# II Metric Spaces:

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

Reference: Analysis by Walter Rudin, Third Edition.

Chapter II Metric Spaces:

V: Basic Concepts:

Section 5.1: Definitions and examples:

One of the most important operations in mathematical analysis is the taking of limits. Here what matters is not so much as the algebraic nature of the real numbers (that is, the fact that real numbers form a field), but rather the fact that distance from one point to another on the real line(or, in two or three dimensional space) is well-defined and has certain properties. Roughly speaking, a metric space is a set equipped with a distance (or, ‘metric’) which has these same properties. More exactly, we have:

Definition 1:

By a metric space is meant a pair $(X, \rho)$ consisting of a set X and a distance $\rho$, that is, a single valued, nonnegative, real function $\rho(x,y)$ defined for all $x, y \in X$ which has the following three properties:

1. $\rho(x,y)=0$ if and only if $x=y$
2. Symmetry: $\rho(x,y)=\rho(y,x)$
3. Triangle Inequality: $\rho(x,z) \leq \rho(x,y)+\rho(y,z)$

We will often refer to the set X as a “space” and its elements x, y, …, as “points.” Metric spaces are usually denoted by a single letter, like

$R=(X,\rho)$

or, even by the same letter X as used for the underlying space, in cases where there is no possibility of confusion.

Example 1:

Setting $\rho(x,y)=0$, if $x=y$ and $\rho(x,y)=1$, when $x \neq y$, where x and y are elements of an arbitrary set X, we obviously get a metric space, which might be called a “discrete space” or a “space of isolated points.”

Check: does this satisfy all the three axioms of a metric space: clearly, the first axiom is true. So, also the second axiom is true because $\rho(x,y)=\rho(y,x)$ is zero when $x = y$ and is 1 when $x \neq y$ or $y \neq x$.

Now we have to check: $\rho(x,z) \leq \rho(x,y) + \rho(y,z)$. Case I: if $x=z$, LHS is zero and again RHS could be zero or one depending on y and z. In all cases, the inequality holds. Case II: If $x \neq z$, then LHS is 1. Now, $x \neq y$, then $\rho(x,y)$ is 1 and depending on z, $\rho(y,z)$ is zero or 1. So, we get LHS = RHS = 1 or LHS=1 less than RHS, which is 2.

So, yes indeed this is a well-defined metric function.

Some remarks: To think further:  Suppose we are given the following function : $f(x)=1$ when $x \in \mathscr{Q}$ and $f(x)=0$ when $x \in \mathscr{Q^{'}}$. Can what can we say about this function with respect to the above discrete space ? What are the limit points of such a function ? Is such a function continuous (if so, at which points) in this metric space? Is it dense in this metric space?

Example 2:

The set of all real numbers with distance $\rho(x-y) = |x-y|$ is a metric space, which we denote by $R^{1}$ —- one dimensional real line.

Check: is this a well-defined metric ? Clearly, axiom 1 holds true because $|x-x|=0$ and axiom 2 is true because $\rho(x-y) = |x-y| = |-(x-y)|=|y-x|=\rho(y-x)$. Now, we need to check: $\rho(x,z) \leq \rho(x,y) + \rho(y,z)$. Here, LHS is $|x-z|=|x-y+y-z| = |(x-y) + (y-z)| \leq |x-y| + |y-z| = \rho(x,y) + \rho(y,z)$ where we have used the triangle inequality. So, axiom 3 holds true.

So, this is indeed a well-defined metric.

Example 3:

The set of all ordered n-tuples $x=(x_{1}, x_{2}, \ldots, x_{n})$ of real numbers $x_{1}, x_{2}, \ldots, x_{n}$ with distance

$\rho(x,y)= \sqrt{\Sigma_{k=1}^{n}(x_{k}-y_{k})^{2}}$ ….call this relation I

is a metric space denoted by $R^{n}$ and called n-dimensional Euclidean space (or, simply Euclidean n-space). The distance (I) obviously satisfies axioms 1 and 2 of definition of a metric. Moreover, it can be seen that (I) satisfies the third axiom also:

In fact, let $x=(x_{1}, x_{2}, \ldots, x_{n})$, $y=(y_{1}, y_{2}, y_{3}, \ldots, y_{n})$, $z=(z_{1}, z_{2}, \ldots, z_{n})$ be three points in $R^{n}$.

Futher, let $a_{k}=x_{k}-y_{k}$ and $b_{k}=y_{k}-z_{k}$ when $k=1,2,\ldots, n$.

Then, the triangle inequality takes the form:

$\sqrt{\Sigma_{k=1}^{n}(x_{k}-z_{k})^{2}} \leq \sqrt{\Sigma_{k=1}^{n}(x_{k}-y_{k})^{2}} + \sqrt{\Sigma_{k=1}^{n}(y_{k}-z_{k})^{2}}$….let us call this Relation II.

Or equivalently,

$\sqrt{\Sigma_{k=1}^{n}(a_{k}+b_{k})^{2}} \leq \sqrt{\Sigma_{k=1}^{n}a_{k}^{2}} + \sqrt{\Sigma_{k=1}^{n}b_{k}^{2}}$….call this as relation II’

It follows from the Cauchy-Schwarz inequality that $(\Sigma_{k=1}^{n}a_{k}b_{k})^{2} \leq \Sigma_{k=1}^{n}a_{k}^{2} \times \Sigma_{k=1}^{n}b_{k}^{2}$ …call this as relation III.

so that we have now,

$\Sigma_{k=1}^{n}(a_{k}+b_{k})^{2} = \Sigma_{k=1}^{n}a_{k}^{2} + 2 \Sigma_{k=1}^{n}a_{k}b_{k} + \Sigma_{k=1}^{n}b_{k}^{2} \leq \Sigma_{k=1}^{n}a_{k}^{2}+2\sqrt{\Sigma_{k=1}^{n}a_{k}^{2}\times \Sigma_{k=1}^{n}b_{k}^{2}} + \Sigma_{k=1}^{n}b_{k}^{2} = (\sqrt{\Sigma_{k=1}^{n}a_{k}^{2}} + \sqrt{\Sigma_{k=1}^{n}}b_{k}^{2})^{2}$.

Taking square roots, we get II’ and hence, II.

QED.

Example 4:

Take the same set of ordered tuples as in preceding example $x = (x_{1}, x_{2}, \ldots, x_{n})$, but this time define the distance function by $\rho_{1}(x,y)= \Sigma_{k=1}^{n}|x_{k}-y_{k}|$….call this as relation IV.

It is clear that this is also a well-defined metric function.

Check: Axiom 1 is obvious. So, also axiom 2 because $|a-b|=|b-a|$. Axiom 3 holds true because the following general inequality holds true: $|\pm x_{1} \pm x_{2} \ldots \pm x_{n}| \leq |x_{1}|+|x_{2}|+ \ldots + |x_{n}|$.

Example 5:

Take the same set as in the previous two examples, but now let us define the distance to between two points $x=(x_{1},x_{2}, \ldots, x_{n})$ and $y=(y_{1}, y_{2}, \ldots, y_{n})$ to be $\rho_{0}(x,y)= \max_{1 \leq k \leq n}|x_{k}-y_{k}|$….call this V.

This is also a well-defined metric function.

This space, denoted by $R_{0}^{n}$ is often as usual the Euclidean space $R^{n}$.

Remark: The last three examples show that it is sometimes important to use a different notation for a metric space than for the underlying set of points in the space, since the latter can be “metrized” in a variety of different ways.

Example 6:

The set $C_{[a,b]}$ of all continuous functions defined on the closed interval $[a,b]$ with distance $\rho(f,g)=-\max_{a \leq t \leq b}|f(t)-g(t)|$…call this VI. This is a metric space of great importance in analysis.

Let us verify so:

*****

****

***

This metric space and the underlying set of points in the space will both be denoted by $C_{[a,b]}$. Instead of $C_{[0,1]}$ we just write C. A space like $C[a,b]$ is often called a “function space” to emphasize that its elements are functions.

Example 7:

Let $I_{2}$ be the set of all “infinite” sequences : $x=(x_{1}, x_{2}, \ldots, x_{k}, \ldots)$ of real numbers $x_{1}, x_{2}, \ldots, x_{k}, \ldots$ satisfying the convergence condition:

$\Sigma_{k=1}^{\infty}x_{k}^{2} \leq \infty$

Note: the infinite sequence with the general term $x_{k}$ can be written as $\{ x_{k}\}$ or simply as $x_{1}, x_{2}, \ldots, x_{k}, \ldots$  (this notation is familiar from calculus). It can also be written in “point notation” as $x=(x_{1}, x_{2}, \ldots, , x_{k}, \ldots)$, that is, as an “ordered $\infty-$tuple” generalizing the notion of an ordered n-tuple. (In writing $x_{k}$) we have another use of curly brackets, but the context will always prevent any confusion between the sequence $x_{k}$  and the set whose only element is $x_{k}$).

Where distance between points is defined by

$\rho(x,y)=\sqrt{\Sigma_{k=1}^{\infty}(x_{k}-y_{k}^{2})}$…call this VII.

Clearly, VII makes sense for all $x, y \in l_{2}$ since it follows from the elementary inequality

$(x_{k} \pm y_{k})^{2}\leq 2(x_{k}^{2}+y_{k}^{2})$ that the convergence of the two series $\Sigma_{k=1}^{n}x_{k}^{\infty}$ and $\Sigma_{k=1}^{\infty}y_{k}^{2}$ also implies the convergence of the series $\Sigma_{k=1}^{\infty}(x_{k}-y_{k})^{2}$.

At the same time, we find that if the points $(x_{1}, x_{2}, \ldots, x_{k}, \ldots)$ and $(y_{1}, y_{2}, \ldots, y_{k}, \ldots)$ both belong to $l_{2}$, then so does the point:

$(x_{1}+y_{1}, x_{2}+y_{2}, \ldots, x_{k}+y_{k}, \ldots)$

(since the lim of a sum of two sequences is the sum of the individual limits)

The function VII obviously has the first two defining properties of a distance. To verify the triangle inequality, which takes the form:

$\sqrt{\Sigma_{k=1}^{\infty}(x_{k}-z_{k})^{2}} \leq \sqrt{\Sigma_{k=1}^{\infty}(x_{k}-y_{k})^{2}} + \sqrt{\Sigma_{k=1}^{\infty}(y_{k}-z_{k})^{2}}$ ….call this VIII.

for the metric VII, we first note that all three series converge, for the reason just given. Moreover, the inequality:

$\sqrt{\Sigma_{k=1}^{n}(x_{k}-z_{k})^{2}} \leq \sqrt{\Sigma_{k=1}^{n}(x_{k}-y_{k})^{2}} + \sqrt{\Sigma_{k=1}^{n}(y_{k}-z_{k})^{2}}$…call this IX, holds for all n, (as proved in Example 3 above). Taking the limit as $n \rightarrow \infty$ in IX, we get VIII, thereby satisfying the triangle inequality in $l_{2}$. Therefore, $l_{2}$ is a metric space.

Example 8:

As in Example 6, consider the set of all functions continuous on the interval $[a,b]$, but now let us define the metric by the formula:

$\rho(x,y) = (\int_{a}^{b}[x(t)-y(t)]^{2}dt)^{\frac{1}{2}}$….call this X.

The resulting metric space will be denoted by $C_{[a,b]}^{2}$. The first two axioms of the metric clearly hold, and the fact that X satisfies the triangle inequality is an immediate consequence of the following Schwarz’s inequality:

$(\int_{a}^{b}x(t)y(tdt))^{2} \leq \int_{a}^{b}x^{2}(t)dt \times \int_{a}^{b}y^{2}(t)dt$

(see Problem 3 in the exercises below), by the continuous analogue of the argument given in example 3 above.

Example 9:

Next consider the set of all bounded infinite sequences of real numbers $x=(x_{1},x_{2}, \ldots, x_{k}, \ldots)$ and let

$\rho(x,y) = \sup_{k} {|x_{k}-y_{k}|}$….call this XII.

This gives a metric space which we denote by m. The fact that XII satisfies axioms 1 and 2 of a metric space is obvious by the definition of a supremum.

Axiom 3 can be verified as follows:

***

***

Example 10:

As in example 3, consider the set of all ordered n-tuples, $x=(x_{1}, x_{2}, \ldots, x_{k}, \ldots)$, but now let the metric be given by the more general formula as follows:

$\rho_{p}(x,y)=(\Sigma_{k=1}^{n}|x_{k}-y_{k}|^{p})^{\frac{1}{p}}$….call this XIII.

where p is a fixed real number greater than or equal to 1. (Examples 3 and 4 correspond to the cases $p=2$ and $p=1$, respectively.) This gives a metric space, which we denote by $R_{p}^{n}$.

It is obvious that $\rho_{p}(x,y) = 0$ if and only if $x=y$.

It is obvious that $\rho_{p}(x,y)=\rho_{p}(y,x)$.

But, verification of the third axiom of the definition of a metric (XIII) (that is, the triangle inequality) requires a little work as follows:

Let $x=(x_{1}, x_{2}, \ldots, x_{n})$, $y=(y_{1}, y_{2}, \ldots, y_{n})$, $z=(z_{1}, z_{2}, \ldots, z_{n})$ be three points in $R_{p}^{n}$, and let:

$a_{k}=x_{k}-y_{k}$, $b_{k}=y_{k}-z_{k}$ for $k=1,2,\ldots, n$ just as in example 3. Then, the triangle inequality

$\rho_{p}(x,z) \leq \rho_{p}(x,y)+\rho_{p}(y,z)$

takes the form of Minkowski’s inequality:

$(\Sigma_{k=1}^{\infty}|a_{k}+b_{k}|^{p})^{\frac{1}{p}} \leq (\Sigma_{k=1}^{\infty}|a_{k}|^{p})^{\frac{1}{p}} + (\Sigma_{k=1}^{\infty}|b_{k}|^{p})^{\frac{1}{p}}$

Call the above inequality as XIV in the current blog article.

PS: I think the proof of the Minkowski inequality can be found in any standard text on Inequalities by B. J. Venkatachala, for example;; or, in wikipedia.

The above inequality holds true clearly for $p=1$, and hence, we assume the case $p \geq 1$.

The proof of XIV in turn again is based on Holder’s inequality:

$\Sigma_{k=1}^{n}|a_{k}b_{k}| \leq (\Sigma_{k=1}^{n}|a_{k}|^{p})^{\frac{1}{p}} (\Sigma_{k=1}^{n}|b_{k}|^{q})^{\frac{1}{q}}$

Call the above as inequality XV.

Where the numbers $p>1$, $q >1$ satisfy the condition:

$\frac{1}{p} + \frac{1}{q} =1$….call this as XVI.

We begin by observing that the inequality XV is homogeneous, that is, if it holds for any two points $(a_{1}, a_{2}, \ldots, a_{n})$ and $(b_{1}, b_{2}, \ldots, b_{n})$ then it holds for the two points $(\lambda a_{1}, \lambda a_{2}, \ldots, \lambda a_{n})$ and $(\mu b_{1}, \mu b_{2}, \ldots, \mu b_{n})$ where $\lambda$ and $\mu$ are any two real numbers. Therefore, we need only prove XV for the following case:

$\Sigma_{k=1}^{n}|a_{k}|^{p}=\Sigma_{k=1}^{n}|b_{k}|^{p}$….call this relation XVII.

Thus, assuming that XVII holds, we now have to prove that: $\Sigma_{k=1}^{n}|a_{k}b_{k}| \leq 1$…call this XVIII.

Consider the two areas $S_{1}$ and $S_{2}$, associated with the curve defined in the $\xi\eta$-plane and given by the equation:

$\eta = \xi^{p-1}$, or equivalently by the equation:

$\xi = \eta^{q-1}$

Then, clearly $S_{1}=\int_{0}^{a}\xi^{p-1}d\xi = \frac{a^{p}}{p}$, and $S_{2}=\int_{0}^{b}\eta^{q-1}d\eta = \frac{b^{q}}{q}$

Moreover, it is apparent (if we draw the figure suitably) that $S_{1}+S_{2} \geq ab$ for arbitrary positive a and b. It follows that $ab \leq \frac{a^{p}}{q} + \frac{b^{q}}{q}$…call this relation (19 or XIX).

Setting $a = |a_{k}|$, $b=|b_{k}|$, summing over k from 1 to n, and taking account of (16, or XVI) and (17, or XVII), we get the desired inequality (18, or XVIII). This proves Holder’s inequality (15 or XV). Note that (15 or XV) reduces to Schwarz’s inequality if $p=2$.

It is now an easy matter to prove Minkowski’s inequality (14 or XIV), starting from the identity

$(|a|+|b|)^{p} = (|a|+|b|)^{p-1}|a|+(|a|+|b|)^{p-1}|b|$.

In fact, putting $a=a_{k}$, $b=b_{k}$ and summing over k from 1 to n, we obtain

$\Sigma_{k=1}^{n}(|a_{k}|+|b_{k}|)^{p}=\Sigma_{k=1}^{n}(|a_{k}|+|b_{k}|)^{p-1}|a_{k}|+\Sigma_{k=1}^{n}(|a_{k}|+|b_{k}|)^{p-1}|b_{k}|$

Next, we apply Holder’s inequality (15 or XV) to both sums on the right, bearing in mind that $(p-1)q=p$:

$\Sigma_{k=1}^{n}(|a_{k}|+|b_{k}|)^{p} \leq (\Sigma_{k=1}^{n}(|a_{k}|+|b_{k}|)^{p})^{\frac{1}{q}}([\Sigma_{k=1}^{n}|a_{k}|^{p}]^{\frac{1}{p}}+[\Sigma_{k=1}^{n}|b_{k}|^{p}]^{\frac{1}{p}})$

Dividing both sides of this inequality by

$(\Sigma_{k=1}^{n}(|a_{k}|+|b_{k}|)^{p})^{\frac{1}{a}}$

we get

$(\Sigma_{k=1}^{n}(|a_{k}|+|b_{k}|)^{p})^{\frac{1}{p}} \leq (\Sigma_{k=1}^{n}|a_{k}|^{p})^{\frac{1}{p}} + (\Sigma_{k=1}^{n}|b_{k}|^{p})^{\frac{1}{p}}$

which immediately implies (14 of XIV), thereby proving the triangle inequality in $R_{p}^{n}$.

QED.

Example 11:

Finally, let $l_{p}$ be the set of all infinite sequences $x = (x_{1}, x_{2}, \ldots, x_{k}, \ldots)$ of real numbers satisfying the convergence condition

$\Sigma_{k=1}^{n}x_{k}^{p} < \infty$ for some fixed number $p \geq 1$, where distance between points is defined by

$\rho(x,y) = (\Sigma_{k=1}^{\infty}|x_{k}-y_{k}|^{p})^{\frac{1}{p}}$….call this (20 or XX)

(the case $p=2$ has already been considered in Example 7). It follows from Minkowski’s inequality (14 or XIV) that

$(\Sigma_{k=1}^{n}|x_{k}-y_{k}|^{p})^{\frac{1}{p}} \leq (\Sigma_{k=1}^{n}|x_{k}|^{p})^{\frac{1}{p}} + (\Sigma_{k=1}^{n}|y_{k}|^{p})^{\frac{1}{p}}$….call this (21 or XXI) for any n.

Since the series $\Sigma_{k=1}^{\infty}|x_{k}|^{p}$, and $\Sigma_{k=1}^{\infty}|y_{k}|^{p}$ converge, by hypothesis, we can take the limit as $n \rightarrow \infty$ in (21 or XXI) obtaining

$(\Sigma_{k=1}^{\infty}|x_{k}-y_{k}|^{p})^{\frac{1}{p}} \leq (\Sigma_{k=1}^{\infty}|x_{k}|^{p})^{\frac{1}{p}} + (\Sigma_{k=1}^{\infty}|y_{k}|^{p})^{\frac{1}{p}} < \infty$.

This shows that (20 or XX) actually makes sense for arbitrary $x, y \in l_{p}$. At the same time, we have verified that the triangle inequality holds in $l_{p}$ (the other two properties of a metric space are obviously satisfied). Therefore, $l_{p}$ is a metric space.

QED.

Remarks:

If $R = (X, \rho)$ is a metric space and M is any subset of X, then obviously $R^{*} = (M, \rho)$ is again a metric space, called a subspace of the original metric space R. This device gives us infinitely more examples of metric spaces.

Example 12:

For $x, y \in R^{1}$, define:

12a) $\rho(x,y) = (x-y)^{2}$

12b) $\rho(x,y) = \sqrt{|x-y|}$

12c) $\rho(x,y) = |x^{2}-y^{2}|$

12d) $\rho(x,y) = |x-2y|$

12e) $\rho(x,y) = \frac{|x-y|}{1+|x-y|}$

Determine for each of these, whether it is a metric or not.

Solution 12a:

Axioms 1 and 2 are clearly satisfied. We have to verify if the following holds true:

$(x-z)^{2} \leq (x-y)^{2} + (y-z)^{2}$. Clearly, RHS is $x^{2} + 2y^{2}+ z^{2} -2xy -2yz$ whereas LHS is $x^{2}+y^{2}-2xy$ so it may not always be true that LHS is lesser than or equal to RHS.

Hence, this is not a metric function.

Solution 12b:

this also satisfies the first two axioms of the definition of a metric.

So, we have to verify if the following is true:

$\rho(x,z) \leq \rho(x,y) + \rho(y,z)$, that is, TPT:

$\sqrt{|x-z|} \leq \sqrt{|x-y|} + \sqrt{|y-z|}$.

Consider the following:

$|x-z|=|x-y+y-z| \leq |x-y|+|y-z|$

Also, $(\sqrt{|x-y|} + \sqrt{|y-z|})^{2} = |x-y|+|y-z|+2\sqrt{|x-y|\times|y-z|}$

So the third axiom holds true in this case. So, the given function is a metric.

Solution 12c:

Once again, the first two axioms clearly hold.

We have to verify if the following holds true:

$\rho(x,z) \leq \rho(x,y) = \rho(y,z)$, that is, to prove that:

$|x^{2}-z^{2}| \leq |x^{2}-y^{2}| + |y^{2}-z^{2}|$, which is obviously true by triangle inequality of real numbers. So, the given function is a metric.

Solution 12d:

Clearly, the first two axioms do not hold. It can easily be checked that the third axiom also does not hold. So, the given function is not a metric.

Solution 12e:

The first axiom holds true.

To check the second axiom consider and compare:

$\rho(x,y) = \frac{|x-y|}{1+|x-y|}$ whereas $\rho(y,x) = \frac{|y-x|}{1+|y-x|}=\frac{|x-y|}{1+|x-y|} = \rho(x,y)$ clearly again.

To verify the third axiom, we have to check if the following is true:

$\rho(x,z) \leq \rho(x,y) + \rho(y,z)$, that is, to prove that:

$\frac{|x-z|}{1+|x-z|} \leq \frac{|x-y|}{1+|x-y|} + \frac{|y-z|}{1+|y-z|}$. A little algebraic work shows that this is aot always possible. Hence, the given function is not a metric.

This site uses Akismet to reduce spam. Learn how your comment data is processed.