Heine-Borel Theorem (Compact Equivalences)

Heine-Burel Theorem

Let $K \subseteq R$ . The following are equivalent:

$K$ is compact.
$K$ is closed and bounded.
Every Open Cover of $K$ has a Finite Subcover.

Proof

Since Characterization of Compactness in R gives $(1) ⟺ (2)$ then all we have to show is $(3) \Rightarrow (2)$ and then $(3) \Leftarrow (2)$ .

( $\Rightarrow$ ): Suppose every open cover of $K$ has a finite subcover. To show that $K$ is bounded, we construct an open cover for $K$ by defining $O_{x}$ to be an open interval of radius 1 around each point $x \in K$ . In the language of Epsilon-Neighborhoods, $O_{x} = V_{1} (x)$ . The open cover ${O_{x} : x \in K}$ then must have a finite subcover (see the supposition) ${O_{x_{1}}, \dots, O_{x_{n}}}$ . Because $K$ is contained in a finite union of bounded sets (ie: $K \subseteq ⋃_{i = 1}^{n} O_{x_{i}}$ ) then $K$ itself must be bounded (just take the right boundary of the largest $x_{i}$ ).

To show that $K$ is closed, there's more work. We'll argue it by contradiction. Let $(y_{n})$ be a Cauchy Sequence contained in $K$ where $y_{n} \to y$ . To show that $K$ is closed it is enough to show that $y \in K$ .

Assume for contradiction that $y \notin K$ . Then $\forall x \in K$ , then each $x$ is some positive distance away from $y$ (otherwise then $x = y \Rightarrow y \in K$ ). Thus, we can try to construct an open cover by taking $O_{x}$ to be an interval of radius $\frac{| x - y |}{2}$ around each point $x \in K$ . Because we suppose (3), the resulting open cover ${O_{x} : x \in K}$ must have some finite subcover ${O_{x_{1}}, \dots, O_{x_{n}}}$ . The contradiction comes from noticing that this finite subcover cannot contain all of the elements of the sequence $(y_{n})$ . To show this, set:

ε_{0} = min {\frac{| x_{i} - y |}{2} : 1 \leq i \leq n}

Because $(y_{n}) \to y$ and $ε_{0} > 0$ (remember, all the distances are positive), then $\exists N \in N$ such that:

| y_{n} - y | < ε_{0} = min {\frac{| x_{i} - y |}{2} : 1 \leq i \leq n}

for $n \geq N$ . Namely just choosing $n = N$ here works, because notice that:

x \in O_{x_{i}} ⟺ \frac{| x_{i} - x |}{2} < \frac{| x_{i} - y |}{2}

Notice the inequality never holds for $y$ so then $y \notin O_{x_{i}}$ for all $1 \leq i \leq n$ . Thus, $y_{N} \notin O_{x_{1}}, \dots, O_{x_{n}}$ , so then:

y_{N} \notin ⋃_{i = 1}^{n} O_{x_{i}}

So our cover doesn't actually cover all of $K$ , which is a contradiction. Thus $y \in K$ and thus $K$ is closed and bounded.

( $\Leftarrow$ ): We'll use the following lemma:

DeMorgan's For Infinite Sets

Given a collection of sets ${E_{λ} : λ \in Λ}$ :

{(⋃_{λ \in Λ} E_{λ})}^{c} = ⋂_{λ \in Λ} E_{λ}^{c}, {(⋂_{λ \in Λ} E_{λ})}^{c} = ⋃_{λ \in Λ} E_{λ}^{c}

Proof

(1) If $x \in {(⋃_{λ \in Λ} E_{λ})}^{c}$ then $x \notin ⋃_{λ \in Λ} E_{λ}$ , meaning $x \notin E_{λ}$ for any $λ \in Λ$ . This implies that $x \in E_{λ}^{c}$ for all $λ$ . So then $x \in ⋂_{λ \in Λ} E_{λ}^{c}$ Thus giving $\subseteq$ for (1).

For $\supseteq$ , let $x \in ⋂_{λ \in Λ} E_{λ}^{c}$ . Then $x \notin E_{λ}$ for all $λ$ meaning $x \notin ⋃_{λ \in Λ} E_{λ}$ so then $\supseteq$ .

(2) Comes from doing a very similar argument to (1).

☐

We'll show closure inductively. Let $F = F_{1} \cup F_{2}$ . Let $x_{n} \in F$ and $x = lim_{n \to \infty} x_{n}$ . Let $(x_{n_{k}})$ be a subsequence of $(x_{n})$ fully contained in $F_{1}$ or $F_{2}$ . The subsequence must also converge to $x$ , so $x \in F_{1} \cup F_{2} = F$ . This is the base case.

For the inductive step, let $F = ⋂_{λ \in Λ} F_{λ}$ . Then:

F^{c} = ⋃_{λ \in Λ} F_{λ}^{c}

Each $F_{λ}^{c}$ is closed, thus $F^{c}$ is open, so then $F = (F^{c})^{c}$ is closed.

☐

Lecture Proof

We did the proof of this in class, but I prefer to use the textbook's proof of the idea (it's just easier to follow). If you want to view the proof in another context though see the following.

Scratch:

For ( $2 \to 3$ ) it's kind-of a mess. But the idea is that we can get a suprema $s$ and then actually say that we could go even futher past the suprema to indicate a finite subcover.

Proof

We'll show $(1) \Rightarrow (2) \Rightarrow (3) \Rightarrow (1)$ .

(1 $\to$ 2): See Characterization of Compactness in R.

(2 $\to$ 3): Let's prove a useful lemma really quick:

Let

A \subseteq R

be nonempty and bounded. Then

\exists (a_{n}) \subseteq A

and

\exists (b_{n}) \subseteq A

such that

a_{n} \to sup A

and

b_{n} \to inf A

Proof

For $n \in N$ choose $a_{n} \in A$ such that $sup A - \frac{1}{n} < a_{n} \leq sup A$ . Using the Squeeze Theorem then we can show $a_{n} \to sup A$ . The same can be done for the $inf$ by using $inf A \leq b_{n} < inf A + \frac{1}{n}$ .

☐

Let's start the proof. Suppose $K$ is closed and bounded. Then all limit points of $K$ are in $K$ and any sequence $(x_{n}) \subseteq K$ is bounded. If $K$ is empty then we are done so suppose $K \neq \emptyset$ . Because of this then $sup K$ and $inf K$ exist, so then let $a = sup K$ and $b = inf K$ . Therefore $K \subseteq [a, b]$ . Note that $a, b \in K$ since $K$ is closed.

Now let ${U_{λ} | λ \in Λ}$ be an open cover of $K$ . Let $S = {c \in [a, b] | K \cap [a, c] has a finite subcover}$ . Notice that since $S \subseteq [a, b]$ then $S$ is bounded. Further $S \neq \emptyset$ since $a \in S$ for example. Thus we can take $s = sup S$ . Since $K$ is closed, then $s \in K$ (see the Lemma we just proved). Thus then $\exists U_{λ_{0}}$ such that $s \in U_{λ_{0}}$ . Therefore $\exists ϵ > 0$ such that $s \in V_{ϵ} (s) \subseteq U_{λ_{0}}$ .

Since $s - ϵ < s$ then $\exists s_{1} \in S$ such that $s - ϵ < s_{1} \leq s$ . Therefore $\exists$ a finite subcover of $K \cap [a, s_{1}]$ . Let's call that subcover $U_{λ_{1}}, \dots, U_{λ_{n}}$ . Then taking:

U_{λ_{0}}, U_{λ_{1}}, \dots, U_{λ_{n}}

certainly still covers $K \cap [a, s]$ . Therefore $s \in S$ .

Now assume for contradiction that $s < b$ . Then $\exists s_{2}$ such that $s < s_{2} \leq b$ and $s_{2} < s + ϵ$ and $K \cap [a, s_{2}]$ is covered by $U_{λ_{0}}, \dots, U_{λ_{n}}$ . Therefore $s_{2} \in S$ is a contradiction $s_{2} > s$ which contradicts $s = sup S$ .

(3 $\to$ 1): Suppose every open cover of $K$ has a finite subcover. Now if you consider:

⋃_{n \in N} (- n, n) = R \supseteq K

Thus we found a finite subcover $(- n_{1}, n_{1}), \dots, (- n_{ℓ}, n_{ℓ})$ . Let $M = max (n_{i} | i = 1, \dots, ℓ)$ . Then $M$ is a bound for $K$ , so $\forall x \in K$ then $| x | \leq M$ .

Now let $(x_{n}) \subseteq K$ . By the Balzano-Weierstrass Theorem then $\exists$ a subsequence $x_{n_{k}}$ where $x_{n_{k}} \to x$ .

We need to show that $x \in K$ to get (1), so assume for contradiction that $x \notin K$ . For $y \in K$ let:

U_{y} = V_{\frac{| y - x |}{2}} (y)

forms an open over of $K$ , so then $\exists$ a finite subcover $U_{y_{1}}, \dots, U_{y_{n}}$ . Now let $ϵ = min {\frac{| y - x |}{2} | i = 1, \dots, n}$ . None of the $U_{y}$ epsilon-neighborhoods get into distance of $x$ , since every $y \in K$ is a distance $\geq ϵ$ from $x$ . This contradicts $x$ being a limit.

☐