VI Convergence. Open and Closed Sets.

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 \epsilonneighbourhood 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:

  1. If M \subset N, then [M] \subset [N]
  2. [[M]] = [M]
  3. [M \bigcup N] = [M] \bigcup [N]
  4. [\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.

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