon_{1} + (\epsilon - \epsilon_{1}) = \epsilon$ ,

That is, $z \in O_{\epsilon}(x)$ . Since $x_{1} \in [M]$ , there is a point $x_{2} \in M$ in $O_{\epsilon_{1}}(x)$ . But then $x_{2} \in O_{\epsilon}(x)$ and hence, $x \in [M]$ , since $O_{\epsilon}(x)$ is an arbitrary neighbourhood of x. Therefore, $[[M]] \subset [M]$ . But, obviously $[M] \subset [[M]]$ and hence, $[[M]] = [M]$ , as required.

Proof of property 3:

Let $x \in [M \bigcup N]$ and suppose $x \notin [M] \bigcup [N]$ . Then, $x \notin [M]$ and $x \notin [N]$ . But then there exist neighbourhoods $O_{\epsilon_{1}}(x)$ and $O_{\epsilon_{2}}(x)$ such that $O_{\epsilon_{1}}(x)$ contains no points of M while $O_{\epsilon_{1}}(x)$ contains no points of N. It follows that the neighbourhood $O_{\epsilon}(x)$ , where $\epsilon=\min (\epsilon_{1}, \epsilon_{2})$ , contains no points of either M or N, and hence no points of $M \bigcup N$ , contrary to the assumption that $x \in [M \bigcup N]$ . Therefore, $x \in [M] \bigcup [N]$ , and hence,

$[M \bigcup N] = [M] \bigcup [N]$ …call this I

since x is an arbitrary point of $[M \bigcup N]$ . On the other hand, since $M \subset M \bigcup N$ and $N \subset M \bigcup N$ , it follows that from property (i) that $[M] \subset [M \bigcup N]$ and $[N] \subset [M \bigcup N]$ . But then,

$[M] \bigcup [N] \subset [M \bigcup N]$

which together with (i) implies that $[M \bigcup N] = [M] \bigcup [N]$ .

Proof of property 4:

We observe that given any $M \subset R$ ,

$[M] = [M \bigcup \phi] = [M] \bigcup [\phi]$ by property (iii).

It follows that $[\phi] \subset [M]$ . But this is possible for arbitrary M only if $[\phi] = \phi$ . (Alternatively, the set with no elements can have no contact points!). QED.

Definition:

A point $x \in R$ is called a limit point of a set $M \subset R$ if every neighbourhood of x contains infinitely many points of M. The limit point may or may not belong to M. For example, if M is the set of rational numbers in the interval $[0,1]$ , then every point of $[0,1]$ , rational or irrational, is a limit point of M. (It is said that the rationals are dense in R; we will discuss this in detail later).

A point x belonging to a set M is called an isolated point of M if there is a (“sufficiently small”) neighbourhood of x containing no points of M other than x itself.

6.2 Convergence and limits.

A sequence of points $\{x_{n}, n \in N \}$ in a metric space R is said to converge to a point $x \in R$ if every neighbourhood $O_{\epsilon}(x)$ of x contains all points $x_{n}$ starting from a certain index (more exactly, if, given any $\epsilon >0$ , there is an integer $N_{\epsilon}$ such that $O_{\epsilon}(x)$ contains all points $x_{n}$ with $n > N_{\epsilon}$ ). The point x is called the limit of the sequence $\{ x_{n}\}$ , and we write $x_{n} \rightarrow x$ (as $n \rightarrow \infty$ ). Clearly, $\{ x_{n}\}$ converges to x if and only if

$\lim_{n \rightarrow \infty} \rho(x,x_{n}) = 0$ .

It is an immediate consequence of the definition of a limit of a sequence that

(1) No sequence can have two distinct limits.

(2) If a sequence $\{ x_{n}\}$ converges to a point x, then so does every subsequence of $\{ x_{n}\}$ .

(Details are filled in here: I believe it is helpful to write little proofs of little details…I call it playing with small pearls of math…: )

Proof of (1) : Let, if possible, a sequence $\{ x_{n}\}$ have two distinct limit points, namely, $x_{0}$ and $x_{0}^{'}$ . Then, there exists a neigbourhood or open ball $O_{\epsilon/2}(x)$ which contains all points of the given sequence after some $N_{0} \in \mathscr{N}$ . So also there exists a neighbourhood (again of radius $\epsilon/2$ about $x_{0}^{'}$ such that it contains all points of the given sequence after some natural number $N_{0}^{'} \in \mathscr{N}$ .

But, we know that $\rho(x_{0},x_{0}^{'}) \leq \rho(x_{0},x) + \rho(x_{0}^{'},x) \leq \epsilon/2+\epsilon/2=\epsilon$ . Which in turn means that the two “distinct limit points” lie in the same open ball. Hence, they cannot be distinct. Hence, a sequence can have only one limit point, and it is unique. QED.

Proof of (2):

Definition of subsequence: Given a sequence $\{ x_{n}\}$ , where $n \in \mathscr{N}$ , consider a strictly increasing sequence of positive integers (that is, $n_{1}, n_{2}, n_{3}, \ldots$ are positive integers and $n_{1}<n_{2}<n_{3}< \ldots$ ) : then the sequence $x_{n_{1}}, x_{n_{2}}, x_{n_{3}}, \ldots$ is called a subsequence of the original sequence $\{ x_{n}\}$ .

Now, the claim is that: if a given sequence $\{ x_{n}\}$ converges to a limit point, so does every subsequence of the given sequence: —-

So, it is given that $\rho(x_{n},x) < \epsilon$ when $n > N$ for some natural number $N \in \mathscr{N}$ . Since $n_{m} \geq m$ ( here m is some dummy variable, some appropriate positive integer) for all positive integers m, we have $\rho(x, x_{n_{m}})<\epsilon$ whenever $m > N$ . This means that $\rho(x_{n_{m}},x)<\epsilon$ , which is what we wanted to prove. QED.

THEOREM 2:

A necessary and sufficient condition for a point x to be a contact point of a set M is that there exists a sequence $\{ x_{n}\}$ of points of M converging to x.

PROOF 2:

The condition is necessary since if x is a contact point of M, then every neighbourhood $O_{\frac{1}{n}}(x)$ contains at least one point $x_{n} \in M$ , and these points form a sequence $\{ x_{n}\}$ converging to M. The sufficiency is obvious. QED.

THEOREM 2′:

A necessary and sufficient condition for a point x to be a limit point of a set M is that there exists a sequence $\{ x_{n}\}$ of distinct points of M converging to x.

Proof 2′:

Clearly, if x is a limit point of M, then the points $x_{n} \in O_{\frac{1}{n}}(x) \bigcap M$ figuring in the proof of Theorem 2 (just above) can be chosen to be distinct. This proves the necessity and the sufficiency is again obvious. QED.

6.3 Dense subsets, Separable spaces:

Let A and B be two subsets of a metric space R. Then, A is said to be dense in B if $[A] \supset B$ . In particular, A is said to be everywhere dense (in R) if $[A] = R$ . A set A is said to be nowhere dense if it is dense in no (open) sphere at all. (Note; it is w.r.t. to the metric function defined in the space R.) (Famous example: The rationals are dense in $\Re$ ).

Example 1:

The set of all rational points is dense in the real line $\Re$ .

proof 1:

We want to prove that $[Q] \supset \Re$ . Clearly, as $Q \subset R$ , the set of rationals are dense in rationals themselves. We only need to show that any irrational number is also a limit point of a sequence of rationals. For example, this is crystal clear: consider the sequence of decimal numbers (finitely many decimals) that we get by extracting square root of 2 by long division. The same applies to any other surd or irrational number. Hence, we can always find a rational sequence of numbers converging to an irrational number. Hence, the rationals are dense in the reals. QED.

Example 2:

The set of all points $x = (x_{1}, x_{2}, x_{3}, \ldots, x_{n})$ with rational co-ordinates is dense in each of the spaces $R^{n}$ , $R_{0}^{n}$ , $R_{1}^{n}$ introduced in the following examples (in earlier blog, and presented here again for handy reference):

2a’) the set of all ordered n-tuples $x=(x_{1}, x_{2}, x_{3}, \ldots, x_{n})$ of real numbers $x_{1}, x_{2}, x_{3}, \ldots, x_{n}$ with distance $\rho(x,y) = \sqrt{\Sigma_{k=1}^{n}(x_{k}-y_{k})^{2}}$ is a metric space denoted by $R^{n}$ and called n-dimensional Euclidean space (or, simply Euclidean n-space).

proof 2a’:

same proof as example 1. QED.

2b) the set of all ordered n-tuples $x=(x_{1}, x_{2}, x_{3}, \ldots, x_{n})$ of real numbers $x_{1}, x_{2}, x_{3}, \ldots, x_{n}$ with distance $\rho_{1}(x,y) = \Sigma_{k=1}^{n}|x_{k}-y_{k}|$ is a metric space denoted by $R_{1}^{n}$ .

proof 2b:

same proof as example 1. QED.

2c) The set of all ordered n-tuples $x=(x_{1}, x_{2}, x_{3}, \ldots, x_{n})$ of real numbers $x_{1}, x_{2}, x_{3}, \ldots, x_{n}$ with distance $\rho_{0}(x,y) = \max_{1 \leq k \leq n}|x_{k}-y_{k}|$ is a metric space denoted by $R_{0}^{n}$ .

proof 2c:

once again same proof as example 1 with minor differences. QED.

Example 3:

The set of all points $x = (x_{1}, x_{2}, x_{3}, \ldots, x_{k}, \ldots)$ with only finitely many non-zero co-ordinates, each a rational number, is dense in the space $l_{2}$ introduced in earlier blog (and presented here again for handy reference):

Definition of $l_{2}$ metric space: Let $l_{2}$ be the set of all infinite sequences $x = (x_{1}, x_{2}, x_{3}, \ldots, x_{k}, \ldots)$ of real numbers $x_{1}, x_{2}, \ldots, x_{k}, \ldots$ satisfying the convergence condition $\Sigma_{k=1}^{\infty} x_{k}^{2}< \infty$ , where distance between points is defined by $\rho(x,y) = \sqrt{\Sigma_{k=1}^{\infty}(x_{k}-y_{k})^{2}}$ .

Again, some detailed steps show that the essence of this proof is also the fact that the rationals are dense in the reals.

Example 4: So, also, similarly, the set of all polynomials with rational coefficients is dense in both spaces $C_{[a,b]}$ and $C_{[a,b]}^{2}$ . Let me reproduce the definitions of these spaces: The set $C_{[a,b]}$ of all continuous functions defined on the closed interval $[a,b]$ with distance $\rho(f,g)= \max_{a \leq t \leq b}|f(t)-g(t)|$ is a metric space. The set of all functions continuous on the interval $[a,b]$ with distance function defined by $\rho(x,y) = (\int_{a}^{b}[x(t)-y(t)]^{2}dt)^{\frac{1}{2}}$ is also a metric space denoted by $C_{[a,b]}^{2}$ .

Definition:

A metric space is said to be separable if it has a countable everywhere dense subset.

Example 5:

The spaces $\Re^{1}$ , $\Re^{n}$ , $\Re_{0}^{n}$ , $l_{2}$ , $C_{[a,b]}$ and $C_{[a,b]}^{2}$ are all separable since the sets in Examples 1 to 4 above are all countable. (this can be easily verified).

Example 6:

The “discrete space” M described in Example 1 (reproduced here again) contains a countable everywhere dense subset and hence is separable if and only it is itself a countable set, since clearly $[M]=M$ in this case. (Put $\rho(x,y)=1$ , when $x=y$ and $\rho(x,y)=0$ if $x \neq y$ . Then, this space M is called the discrete metric space).

Example 7:

There is no countable everywhere dense set in the space m of all bounded sequences, introduced in an earlier blog ( To help recall, we present it here again: Consider the set of all bounded infinite sequences of real numbers $x = (x_{1}, x_{2}, \ldots, x_{n}, \ldots)$ , and let $\rho(x,y) = \sup_{k}|x_{k}-y_{k}|$ . This gives a metric space which we denote by m. The fact that it has the three properties of a metric is almost obvious.)

In fact, consider the set E of all sequences consisting exclusively of zeros and ones. Clearly, E has the power of continuum since there is a one-to-one correspondence between E and the set of all subsets of the set $\mathcal{Z_{+}} = \{1,2, \ldots, n, \ldots \}$ (what is the correspondence here? answer:) Now, the distance between any two points of E equals 1. Suppose we surround each point of E by an open sphere of radius $\frac{1}{2}$ , thereby obtaining an uncountably infinite family of pairwise disjoint spheres. Then, if some set M is everywhere dense in m, there must be at least one point of M in each of the spheres. It follows that M cannot be countable and hence, that m cannot be separable.

Closed Sets:

We say that a subset M of a metric space R is closed if it coincides with its own closure, that is, if $[M]=M$ . In other words, a set is called closed if it contains all its limit points (we will solve this problem as an exercise).

Example 1:

The empty set $\phi$ and the whole space R are closed sets.

Example 2:

Every closed interval $[a,b]$ on the real line is a closed set.

Example 3:

Every closed sphere in a metric space is a closed set. In particular, the set of all functions f in the space $C_{[a,b]}$ such that $|f(t)|\leq K$ (where K is a constant) is closed.

Example 4:

The set of all functions f in $C_{[a,b]}$ such that $|f(t)|<K$ (an open sphere) is not closed. The closure of the set is the closed sphere in the preceding example.

Example 5:

Any set consisting of a finite number of points is closed.

Theorem 3:

The intersection of arbitrary number of closed sets is closed. The union of a finite number of closed sets is closed.

Proof 3:

Given arbitrary sets $F_{\alpha}$ indexed by a parameter $\alpha$ , let x be a limit point of the intersection $F = \bigcap_{\alpha}F_{\alpha}$

Then, any neighbourhood $O_{\epsilon}(x)$ contains infinitely many points of F, and hence, infinitely many points of each $F_{\alpha}$ . Therefore, x is a limit point of each $F_{\alpha}$ and hence belongs to each $F_{\alpha}$ , since the sets $F_{\alpha}$ are all closed. It follows that $x \in F_{\alpha}$ and hence that F itself is closed.

Next, let $F = \bigcup_{k=1}^{\infty}F_{k}$ be the union of a finite number of closed sets $F_{k}$ , and suppose x does not belong to F. Then, x does not belong to any of the sets $F_{k}$ , and hence, x cannot be a limit point of any of them. But then, for every k, there is a neighbourhood $O_{\epsilon_{k}}(x)$ containing no more than a finite number of points of $F_{k}$ . Choosing

$\epsilon = \min (\epsilon_{1}, \epsilon_{2}, \ldots, \epsilon_{n})$

we get a neighbourhoood $O_{\epsilon}(x)$ containing no more than a finite number of points of F, so that x cannot be a limit point of F. This proves that a point $x \in F$ cannot be a limit point of F. Therefore, F is closed. QED.

6.5 Open sets.

A point x is called an interior point of a set M if x has a neighbourhood $O_{\epsilon}(x) \subset M$ , that is, a neighbourhood consisting entirely of points of M. A set is said to be open if all points are interior points (with respect to that metric).

Example 1:

Every open interval $(a,b)$ on the real line is an open set. In fact, if $a < x < b$ , choose $\epsilon = \min(x-a, b-x)$ . Then, clearly $O_{\epsilon}(x) \subset (a,b)$ .

Example 2:

Every open sphere $S(a,r)$ in a metric space is an open set. In fact, $x \in S(a,r)$ implies that $\rho(a,x)<r$ . Hence, choosing $\epsilon = r - \rho(a,x)$ we have $O_{\epsilon}(x)=S(x,\epsilon) \subset S(a,r)$ .

Example 3:

Let M be the set of of all functions f in $C_{[a,b]}$ such that $f(t) < g(t)$ , where g is a fixed function in $C_{[a,b]}$ . Then, M is an open subset of $C_{[a,b]}$ .

Theorem 4:

A subset M of a metric space R is open if and only its complement $R-M$ is closed.

Proof of theorem 4:

If M is open, then every point $x \in M$ has a neighbourhood (entirely) contained in M. Therefore, no point $x \in M$ can be a contact point of $R - M$ . In other words, if x is a contact point of $R-M$ then $x \in R-M$ , that is, $R-M$ is closed.

Conversely, if $R-M$ is closed, then any point $x \in M$ must have a neighbourhood contained in M; since, otherwise every neighbourhood of x would contain points of $R-M$ , that is, x would be a contact point of $R-M$ , not in $R-M$ . Therefore, M is open.

QED.

corollary to theorem 4:

The empty set $\phi$ and the whole space R are open sets.

Proof:

An immediate consequence of Theorem 4 and Example 1 in the previous section.

QED.

THEOREM 5:

The union of an arbitrary number of open sets is open. The intersection of a finite number of open sets is open.

Proof of theorem 5:

this is the dual of theorem 3. The proof is an immediate consequence of theorem 4 and DeMorgan laws. QED.

6.6 Open and closed sets on the real line:

The structure of open and closed sets in a given metric space can be quite complicated. This is true even for open and closed sets in a Euclidean space of two or more dimensions. In the one-dimensional case, however, it is an easy matter to give a complete description of all open sets (and, hence, of all closed sets).

Theorem 6:

Every open set G on the real line is the union of a finite or countable system of pairwise disjoint open intervals.

Proof of theorem 6:

Let x be an arbitrary point of G. By the definition of an open set, there is at least one open interval containing x and contained in G. Let $I_{x}$ be the union of all such open intervals. Then, as we now show,

$I_{x}$ is itself an open interval.

proof:

In fact, let $a = \inf {I_{x}}$ and $b = \sup {I_{x}}$ where we allow the cases $-\infty=a$ and $+\infty=b$ .

Then, obviously, $I_{x} \subset (a,b)$ ….call this II.

Moreover, let us assume that y is an arbitrary point of $(a,b)$ distinct from x, where, to be explicit, we assume that $a<y<x$ . Then, there is a point $y^{'}\in I_{x}$ such that $a < y^{'}<y$ . (HW: think, why!) Hence, G contains an open interval containing the points $y^{'}$ and $x$ . But, then this interval also contains y, that is, $y \in I_{x}$ . (The case $y>x$ is treated similarly. )

Moreover, $x \in I_{x}$ , by hypothesis. It hence follows that $I_{x} \supset (a,b)$ , and hence, by (II) that $I_{x}=(a,b)$ . Thus, $I_{x}$ is itself an open interval, as asserted, in fact the open interval $(a,b)$ .

Next, we want to show that $I_{x}$ is a union of countable, pairwise, disjoint open intervals. Let us proceed as follows:

By its very construction, the interval $(a,b)$ is contained in G and is not a subset of a larger interval contained in G. Moreover, it is clear that two intervals $I_{x}$ and $I_{x^{'}}$ corresponding to distinct points x and $x^{'}$ either coincide or else are disjoint (otherwise, $I_{x}$ and $I_{x^{'}}$ would both be contained in a larger interval $I_{x} \bigcup I_{x^{'}} = I \subset G$ . There are no more than countably many such pairwise disjoint intervals $I_{x}$ . In fact, choosing an arbitrary rational point in each $I_{x}$ we establish a one-to-one correspondence between the intervals $I_{x}$ and a subset of the rational numbers. Finally, it is obvious that

$G = \bigcup_{x}I_{x}$ .

QED.

COROLLARY TO PROOF 6:

Every closed set on the real line can be obtained by deleting a finite or countable system of pairwise disjoint intervals from the line.

EXAMPLE 1:

Every closed interval $[a,b]$ is a closed set (here, a and b are necessarily finite).

EXAMPLE 2:

Every single element set $\{ x_{k}\}$ is closed.

EXAMPLE 3:

The union of a finite number of closed intervals and single element sets is a closed set.

EXAMPLE 4 (The CANTOR Set):

A more interesting example of a closed set on the real line can be constructed as follows:

Delete the open interval $(\frac{1}{3},\frac{2}{3})$ from the closed interval $F_{0}=[0,1]$ and let $F_{1}$ denote the remaining closed set, consisting of two closed intervals. Then, delete the open intervals $(\frac{1}{9}, \frac{2}{9})$ and $(\frac{7}{9}, \frac{8}{9})$ from $F_{1}$ and let $F_{2}$ denote the remaining closed set consisting of four closed intervals. Then, delete the “middle third” from each of these four intervals, getting a new closed set $F_{3}$ , and so on. Continuing this process indefinitely, we get a sequence of closed sets $F_{n}$ such that

$F_{0} \supset F_{1} \supset F_{2} \supset \ldots \supset F_{n} \supset \ldots$

(such a sequence is said to be decreasing). The intersection

$F = \bigcap_{n=0}^{\infty}F_{n}$

of all these sets is called the CANTOR SET.

Clearly, F is closed, and is obtained from the unit interval $[0,1]$ by deleting a countable number of open intervals. In fact, at the nth stage of construction, we delete $2^{n-1}$ intervals, each of length $\frac{1}{3^{n}}$ .

To describe the structure of the set $F_{n}$ we first note that F contains the points, 0,1, $\frac{1}{3}$ , $\frac{2}{3}$ , $\frac{1}{9}$ , $\frac{2}{9}$ , $\frac{7}{9}$ , $\frac{8}{9}$ , $\ldots$ ,….call this III.

that is, the end points of the deleted intervals (together with the points 0 and 1)

However, F contains many other points.

In fact, given any $x \in [0,1]$ , suppose we write x in ternary notation, representing x as a series :

$x=\frac{a_{1}}{3}+\frac{a_{2}}{3^{2}}+\frac{a_{3}}{3^{3}}+\ldots + \frac{a_{n}}{3^{n}}+\ldots$ …call this IV.

where each of the numbers $a_{1}, a_{2}, \ldots$ is 0, 1 or 2 in radix-3. Then, it is easy to see that x belongs to F if and only if x has a representation (IV) such that none of the numbers $a_{1}, a_{2}, \ldots, a_{n}, \ldots$ equals 1 (you need to think about this as HW!!)

Remarkably enough, the set F has the power of the continuum, that is, there are as many points in F as in the whole interval $[0,1]$ , despite the fact that the sum of the lengths of the deleted intervals equals

$\frac{1}{3} + \frac{2}{3^{2}} + \frac{2^{2}}{3^{3}} + \ldots =1$

To see this, we associate a new point

$y=\frac{b_{1}}{2} + \frac{b_{2}}{2^{2}}+\ldots + \frac{b_{n}}{2^{n}}+\ldots$

with each point IV, where

$b_{n}=0$ , if $a_{n}=0$ ; $b_{n}=1$ ; if $a_{n}=2$ .

In this way, we set up a one-to-one correspondence between F and the whole interval $[0,1]$ . It follows that F has the power of the continuum, as asserted. Let $A_{1}$ be the set of points III. Then,

$F = A_{1} \bigcup A_{2}$ where the set $A_{2}=F-A_{1}$ is uncountable, since $A_{1}$ is countable and F itself is not. The points of $A_{1}$ are often called “points of F of the first kind,” while those of $A_{2}$ are called “points of the second kind.”

QED.

Solutions to System of Sets:

Posted on January 15, 2020January 24, 2020 by Nalin Pithwa

Reference: Introductory Real Analysis, Kolmogorov and Fomin. Dover Publications. Translated and edited by Richard A. Silverman:

Problem 1:

Let X be an uncountable set, and let $\mathscr{R}$ be the ring consisting of all finite subsets of X and their complements. Is $\mathscr{R}$ a $\sigma$ -ring?

Solution 1:

Recall the definition: A ring of sets is called $\sigma$ -ring if it contains the union $S= \bigcup_{n=1}^{\infty}A_{n}$ whenever it contains the sets $A_{1}, A_{2}, \ldots, A_{n}, \ldots$ .

There are countably many finite subsets of X (given uncountable set) and their complements (also especially) so that the union of all these is the whole set X. Hence, X is a $\sigma$ -ring as it is closed under countably many set unions. QED.

Problem 2:

Are open intervals Borel sets?

Solution 2:

Recall the following:

(1) Borel sets or B-sets are the subsets of the real line belonging to the minimal B-algebra generated by the set of all closed intervals $[\alpha, \beta]$ .

Problem 3:But a B-algebra is also a ring and is closed under differences/complements. Open intervals are complements of closed intervals. And, they belong hence to B-algebra. Note we still have to answer the question whether open sets are Borel sets and so we have to check if they are contained in a minimal B-algebra. As per the following theorem: Given any non-empty system of sets $\mathscr{S}$ , there is a unique irreducible (minimal) B-algebra $\mathscr{B}(\mathscr{P})$ generated by the system $\mathscr{S}$ or the Borel closure of $\mathscr{P}$ . So, yes indeed open sets are Borel sets. QED.

Problem 3A:

Let $y=f(x)$ be a function defined on a set M and taking values in a set N. Let $\mathscr{M}$ be a system of subsets of M, and let $f(\mathscr{M})$ denote the system of all images $f(A)$ of sets $A \in \mathscr{M}$ . Moreover, let $\mathscr{N}$ be a system of subsets of N, and let $f^{-1}(\mathscr{N})$ denote the system of all preimages $f^{-1}(B)$ of sets $B \in \mathscr{N}$ . Prove that:

i) If $\mathscr{N}$ is a ring, so is $f^{-1}(\mathscr{N})$ .

ii) If $\mathscr{N}$ is an algebra, so is $f^{-1}(\mathscr{N})$

iii) If $\mathscr{N}$ is a B-algebra, so is $f^{-1}(\mathscr{N})$

iv) $\mathscr{R}(f^{-1}(\mathscr{N})) = f^{-1}(\mathscr{R}(\mathscr{N}))$ where $\mathscr{R}$ stands for a ring.

v) $\mathscr{B}(f^{-1}(\mathscr{N})) = f^{-1}(\mathscr{B}(\mathscr{N}))$ where $\mathscr{B}$ is irreducible or minimum Borel algebra generated by a non-empty system of sets $\mathscr{S}$ .

Problem 3B:

Prove the following:

i) If $\mathscr{M}$ is a ring, so is $f(\mathscr{M})$ .

ii) If $\mathscr{M}$ is an algebra, so is $f(\mathscr{M})$ .

iii) If $\mathscr{M}$ is Borel-algebra, so is $f(\mathscr{M})$ .

iv) $\mathscr{R}(f(\mathscr{M}))=f(\mathscr{R}(\mathscr{M}))$

v) $\mathscr{B}(f(M))=f(\mathscr{B}(\mathscr{M}))$ where $\mathscr{B}$ is irreducible minimum Borel algebra generated by a non-empty system of sets $\mathscr{S}$ .

Solution 3A:

Part 1: Given mapping $f: M \rightarrow N$ , that is, $y=f(x)$ , Let $Y_{1}, Y_{2}, Y_{3}, \ldots$ be subsets of N and let $X_{1}, X_{2}, X_{3}, \ldots$ be subsets of M. Then, as it is given that $\mathscr{N}$ is a ring then :

$Y_{1} \bigcap Y_{2} \in \mathscr{N}$

$Y_{1}\bigcup Y_{2} \in \mathscr{N}$

$Y_{1}-Y_{2} \in \mathscr{N}$

$Y_{1} \triangle Y_{2} = (Y_{1}-Y_{2}) \bigcup (Y_{2}-Y_{1}) \in \mathscr{N}$

The above implies:

$f^{-1}(Y_{1}\bigcap Y_{2}) \in f^{-1}(\mathscr{N})$ , that is, $f^{-1}(Y_{1}) \bigcap f^{-1}(Y_{2}) \in f^{-1}(\mathscr{N})$ , that is, $X_{1} \bigcap X_{2} \in \mathscr{M}$ .

Also, $Y_{1} \bigcup Y_{2} \in \mathscr{N}$ so that $f^{-1}(Y_{1} \bigcup Y_{2}) \in f^{-1}(\mathscr{N})$ , that is, $f^{-1}(Y_{1}) \bigcup f^{-1}(Y_{2}) \in f^{-1}(\mathscr{N})$ , that is, we conclude $X_{1} \bigcup X_{2} \in \mathscr{M}$ .

Also, $Y_{1}-Y_{2} \in \mathscr{N}$ as $\mathscr{N}$ is a ring; so that $f^{-1}(Y_{1}-Y_{2}) \in f^{-1}(\mathscr{N})$ so that for some $X_{0}$ , which is $f^{-1}(Y_{1}-Y_{0}) \longrightarrow f^{-1}(Y_{1}) - f^{-1}(Y_{0}) \in \mathscr{M}$

So, we have shown that the system of sets $\mathscr{M}$ is closed under set union, set intersection, set difference, and hence, also set symmetric difference. Hence, it is is a ring. QED.

Part ii:

We have already shown that $f^{-1}(\mathscr{N})$ is a ring, now $\mathscr{N}$ is also an algebra meaning a ring of sets with a unit element. Let it contain a unit E $\in \mathscr{N}$ such that if $Y_{0}$ is any set belonging to $\mathscr{N}$ , then $Y_{0} \bigcap E = Y_{0}$ . so that $f^{-1}(Y_{0} \bigcap E) = f^{-1}(Y_{0})$ , hence, $f^{-1}(Y_{0}) \bigcap f^{-1}(E) = f^{-1}(Y_{0})$ , which in turn implies that $f^{-1}(E)$ is a unit of $f^{-1}(\mathscr{N})$ . Hence, $f^{-1}(\mathscr{N})$ is a ring of sets with a unit element, or in other words, an algebra of sets.

Part iii:

To prove that: if $\mathscr{N}$ is a Borel algebra, so is $f^{-1}(\mathscr{N})$ . From parts i and ii above, $f^{-1}(\mathscr{N})$ is a ring with a unit or in other words, $f^{-1}(\mathscr{N})$ is an algebra. Moreover, if it contains $\bigcup_{k=1}^{\infty}Y_{k}$ whenever it contains the sets $Y_{1}, Y_{2}, Y_{3}, \ldots, Y_{n}, \ldots$ all of which are members of $\mathscr{N}$ .

But for an arbitrary (finite or infinite) collection of sets:

$f^{-1}(\bigcup_{\alpha}A_{\alpha}) = \bigcup_{\alpha}f^{-1}(A_{\alpha})$

Now, $\bigcup_{k=1}^{\infty}Y_{k} \in \mathscr{N}$

hence $f^{-1}(\bigcup_{k=1}^{\infty}Y_{k}) \in f^{-1}(\mathscr{N})$

Consequently, $\bigcup_{k=1}^{\infty}f^{-1}(Y_{k}) \in f^{-1}(\mathscr{N})$ , which in turn implies that $\mathscr{N}$ is a Borel algebra also. QED.

Part iv and part v:

Using parts i and iii above, these can be very easily proved.

Solutions 3b:

i) Prove: If $\mathscr{M}$ is a ring, so is $f(\mathscr{M})$ . Note that $f: M \rightarrow N$ , $y=f(x)$ . Proof (i): As $\mathscr{M}$ is a system of subsets of M. and it is a ring, so if $A_{1} \in \mathscr{M}$ , and $A_{2} \in \mathscr{M}$ , it implies that $A_{1} \bigcup A_{2} \in \mathscr{M}$ ; $A_{1} \bigcap A_{2} \in \mathscr{M}$ ; $(A_{1}-A_{2}) \in \mathscr{M}$ . We know that $f(A_{1}\bigcup A_{2}) = f(A_{1}) \bigcup f(A_{2}) \in f(\mathscr{M})$ and $f(A_{1} \bigcap A_{2}) \subset f(A_{1}) \bigcap f(A_{2}) \in f(\mathscr{M})$ . And, quite obviously, $f(A_{1}-A_{2}) = f(A_{1}) - f(A_{2}) \in f(\mathscr{M})$ so that it is true that if $\mathscr{M}$ is a ring, then $f(\mathscr{M})$ is also a ring.

ii) Check if it is true or false: if $\mathscr{M}$ is an algebra, so is $f(\mathscr{M})$ . Argument: As $\mathscr{M}$ is an algebra, it is already a ring. But additionally, $\mathscr{M}$ contains an element E such that $A_{1} \bigcap E= A_{1}$ (whenever $A_{1} \in \mathscr{M}$ but $f(A_{1} \bigcap E) \subset f(A_{1}) \bigcap f(E) \in f(\mathscr{M})$ but it need not be true that $f(E)$ is a unit element of $f(\mathscr{M})$ . So, we can say that in case $\mathscr{M}$ is an algebra, then $f(\mathscr{M})$ may or may not be an algebra. QED.

iii) Check if it is true or false: If $\mathscr{M}$ is a Borel algebra, then so also $f(\mathscr{M})$ is also a Borel algebra. Argument: As in previous case, $f(\mathscr{M})$ may not have a unit element. So, $f(\mathscr{M})$ need not be a Borel algebra. QED.

iv) $\mathscr{R}(f(\mathscr{M})) = f(\mathscr{R}(\mathscr{M}))$ . Check if this is true or false. Argument: Let $f(A_{1}) \in f(\mathscr{M})$ , $f(A_{2}) \in \mathscr{M}$ . By definition of a ring, $f(A_{1}) \bigcup f(A_{2}) \in f(\mathscr{M})$ , $f(A_{1}) \bigcap f(A_{2}) \in f(\mathscr{M})$ , $f(A_{1}) - f(A_{2}) \in f(\mathscr{M})$ , hence we get the following: $f(A_{1} \bigcup A_{2}) \subset f(\mathscr{M})$ ; $f(A_{1} \bigcap A_{2}) \subset f(A_{1}) \bigcap f(A_{2}) \in f(\mathscr{M})$ and $f(A_{1}-A_{2}) = f(A_{1}) - f(A_{2}) \in f(\mathscr{M})$ . This in turn means that $A_{1} \bigcup A_{2} \in \mathscr{M}$ , $A_{1} \bigcap A_{2} \in \mathscr{M}$ and $A_{1} - A_{2} \in \mathscr{M}$ . Hence, $LHS \subset RHS$ . Similarly, we can show that $RHS \subset LHS$ . QED.

v) $\mathscr{B}(f(\mathscr(M))) = f(\mathscr{B}(\mathscr{M}))$ . Check if this is true or false. (We will discuss this later; meanwhile, if you wish, please try as homework and let me know).

Cheers,

Nalin Pithwa.

Comments are most welcome.

Introductory Real Analysis: Exercise 3 solutions: related to set theory

Posted on January 15, 2020January 15, 2020 by Nalin Pithwa

Problem 1:

Exhibit both a partial ordering and a simple ordering on the set of complex numbers.

Solution I:

Recall the definition of a partial ordering: A binary relation R on a set M is said to be a partial ordering (and the set M itself is said to be partially ordered) if:

(i) R is reflexive (aRa for every $a \in M$ ) (ii) R is transitive (aRb and bRc together imply aRc) (iii) aRb and bRa together imply $a=b$ (anti symmetric).

Recall the definition of a simple ordering or totally ordered or (just) ordered set: a set M is said to be ordered if it is a partially ordered set and if, given any two distinct elements $a, b \in M$ , either $a < b$ or $b < a$ .

Let us consider the following relation R on the set of complex numbers, C:

consider any two elements $z_{1}, z_{2} \in C$ such that $z_{1}=x_{1}+i y_{1}$ and $z_{2} = x_{2}=iy_{2}$ . We define a relation $z_{1}Rz_{2}$ if and only if $x_{1} \leq x_{2}$ and $y_{1} \leq y_{2}$ . Clearly, this is both a partial ordering on C as well as total ordering on C.

Now consider the following relation K on the set of complex numbers:

We define the relation K as: $z_{1}Kz_{2}$ if and only if $\frac{x_{1}}{y_{1}} = \frac{x_{2}}{y_{2}}$ where $x_{1}, y_{1}, x_{2}, y_{2} \in \Re$ . Clearly, if any one of $y_{1}$ or $y_{2}$ is zero, then the complex numbers $z_{1}$ and $z_{2}$ are non-comparable. So this could be a partial ordering but not a total ordering on the set of complex numbers.

Problem 2:

What is the minimal element of the set of all subsets of a given set X partially ordered by set inclusion. What is the maximal element?

Solution 2:

Recall the definition: An element “a” of a partially ordered set is said to be maximal if $aRb$ implies $b=a$ and minimal if $bRa$ implies b=a.

Note that in the above definition, element “a” is our given/chosen element to be compared with other elements of the set. So, clearly, if the partial ordering is set inclusion, the minimal element is the null set $\phi$ and the maximal element is the given set X itself.

Problem 3:

A partially ordered set M is said to be a directed set if, given any two elements $a, b \in M$ , there is an element $c \in M$ such that $a \leq c$ and $b \leq c$ . Are the following partially ordered sets directed sets also ?

PS: in the above we use the notation $aRb$ or equivalently, $a \leq b$ to mean one and the same thing.

Question 3a: Any set M can be trivially partially ordered by setting $a \leq b$ if and only if $a=b$ .

Answer 3a: Clearly, this is a trivial directed set.

Question 3b: Let M be the set of all continuous functions $f, g, \ldots$ defined in a closed interval $[\alpha, \beta]$ . Then, we get a partial ordering by setting $f \leq g$ if and only if $f(t) \leq g(t)$ for every $t \in [\alpha, \beta]$ . Is this also a directed set ?

Answer 3b: We need to choose two arbitrary elements say $f_{1}$ and $f_{2}$ of the given set of continuous functions on $[\alpha, \beta]$ . Our desired “element” could be $f_{3}= |f_{1}(t)|+|f_{2}(t)|$ again defined on $[\alpha, \beta]$ . Then, the given set becomes a directed set.

Note that the function $y=f(x) = |x|$ is continuous at zero also (but not differentiable at zero).

Question 3c:

Set inclusion.

Answer 3: Clearly, given any two arbitrary subsets, both are contained always in the given big set M or X. So, this is again a directed set.

Question 3d: The set of all integers greater than 1 is partially ordered if $a \leq b$ means that “b is divisble by a.”

Answer 3: So let us consider two arbitrary positive integers, both greater than 1, call them a and b. We want to know if there exist a positive integer, also greater than 1, call it c such that $a \leq c$ and $b \leq c$ ? That is, “c is divisible by a” and “c is divisible by b”. Clearly, c can be least common multiple of a and b. So, yes, this is also a directed set.

Problem 4:

By the greatest lower bound of two elements a and b of a partially ordered set M, we mean an element $c \in M$ such that $c \leq a$ and $c \leq b$ and there is no element $d \in M$ such that $c < d \leq a$ , $d \leq b$ . Similarly, by the least upper bound of a and b, we mean an element $c \in M$ such that $a \leq c$ , $b \leq c$ , and there is no element $d \in M$ such that $a \leq d < c$ and $b \leq d$ . By a lattice is meant a partially ordered set any two elements of which have both a greatest lower bound and a least upper bound. Prove that the set of all subsets of a given set M, partially ordered by set inclusion, is a lattice. What is the set theoretic meaning of the greatest lower bound and least upper bound of two elements of this set ?

Solution 4:

It can be easily checked that the null set is the greatest lower bound and the set M itself if the least upper bound. (PS: these are the minimal and maximal elements of this set M respectively.)

Problem 5:

Prove that an order-preserving mapping of one ordered set onto another is automatically an isomorphism.

Solution 5:

By definition.

Problem 6:

Prove that ordered sums and products of ordered sets are associative, that is, prove that if $M_{1}, M_{2}$ and $M_{3}$ are ordered sets, then

$(M_{1}+M_{2})+M_{3}=M_{1}+(M_{2}+M_{3})$

$(M_{1}.M_{2}).M_{3} = M_{1}.(M_{2}.M_{3})$

PS: comment: this allows us to drop the parentheses in writing ordered sums and products.

where the operations are as defined below:

Let $M_{1}$ and $M_{2}$ be two ordered sets of type $\theta_{1}$ and $\theta_{2}$ respectively. Then, we can introduce an ordering in the union $M_{1} \bigcup M_{2}$ of the two sets by assuming that: (i) a and b have the same ordering as in $M_{1}$ if $a, b \in M_{1}$ (ii) a and b have the same orderiing as in $M_{2}$ if $a, b \in M_{2}$ (iii) $a \leq b$ if $a \in M_{1}$ and $b \in M_{2}$ .

In this case/example, the ordered sum or union of three ordered sets $M_{1}, M_{2}, M_{3}$ would mean that any two elements a, b would have the same ordering if $a, b \in M_{1}$ , or $a, b \in M_{2}$ or $a,b \in M_{3}$ . Also, comparing with ordered sum of two ordered sets, we can say that $a \leq b$ if and only if $a \in M_{1}, b \in M_{2}$ or $a \in M_{1}, b \in M_{3}$ , or $a \in M_{2}, b \in M_{3}$ . In such a case, the ordered sum is also associative.

The operation of ordered product of two ordered sets is defined as follows: $M_{1}. M_{2}$ is the set of all pairs $(a,b)$ where $a \in M_{1}$ and $b \in M_{2}$ ordered in such a way that (i) $(a_{1}, b_{1}) < (a_{2}, b_{2})$ if $b_{1} < b_{2}$ for arbitrary $a_{1}, a_{2}$ , and (ii) $(a_{1},b) < (a_{2}, b)$ if $a_{1} < a_{2}$ .

So, if we have three ordered sets $M_{1}, M_{2}, M_{3}$ , we can define their ordered product $M_{1}.M_{2}.M_{3}$ as consisting of all ordered triples such that (i) $(a_{1}, b_{1}, c_{1}) < (a_{2}, b_{2}, c_{2}) < (a_{3}, b_{3}, c_{3})$ if and only if $b_{1}<b_{2}<b_{3}$ and $c_{1} < c_{2} < c_{3}$ for arbitrary $a_{1}, a_{2}, a_{3}$ , and (ii) $(a_{1}, b, c) < (a_{2}, b, c) < (a_{3}, b, c)$ if $a_{1}<a_{2}<a_{3}$ ; $(a, b_{1}, c) < (a, b_{2}, c) < (a, b_{3}, c)$ if $b_{1}<b_{2}<b_{3}$ ; $(a,b, c_{1})< (a,b,c_{2}) < (a,b, c_{3})$ of $c_{1} < c_{2} < c_{3}$ . In such a case, the product is well-defined and is also associative. (any comments ??)

Problem 7:

Construct well-ordered sets with ordinals

$\omega + n$ , $\omega + \omega$ , $\omega + \omega + n$ , $\omega + \omega + \omega$ , $\ldots$

Show that the sets are all countable.

Solution 7:

$\omega + n$ : $\{ 1,2,3,\ldots, a_{1}, a_{2}, \ldots, a_{n}\}$ . This is countable because there can be bijection from the set $Z^{+}$ to the given set: $1 \rightarrow a_{1}$ , $2 \rightarrow a_{2}$ , $\ldots$ , $n \rightarrow a_{n}$ , $n+1 \rightarrow 1$ , $n+2 \rightarrow 2$ , $\ldots$

$\omega + \omega$ : $\{1,2,3, \ldots, 1,2,3, \ldots \}$ , this is also countable as it can be put in one-to-one correspondence with $Z^{+}$ .

$\omega + \omega + n$ ; $\{ 1,2, 3, \ldots, 1, 2, 3, \ldots, a_{1}, a_{2}, a_{3}, \ldots, a_{n}\}$ . This is also countable in the same manner as the first example.

$\omega + \omega + \omega$ : $\{ 1,2, 3, \ldots, 1,2, 3, \ldots, 1,2,3, \ldots\}$ : this can be put in one-to-one correspondence with $Z^{+}$

Problem 8:

Construct well-ordered sets with ordinals

$\omega.n$ , $\omega^{2}$ , $\omega^{2}.n$ , $\omega^{3}$ , $\ldots$ .

Show that the sets are all countable.

Solution 8:

First let us recall the definition: An ordered set M is said to be well-ordered if every non empty subset A of M has a smallest (or, first) element, that is, an element $\mu$ such that $\mu < a$ for every $a \in A$ .

Ans 1: $\omega.n$ is set : $M= \{(1,a_{1}), (2,a_{2}), (3, a_{3}), \ldots, (n, a_{n}), (n+1, a_{1}), (n+2, a_{2}), \ldots \}$ . This set is ordered clearly by the definition of ordered product of two ordered sets; and this is well-ordered if we define the first element to be $(1, a_{1})$ .

Ans 2: $\omega^{2} = \omega. \omega$ : this set can be constructed as $M= \{ (1,1), (1,2),(1,3), \ldots, (2,1), (2,2), (2,3), \ldots, \}$ and we can define the first element as $(1,1)$ so that it becomes well-ordered under the usual definition of product of two ordered sets.

Ans 3: $\omega^{2}.n = \omega. \omega. n$ : this set can be constructed as follows as an ordered product of three ordered sets: $M = \{ (1,1,a_{1}), (1,1,a_{2}), (1,1,a_{3}), \ldots, (1,1,a_{n}), (1,2,a_{1}), (1,2,a_{2}), (1,2,a_{3}), \ldots, (1, 2, a_{n}), \ldots\}$ . We see that this is also well-ordered if we define the first element to be $(1,1,a_{1})$ and “order or list or count” them as shown. (any comments ? )

Ans 4: $\omega^{3}= \omega.\omega.\omega$ : this can be well-ordered as above with clearly the first element to be $(1,1,1)$ . (any comments?)

PS: the explicit listing shows that all the above sets are clearly countable.

More later,

Cheers,

Nalin Pithwa

Metric Spaces: Tutorial problems: Solutions: Part I

Posted on January 15, 2020April 20, 2020 by Nalin Pithwa

Reference: Introductory Real Analysis by Kolmogorov and Fomin.

Reference: Analysis by Walter Rudin.

Reference: Introduction to Analysis by Rosenlicht.

Reference: Topology by Hocking and Young.

The problems proposed were as follows:

Given a metric space $(X,\rho)$ , prove that (1a) $|\rho(x,z)-\rho(y,u)| \leq \rho(x,y) + \rho(z,u)$ where $x,y,z,u \in X$ ; (1b) $|\rho(x,z) -\rho(y,z)| \leq \rho(x,y)$ where $x,y,z \in X$ .
Verify that $(\Sigma_{k=1}^{n}a_{k}b_{k})^{2} = \Sigma_{k=1}^{n}a_{k}^{2} \times \Sigma_{k=1}^{n}b_{k}^{2} - \frac{1}{2}\Sigma_{i=1}^{n}\Sigma_{j=1}^{n}(a_{i}b_{j}-a_{j}b_{i})^{2}$ . Deduce the Cauchy-Schwarz inequality from the above identity. (refer second last blog from here). For convenience, we reproduce it here again: consider the set of all ordered n-tuples such that $x=(x_{1}, x_{2}, \ldots, x_{n})$ and $y=(y_{1}, y_{2}, \ldots, y_{n})$ and $z=(z_{1}, z_{2}, \ldots, z_{n})$ and let $a_{k}=x_{k}-y_{k}$ , $b_{k}=y_{k}-z_{k}$ where $k=1,2, \ldots, n$ : then Cauchy Schwarz inequality tells us that : $(\Sigma_{k=1}^{n}a_{k}b_{k})^{2} \leq \Sigma_{k=1}^{n}a_{k}^{2}\Sigma_{k=1}^{n}b_{k}^{2}$
Verify that $(\int_{a}^{b}x(t)y(t)dt)^{2} = \int_{a}^{b}x^{2}(t)dt \int_{a}^{b}y^{2}(t)dt -\frac{1}{2}\int_{a}^{b}\int_{a}^{b}(x(s)y(t)-x(t)y(s))^{2}dsdt$ .

Deduce Schwarz’s inequality from this identity. (refer second last blog from here). To aid the memory, we present Schwarz’s inequality below: $(\int_{a}^{b}x(t)y(t))^{2} \leq \int_{a}^{b}x^{2}(t)dt \int_{a}^{b}y^{2}(t)dt$

4. Identify the fallacy in example 10, if $p \leq 1$ ? (refer second last blog from here). Hint: Show that Minkowski’s inequality fails for $p \leq 1$ . To help recall, we are reproducing Minkowski’s inequality below: Consider again the set of all ordered n-tuples, and let $x=(x_{1}, x_{2}, \ldots, x_{n})$ , and $y=(y_{1}, y_{2}, \ldots, y_{n})$ and $z=(z_{1}, z_{2}, \ldots, z_{n})$ and again further let $a_{k}=x_{k}-y_{k}$ and $b_{k}=y_{k}-z_{k}$ , where $k=1,2, \ldots, n$ then Minkowski’s inequality tells us that: $(\Sigma_{k=1}^{n}|a_{k}+b_{k}|^{p})^{\frac{1}{p}} \leq (\Sigma_{k=1}^{n}|a_{k}|^{p})^{\frac{1}{p}} + (\Sigma_{k=1}^{n}|b_{k}|^{p})^{\frac{1}{p}}$

5. Prove that the following metric : Consider two points $x=(x_{1}, x_{2}, \ldots, x_{n})$ and $y=(y_{1}, y_{2}, \ldots, y_{n})$ with the metric function: $\rho_{0}(x,y) = \max_{1 \leq k \leq n}|x_{k}-y_{k}|$

is the limiting case of the following metric: consider the set of all ordered n-tuples $x = (x_{1}, x_{2}, \ldots, x_{n})$ of real numbers with the metric function now defined by: $\rho_{p}(x,y) = (\Sigma_{k=1}^{\infty}|x_{k}-y_{k}|^{p})^{\frac{1}{p}}$

in the sense that

$\rho_{0}(x,y) = \max_{1 \leq k \leq n}|x_{k}-y_{k}| = \lim_{p \rightarrow \infty}(\Sigma_{k=1}^{\infty}|x_{k}-y_{k}|^{p})^{\frac{1}{p}}$

6. Starting from the following inequality: $ab \leq \frac{a^{p}}{p} + \frac{b^{q}}{q}$ , deduce Holder’s integral inequality as given by:

$\int_{a}^{b} x(t)y(t)dt \leq (\int_{a}^{b}|x(t)|^{p}dt)^{\frac{1}{p}}(\int_{a}^{b}|y(t)|^{q})^{\frac{1}{q}}$ where $\frac{1}{p} + \frac{1}{q} =1$ valid for any function $x(t)$ and $y(t)$ such that the integrals on the right exist.

7. Use Holder’s integral inequality to prove Minkowski’s integral inequality:

$(\int_{a}^{b}|x(t)+y(t)|^{p}dt)^{\frac{1}{p}} \leq (\int_{a}^{b}|x(t)|^{p}dt)^{\frac{1}{p}} + (\int_{a}^{b}|y(t)|^{p}dt)^{\frac{1}{p}}$

where $p \geq 1$ .

8. Exhibit an isometry between the spaces $C_{[0,1]}$ and $C_{[1,2]}$ .

9. Verify that the following is a metric space: all bounded infinite sequences $x=(x_{1}, x_{2}, x_{3}, \ldots)$ of elements of $\Re$ , with $\rho(x,y) = lub \{ |x_{1}-y_{1}|,|x_{2}-y_{2}|,|x_{3}-y_{3}|, \ldots\}$

10. Verify that $(\Re \times \Re, \rho)$ is a metric space where $\rho((x,y),(x^{'}, y^{'})) = |y|+|y^{'}|+|x-x^{'}|$ when $x \neq x^{'}$ and is $|y-y^{'}|$ when $x=x^{'}$ . Illustrate by diagrams in the plane $E^{2}$ or $R^{2}$ what the open balls of this metric space are.

11. Show that a one-to-one transformation $f: S \rightarrow T$ of a space S onto a space T is a homeomorphism if and only if both f and $f^{'}$ are continuous.

The solutions are as follows:

Problem 1:

Given a metric space $(X,\rho)$ , prove that (1a) $|\rho(x,z)-\rho(y,u)| \leq \rho(x,y) + \rho(z,u)$ where $x,y,z,u \in X$ ; (1b) $|\rho(x,z) -\rho(y,z)| \leq \rho(x,y)$ where $x,y,z \in X$ .

Solution 1:

Part I: TPT: $|\rho(x,z)-\rho(y,u)| \leq \rho(x,y) + \rho(z,u)$

Proof: We know that: $\rho(x,z) \leq \rho(x,u)+\rho(z,u)$ , which also means that, $\rho(x,z) \leq \rho(x,y) + \rho(y,u) + \rho(z,u)$ , that is, $\rho(x,z)-\rho(y,u) \leq \rho(x,y)+\rho(z,u)$ . On the other hand, we also know that $\rho(y,u) \leq \rho(y,z) + \rho(z,u)$ , which also means that, $\rho(y,u) \leq \rho(x,y) + \rho(x,z) + \rho(z,u)$ , that is, $-\rho(x,z) + \rho(y,u) \leq \rho(x,y) + \rho(z,u)$ . Combining the two smallish results, we get $|\rho(x,z)-\rho(y,u)| \leq \rho(x,y) + \rho(z,u)$ . QED.

Part II: TPT: $|\rho(x,z)-\rho(y,z)| \leq \rho(x,y)$ . Proof : We know that $\rho(x,z) \leq \rho(x,y)+\rho(y,z)$ , that is, $\rho(x,z) - \rho(y,z) \leq \rho(x,y)$ . On the other hand, we know that $\rho(y,z) \leq \rho(y,z)$ , that is, $\rho(y,z) \leq \rho(x,y) +\rho(x,z)$ , that is, $-\rho(x,z)+\rho(y,z) \leq \rho(x,y)$ . Combining the two smallish results, we get $|\rho(x,z) -\rho(y,z)| \leq \rho(x,y)$ where $x,y,z \in X$ . QED.

PS: Now that the detailed proof is presented, try to interpret the above two results “geometrically.”

Problem 2 was:

Verify that $(\Sigma_{k=1}^{n}a_{k}b_{k})^{2} = \Sigma_{k=1}^{n}a_{k}^{2} \times \Sigma_{k=1}^{n}b_{k}^{2} - \frac{1}{2}\Sigma_{i=1}^{n}\Sigma_{j=1}^{n}(a_{i}b_{j}-a_{j}b_{i})^{2}$ . Deduce the Cauchy-Schwarz inequality from the above identity. (refer second last blog from here). For convenience, we reproduce it here again: consider the set of all ordered n-tuples such that $x=(x_{1}, x_{2}, \ldots, x_{n})$ and $y=(y_{1}, y_{2}, \ldots, y_{n})$ and $z=(z_{1}, z_{2}, \ldots, z_{n})$ and let $a_{k}=x_{k}-y_{k}$ , $b_{k}=y_{k}-z_{k}$ where $k=1,2, \ldots, n$ : then Cauchy Schwarz inequality tells us that : $(\Sigma_{k=1}^{n}a_{k}b_{k})^{2} \leq \Sigma_{k=1}^{n}a_{k}^{2}\Sigma_{k=1}^{n}b_{k}^{2}$ .

Proof of problem 2: We present a proof based on the principle of mathematical induction. Let us check the proposition for $n=1$ : Then, $LHS=(a_{1}b_{1})^{2}=a_{1}^{2}b_{1}^{2} = RHS = a_{1}^{2}b_{1}^{2}-\frac{1}{2} \times 0$ . So now let the proposition be true for $m=n, m \in N, m >1$ . Then, the following holds true:

$(\Sigma_{k=1}^{m}a_{k}b_{k})^{2} = \Sigma_{k=1}^{m}a_{k}^{2}\Sigma_{k=1}^{m}b_{k}^{2} -\frac{1}{2}\Sigma_{i=1}^{m}\Sigma_{j=1}^{m}(a_{i}b_{j}-a_{j}b_{i})^{2}$ .

Now, we want to prove the proposition for $n=m+1$ , that is, we need to prove that:

$(\Sigma_{k=1}^{m+1}a_{k}b_{k})^{2}=\Sigma_{k=1}^{m+1}a_{k}^{2}\Sigma_{k=1}^{m+1}b_{k}^{2} - \frac{1}{2}\Sigma_{i=1}^{m+1}\Sigma_{j=1}^{m+1}(a_{i}b_{j}-a_{j}b_{i})^{2}$ .

Now, let us see what we need really; we need to add some(same) terms to both sides of the proposition for $n=m$ and manipulate to derive the proposition for $n=m+1$ :

Further notice that

$(\Sigma_{k=1}^{m+1}a_{k}b_{k})^{2}=(a_{1}b_{1}+a_{2}b_{2}+ \ldots+a_{m}b_{m}+a_{m+1}b_{m+1})^{2}=\Sigma_{k=1}^{m}a_{k}^{2}b_{k}^{2}+a_{m+1}^{2}b_{m+1}^{2}+2(a_{1}b_{1})(a_{2}b_{2}+a_{3}b_{3}+\ldots+a_{m}b_{m}+a_{m+1}b_{m+1})+2(a_{2}b_{2})(a_{3}b_{3}+a_{4}b_{4}+\ldots+a_{m}b_{m}+a_{m+1}b_{m+1}) + 2a_{3}b_{3}(a_{4}b_{4}+a_{5}b_{5}+ \ldots + a_{m+1}b_{m=1})+ \ldots + 2a_{m}b_{m}a_{m+1}b_{m+1}$

Comparing the algebraic statements for proposition for $n=m$ and $n=m+1$ , we can clearly see which terms are to be added to both sides of the proposition for $n=m$ ,

(please fill in the few missing last details…a good deal of algebraic manipulation…some concentrated effort …:-))

From this the Cauchy Schwarz inequality follows naturally.

PS: I have no idea how to solve question 3. If any reader helps, I would be obliged.

PS: I have yet to try the other questions.

Cheers,

Nalin Pithwa

MScMathematics

Category: undergrad analysis

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Cauchy’s Mean Value Theorem and the Stronger Form of l’Hopital’s Rule

VII. Complete Metric Spaces

Problem Set based on VI. Convergence, open and closed sets.

VI Convergence. Open and Closed Sets.

Solutions to System of Sets:

Introductory Real Analysis: Exercise 3 solutions: related to set theory

Metric Spaces: Tutorial problems: Solutions: Part I

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