# Compact sets, perfect sets, connected sets

Reference: Principles of Mathematical Analysis by Walter Rudin

2.31 Definition: By an open cover of a set E in a metric space X we mean a collection $\{ G_{\alpha} \}$ of open subsets of X such that $E \subset \bigcup_{\alpha}G_{\alpha}$.

2.32 Definition A subset K of a metric space X is said to be compact if every open cover of K contains a finite subcover.

More explicitly, the requirement is that if $\{ G_{\alpha}\}$ is an open cover of K, then there are finitely many indices $\alpha_{1}, \alpha_{2}, \ldots, \alpha_{n}$ such that

$K \subset G_{\alpha_{1}} \bigcup G_{\alpha_{2}} \ldots \bigcup G_{\alpha_{n}}$

The notion of compactness is of great importance in analysis, especially in connection with continuity.

It is clear that every finite set is compact. The existence of a large class of infinite compact sets in $R^{k}$ will follow with Theorem 2.41

We observed earlier (section 2.29) that if $E \subset Y \subset X$ then E may be open relative to Y without being open relative to Y. The property of being open thus depends on the space in which it is embedded. Note that the same is true of the property of being closed.

Compactness, however, behaves better as we shall see now. To formulate the next theorem, let us say, temporarily that K is compact relative to X if the requirements of Definition 2.32 are met.

2.33 Theorem: Suppose $K \subset Y \subset X$. Then K is compact relative to X if and only if K is compact relative to Y.

By the virtue of the theorem we are able, in many situations, to regard compact sets as metric spaces in their own right, without any paying any attention to any embedding space. In particular, although it makes little sense to talk of open spaces or of closed spaces (every metric space X is an open subset of itself, and is a closed subset of itself), it does make sense to talk of compact metric spaces.

Proof:

Suppose K is compact relative to X, and let $\{ V_{\alpha}\}$ be a collection of sets, open relative to Y, such that $K \subset \bigcup_{\alpha}V_{\alpha}$. By theorem 2.30 there are sets $G_{\alpha}$ open relative to X, such that $V_{\alpha} = Y \bigcap G_{\alpha}$ for all x, and since K is compact relative to X, we have

(22)…..$K \subset \subset G_{\alpha_{1}} \bigcup \ldots \bigcup G_{\alpha_{n}}$

for some choice of finitely many indices $\alpha_{1}, \alpha_{2}, \ldots, \alpha_{n}$. Since $K \subset Y$, the above equation 22 implies that

(23) $K \subset V_{\alpha_{1}} \bigcup \ldots \bigcup V_{\alpha_{n}}$

This proves that K is compact relative to Y.

Conversely, suppose K is compact relative to Y and let $\{ G_{\alpha}\}$ be a collection of open subsets of X which covers K, and put $V_{\alpha} = Y \bigcap G_{\alpha}$. Then (23) will hold for some choice of $\alpha_{1}$ of $\alpha_{1}, \alpha_{2}, \ldots, \alpha_{n}$; and since $V_{\alpha} \subset G_{\alpha}$ (23) implies (22).

This completes the proof. QED.

2.34 Theorem Compact subsets of metric spaces are closed.

Proof:

Let K be a compact subset of a metric space X. We shall prove that the complement of K is an open subset of qX.

Suppose $p \in X$, and $p \notin K$. If $q \in K$, let $V_{q}$ and $W_{q}$ be neighbourhoods of p and q, respectively, of radius less than $\frac{1}{2}d(p,q)$. Since K is compact, there are finitely many points $q_{1}, \ldots, q_{n}$ in K such that

$K \subset W_{q_{1}} \bigcup \ldots \bigcup W_{q_{n}}$

If $V = V_{q_{1}} \bigcap \ldots V_{q_{n}}$, then V is a neighbourhood of p which does not intersect W. Hence, $V \subset K^{c}$ so that p is an interior point of $K^{c}$. Thus, we have proved the theorem. QED.

2.35 Theorem Closed subsets of compact sets are compact.

Proof: Suppose $F \subset K \subset X$, F is closed (relative to X), and K is compact. Let $\{ V_{\alpha}\}$ be an open cover of F. If $F^{c}$ is adoined to $\{ V_{\alpha}\}$, we obtain an open cover $\Omega$ of K. Since K is compact, there is a finite subcollection $\Phi$ of $\Omega$ which covers K, and hence, F. If $F^{c}$ is a member of $\Phi$, we may remove it from $\Phi$ and still retain an open cover of F. We have thus proved that a finite subcollection of $\{ V_{\alpha}$ covers F.

PS: Remark: Note the technique of proof here.

Corollary: If F is closed and K is compact, then $F \bigcap K$ is compact.

Proof: Theorem 2.24b and 2.34 show that $F \bigcap K$ is closed since $F \bigcap K \subset K$. Theorem 2.35 shows that $F \bigcap K$ is compact. Note: Theorem 2.24b is as follows: For any collection $\{ F_{\alpha}\}$ of closed sets, $\bigcap_{\alpha}F_{\alpha}$ is closed. Theorem 2.34 just proved above says that compact subsets of metric spaces are closed.

2.36 Theorem If $\{ K_{\alpha}\}$ is a collection of compact subsets of a metric space X such that the intersection of every finite subcollection of $\{ K_{\alpha}\}$ is nonempty, then $\bigcap K_{\alpha}$ is nonempty.

Proof:

PS: This proof of Rudin is not v clear to me. If one of my readers can help, I would be obliged.

2.37 Theorem: If E is an infinite subset of a compact set K, then E has a limit point in K.

2.37 Proof: If no point of K were a limit point of E, (proof by contradiction), then each $q \in K$ would have a neighbourhood which contains at most one point of E (namely, q, if $q \in E$). (This follows just from the definition of limit point). It is clear that no finite subcollection of $V_{q}$ can cover E and the same is true of K, since $E \subset K$. This contradicts the compactness of K. (recall definition of a compact set).

2.38 Theorem If $\{ I_{n} \}$ is a sequence of intervals in $R^{1}$, such that $I_{n} \supset I_{n+1}$ where $n \in \mathcal{N}$, then $\bigcap_{1}^{\infty} I_{n}$ is not empty.

2.38 Proof: If $I_{n} = [a_{n}, b_{n}]$ let E be the set of all $a_{n}$. Then E is nonempty and bounded above by $b_{1}$. Let x be the sup of E. If m and n are positive integers, then

$a_{n} \leq a_{m+n} \leq b_{m+n} \leq b_{m}$

so that $x \leq b_{m}$ for each m. Since it is obvious that $a_{m} \leq x$, we see that $x \in I_{m}$ for m=1,2,3….

QED.

2.39 Theorem: Let k be a positive integer. If $\{ I_{n}\}$ is a sequence of k-cells such that $I_{n} \supset I_{n+1}$ (n=1,2,3,…) then $\bigcap_{1}^{n} I_{n}$ is not empty.

Proof: Let $I_{n}$ consist of all points $x = (x_{1}, x_{2}, \ldots, x_{n})$ such that

$a_{n,j} \leq x_{j} \leq b_{n,j}$ ($1 \leq j \leq k$) (where $n = 1,2,3,\ldots$)

and put $I_{n,j} = [a_{n,j}. b_{n,j}]$

For each j, the sequence $\{ I_{n,j}\}$ satisfies the hypotheses of previous theorem 2.38. Hence there are real numbers $x_{j}^{*}$ with $1 \leq j \leq k$ such that

$a_{i,j} \leq x_{j}^{*} \leq b_{n,j}$ with $1 \leq j \leq k$ and $n=1,2,3,\ldots$

Setting $x^{*} = (x_{1}^{*}, \ldots, x_{k}^{*})$ we see that $x^{*} \in I_{n}$ for $n=1,2,3,\ldots$. Thus the theorem is proved. QED.

2.40 Theorem Every k-cell is compact.

Proof:

Let I be a k-cell, consisting of all points $x = (x_{1}, \ldots, x_{k})$ such that $a_{j} \leq x_{j} \leq b_{j}$ with $1 \leq j \leq k$.

Put $\delta = (\Sigma_{1}^{k}(b_{j}-a_{j})^{2})^{1/2}$

Then $|x-y| \leq \delta$ if $x \in I, y \in I$.

Suppose, to get a contradiction, that there exists an open cover $\{ G_{\alpha} \}$ of I which contains no finite subcover of I. Put $c_{j} = \frac{(a_{j}+b_{j})}{2}$. The intervals $[a_{j}, c_{j}]$ and $[c_{j}, b_{j}]$ then determine $2^{k}$ k-cells $Q_{i}$ whose union is I. At least one of these sets $Q_{i}$, call it $I_{1}$, cannot be covered by any finite subcollection of $\{ G_{\alpha}\}$ (otherwise I could be so covered). We next subdivide $I_{1}$ and continue the process. We obtain a sequence $I_{n}$ with the following properties:

(a) $I \supset I_{1} \supset I_{2} \supset I_{3} \supset \ldots$

(b) $I_{n}$ is not covered by any finite subcollection of $\{ G_{\alpha}\}$

(c) if $x \in I_{n}$ and $y \in I_{n}$, then $|x-y| \leq 2^{-n} \delta$

By (a) and Theorem 2.39 there is a point $x^{*}$ which lies in every $I_{n}$. For some x, $x^{*} \in G_{\alpha}$. Since $G_{\alpha}$ is open, there exists $r >0$ such that $|y-x^{*}| implies that $y \in G_{\alpha}$. If n is so large that $2^{-n} \delta (there is such an n, for otherwise $2^{n} \leq \frac{\delta}{r}$ for all positive integers n, which is absurd since R is archimedean), then $(c)$ implies that $I_{n} \subset G_{\alpha}$ which contradicts (b).

This completes the proof.

QED.

The equivalence of (a) and (b) in the next theorem is known as the Heine Borel Theorem.

2.41 Theorem: If a set E in $R^{k}$ has one of the following three properties then it has the other two:

(a) E is closed and bounded;

(b) E is compact.

(c) Every infinite subset of E has a limit point in E.

Proof:

Let us assume (a) is true.

Then $E \subset I$ for some k-cell I, and (b) follows from Theorems 2.40 and 2.35. Theorem 2.37 shows that (b) implies (c). It remains to be shown that (c) implies (a).

If E is not bounded, then E contains points $x_{n}$ with

$|x_{n}| >n$ where $n=1,2,3,\ldots$

The set S consisting of these points $x_{n}$ is infinite and clearly has no limit point in $R^{k}$ hence has none in E. Thus (c) implies that E is bounded.

If E is not closed, then there is a point $x_{0} \in R^{k}$ which is a limit point of E but not a point of E. For $n=1,2,3,\ldots$ there are points $x_{n} \in E$ of E such that $|x_{n}-x_{0}|<\frac{1}{n}$. Let S be the set of these points $x_{n}$. Then S is infinite (otherwise $|x_{n}-x_{0}|$ would have a constant positive value for infinitely many n). S has $|x_{0}|$ as a limit point, and S has no other limit point in $R^{k}$. For if $y \in R^{k}$, $y \neq 0$, then

$|x_{n}-y| \geq |x_{n}-y|-|x_{n}-x_{0}| \geq |x_{n}-y| - \frac{1}{n} \geq \frac{1}{2}|x_{n}-y|$

for all but finitely many n, this shows that y is not a limit point of S (use the following Theorem 2.20: if p is a limit point of E, then every neighbourhood of p contains infinitely many points of E).

Thus, S has no limit points in E, hence E must be closed if (c) holds.

QED.

Remarks: (b) and (c) are equivalent in any metric space (prove this as an exercise ) but that (a) does not, in general imply, (b) and (c) both. Examples are furnished in Exercise 16 of this chapter and by the space $L^{2}$ which is discussed in Chapter 11.

2.42 Theorem (Weierstrass) : Every bounded infinite subset of $R^{k}$ has a limit point in $R^{k}$.

Proof: Being bounded, the set E in question is a subset of a k-cell $I \subset R^{k}$. By Theorem 2.40, I is compact, and so E has a limit point in I, by Theorem 2.37 (If E is an infinite subset of a compact set K, then E has a limit point in K).

Perfect Sets:

2.41 Theorem: Let P be a nonempty perfect set in $R^{k}$. Then P is uncountable.

Proof:

((Note: Definition of a perfect set: E is perfect if E is closed and if every point of E is a limit point of E. ))

Since P has limit points, P must be infinite. Suppose P is countable and denote the points of P by $x_{1}, x_{2}, x_{3}, \ldots$ We shall construct a sequence $\{ V_{n}\}$ of neighbourhoods, as follows:

Let $V_{1}$ be any neighbourhood of $x_{1}$. If $V_{1}$ consists of all $y \in R^{k}$ such that $|y-x_{1}|, the closure $\overline{V_{1}}$ of $V_{1}$ is the set of all $y \in R^{k}$ such that $|y-x| \leq r$. (note this) (this is true or makes sense : we have used the fact that P is also closed, being perfect so the limit point of P can belong to E itself which is the case when $y=x_{1}$).

Suppose $V_{n}$ has been constructed so that $V_{n} \bigcap P$ is not empty. Since every point of P is a limit point of P (note that here we have used the other part of the definition a perfect set), there is a neighbourhood $V_{n+1}$ such that (i) $\overline{V_{n+1}} \subset V_{n}$ (ii) $x_{n} \notin \overline{V_{n+1}}$ (iii) $V_{n+1} \bigcap P$ is not empty. By (iii) $V_{n+1}$ satisfies our induction hypothesis, and the construction can proceed.

Put $K_{n}=\overline{V_{n}} \bigcap P$. Since $\overline{V_{n}}$ is closed and bounded. $\overline{V_{n}}$ is compact. And, by definition of $K_{n}$, it is an infinite set. Since $V_{n}$ is closed and bounded, $\overline{V_{n}}$ is compact. Since $x_{n} \notin K_{n+1}$ no point of P lies in $\bigcap_{1}^{\infty}$. But each $K_{n}$ is nonempty, by (iii) and $K{n} \supset K_{n+1}$ by (i), this contradicts the Corollary to theorem 2.36 (If $\{ K_{n}\}$ is a sequence of nonempty compact sets such that $K_{n} \supset K_{n+1}$ (where $n \in \mathcal{N}$) then $\bigcap_{1}^{\infty}K_{n}$ is not empty. )

Corollary : Every interval $[a,b]$ where $a < b$ is uncountable. In particular, the set of all real numbers is uncountable.

2.44 The Cantor Set: The set which we are now going to construct shows that there exist perfect sets in $R^{1}$ which contain no segment.

Let $E_{0}$ be the interval $[0,1]$.. Remove the segment $(\frac{1}{3}, \frac{2}{3})$ and and let $E_{1}$ be the union of the intervals

$[0,\frac{1}{3}]$ and $[\frac{2}{3}, \frac{1}{1}]$

Remove the middle thirds of these intervals, and let $E_{2}$ be the union of the intervals

$[0,\frac{1}{9}]$ and $[\frac{2}{9}, \frac{3}{9}]$ and $[\frac{6}{9}, \frac{7}{9}]$ and $[\frac{8}{9}, 1]$

Continuing in this way, we obtain a sequence of compact sets $E_{n}$ such that

(a) $E_{1} \supset E_{2} \supset E_{3} \ldots$

(b) $E_{n}$ is the union of $2^{n}$ intervals each of length $3^{-n}$

The set

$P = \bigcap_{n=1}^{\infty}E_{n}$

is called the Cantor set. P is clearly compact, and Theorem 2.36 shows that P is not emtpy. Also, we have shown that P is closed being compact. Now, we have to prove that every point of P is a limit point of P: we can also show that P contains no isolated points: let us therefore look at the construction of P:

No segment of the form

(24) $(\frac{3k+1}{3^{n}}, \frac{3l+2}{3^{n}})$

where k and m are positive integers, has a point in common with P. Since every segment $(\alpha, \beta)$ contains a segment of the form (24) if $3^{-m} < \frac{\beta - \alpha}{6}$, P contains no segment.

To show that P is perfect, it is enough to show that P contains no isolated point. Let $x \in P$ and let S be any segment containing x. Let $I_{n}$ be that interval of $E_{n}$ which contains x. Choose n large enough so that $I_{n} \subset S$. Let $x_{n}$ be an endpoint of $I_{n}$ such that $x_{n} \neq x$.

It follows from the construction of P that $x_{n} \in P$. Hence, x is a limit point of P and P is perfect.

QED.

One of the most interesting properties of the Cantor set is that it provides us with an example of an uncountable set of measure zero. (the concept of measure will be discussed in Chapter 11).

CONNECTED SETS:

2.45 Definition: Two subsets A and B of a metric space X are said to be separated if both $A \bigcap \overline{B}$ and $\overline{A} and \bigcap B$ are both empty; that is, if no point of A lies in the closure of B and no point of B lies in the closure of A.

A set $E \subset X$ is said to be connected if E is not a union of two nonempty separated sets.

2.46 Remark: Separated sets are of course disjoint, but disjoint sets need not be separated. For example, the interval $[0,1]$ and the segment $(1,2)$ are not separated since 1 is a limit point of $(1,2)$. However, the segments $(0,1)$ and $(1,2)$ are separated.

The connected subsets of the line have a particularly simpe structure.

2.47 Theorem: A subset E of the real line $R^{1}$ is connected if and only if it has the following property: if $x \in E$, $y \in E$ and $x, then $z \in E$.

Proof:

If there exist $x \in E$, $y \in E$, and soe $z \in (x,y)$ such that $z \notin E$, then $E = A_{1} \bigcup B$, where

$A_{z} = E \bigcap (-\infty, z)$ and $B_{z} = E \bigcap (z,\infty)$

Since $A_{z} \subset (-\infty, z)$ and $B_{z} \subset (z, \infty)$, they are separated. Hence E is not connected.

To prove the converse (note we prove the contrapositive of the converse) suppose E is not connected. So is separated. Then there are nonempty separated sets A and B such that $A \bigcup B = E$. Choose $x \in A$ and $y \in B$ and assume WLOG that $x < y$. Define

$z = \sup {(A \bigcap [x,y])}$

By theorem 2.28 (Let E be a nonempty set of real numbers which is bounded above Let $y=\sup {E}$. Then $y \in E$ if E is closed), $z \in \overline{A}$ hence $z \notin B$. In particular $x \leq z .

If $z \notin A$, it follows that $x < z and $z \notin E$.

If $z \notin A$, then $z \notin \overline{B}$, hence there exists $z_{1}$ such that $z < z_{1} and $z_{1} \notin B$. Then $x and $z_{1} \notin E$.

QED.

Cheers,

Nalin Pithwa

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