# Verification/Hint for Proof: “Finite Subcovers and the Heine-Borel Theorem”

I hope it's going well. I am hopeful that someone here can assist me with criticizing, critiquing, and correcting my proof for the following problem from Understanding Analysis, ed.2 by S. Abbott.

### Problem

Assume $$K$$ satisfies (i) and (ii) (in the Heine–Borel Theorem), and let $$\{O_{\lambda} : \lambda \in \Lambda\}$$ be an open cover for $$K$$. For contradiction, let’s assume that no finite subcover exists. Let $$I_0$$ be a closed interval containing $$K$$.

1. Show that there exists a nested sequence of closed intervals $$I_0 \supseteq I_1 \supseteq I_2 \supseteq \dotsm$$ with the property that, for each $$n$$, $$I_n \cap K$$ cannot be finitely covered and $$\lim_{n\to \infty}|I_n| = 0$$.
2. Argue that there exists an $$x\in K$$ such that $$x\in I_n$$ for all $$n$$.
3. Because $$x \in K$$, there must exist an open set $$O_{\lambda_0}$$ from the original collection that contains $$x$$ as an element. Explain how this leads to the desired contradiction.

### Definitions Used

Theorem 2.5.2. - Subsequences of a convergent sequence converge to the same limit as the original sequence.

Open Cover - Let $$A \subseteq \mathbb{R}$$. An open cover for A is a (possibly infinite) collection of open sets $$\{O_{\lambda}:\lambda \in \Lambda\}$$ whose union contains the set $$A$$; that is, $$A\subseteq \bigcup_{\lambda \in \Lambda} O_{\lambda}$$.

Finite Subcover - Given an open cover for $$A$$, a finite subcover is a finite sub-collection of open sets from the original open cover whose union still manages to completely contain $$A$$.

Heine–Borel Theorem - Let $$K$$ be a subset of $$\mathbb{R}$$. All of the following statements are equivalent in the sense that any one of them implies the two others:

1. $$K$$ is compact.
2. $$K$$ is closed and bounded.
3. Every open cover for $$K$$ has a finite subcover.

Nested Compact Set Property - If $$I_0 \supseteq I_1 \supseteq I_2 \supseteq I_3 \dotsm$$ is a nested set sequence of nonempty compact sets, then the intersection $$\bigcap_{n=1}^{\infty} = K_n$$ is not empty.

Monotone Convergence Theorem - If a sequence is monotone and bounded, then it converges.

Uniqueness of Limits - The limit of a sequence, when it exists, must be unique.

### My Attempt

1. $$\text{ }$$ Let $$K$$ be a compact set, let $$\{O_{\lambda}:\lambda \in \Lambda\}$$ be an open cover for $$K$$, and let $$I_0 = [a_0, b_0] = \{x \in \mathbb{R}: a_0 \leq x \leq b_0\}$$ be a closed interval containing $$K$$. For the sake of contradiction, assume that no finite subcover exists that contains $$K$$.

$$\text{ }$$ We can begin by letting $$c_0 = \frac{a_0 + b_0}{2}$$ and have from this that $$[a_0, c_0] \cup [c_0, b_0] = I_0$$. Since we assumed that finite subcover for $$K$$ exists, we know that at least one of $$[a_0, c_0] \cap K$$ and $$[c_0, b_0] \cap K$$ does not have a finite subcover for $$K$$. If both $$[a_0, c_0] \cap K$$ and $$[c_0, b_0] \cap K$$ had finite subcovers, $$K \subseteq [a_0, c_0] \cup [c_0, b_0]$$ would imply $$K$$ has a finite subcover, which we assumed is not the case.

$$\text{ }$$ Now we select whichever set $$[a_0, c_0] \cap K$$ or $$[c_0, b_0] \cap K$$ does not have a finite subcover and denote it $$I_1$$. Let $$c_1 = \frac{a_1 + b_1}{2}$$ and have from this that $$[a_1, c_1] \cup [c_1, b_1] = I_1$$. Select whichever set $$[a_1, c_1] \cap K$$ or $$[c_1, b_1] \cap K$$ has no finite subcover and denote it $$I_2$$. Continuing in this fashion, we find that for $$n \in \mathbb{N}$$, $$I_n$$ has at least one half interval whose intersection with $$K$$ has no finite subcover by construction. Thus, for $$n \in \mathbb{N}$$, $$I_n \cap K$$ has no finite subcover.

$$\text{ }$$ In particular, we have that the length of the $$n$$-th interval $$|I_n| = \frac{|I_0|}{2^{n-1}}$$. We have that $$0\leq \frac{|I_0|}{2^{n-1}} \leq a_0 + b_0$$ and that $$\frac{|I_0|}{2^0} > \frac{|I_0|}{2^{1}} > \frac{|I_0|}{2^2} \dotsm > 0$$

This implies that $$(\frac{|I_0|}{2^{n-1}})$$ is bounded and monotone decreasing. By the Monotone Convergence Theorem, we have that $$(\frac{|I_0|}{2^{n-1}}) \to x$$ where $$\frac{|I_0|}{2^0} > x \geq 0$$. Observe that $$(\frac{|I_0|}{2^{2(n-1)}})$$ is a subsequence of $$(\frac{|I_0|}{2^{n-1}})$$. By Theorem 2.5.2. we get that $$(\frac{|I_0|}{2^{2(n-1)}}) \to x$$. Since $$\frac{|I_0|}{2^{2(n-1)}} = \frac{|I_0|}{2^{n-1}} \cdot \frac{|I_0|}{2^{n-1}}$$, then by the Algebraic Limit Theorem, we have $$\lim_{n\to \infty }(\frac{|I_0|}{2^{2(n-1)}}) = \lim_{n\to \infty}\frac{|I_0|}{2^{n-1}} \cdot \lim_{n\to \infty}\frac{|I_0|}{2^{n-1}} = x \cdot x$$ By the Uniqueness of Limits, $$x \cdot x = x^2 = x$$ implies that $$x = 0$$, thus we have that the lengths of the intervals converges to $$0$$, i.e. $$\lim_{n\to \infty} |I_n| = 0$$.

1. For all $$n \in \mathbb{N}$$, we have $$I_n \cap K \subseteq K$$. Since $$K$$ is compact, this implies that $$I_n \cap K$$ is compact. Using the Nested Compact Set Property, we have that $$\bigcap_{n\in N}I_n \cap K \neq \emptyset$$, which means that for each set $$I_n \cap K$$, we know there exists some $$x \in \mathbb{R}$$ where $$x \in K$$ and $$x \in I_n$$.
2. INCOMPLETE

1. $$O_{\lambda_0}$$ covers $$x$$. Pick some $$I_n$$ for some natural number $$n$$. As $$I_n$$ does not have a finite subcover, this implies $$I_n \setminus O_{\lambda_0}$$ does not have a finite subcover and $$I_n \cap O_{\lambda_0}$$ has a finite subcover. Thus $$\exists j > n \; I_{j} \subseteq I_n \setminus O_{\lambda_0} \Rightarrow x \notin I_{j}$$. This is a contradiction.
Once you have $$|I_n| = \frac{|I_0|}{2^{n-1}}$$, you can directly conclude $$\lim_{n \rightarrow \infty} |I_n| = 0$$, by taking limits on both sides, and noting that $$|I_0|$$ is a constant.
Edit made: Note that actually $$I_{n+1}$$ need not necessarily be a subset of $$I_n \setminus O_{\lambda_0}$$, but eventually some $$I_j$$ will. (trivial proof by contradiction)
• Thank you for these notes and suggestions. For your important additional comment, how exactly does the fact that $\lim_{n\to \infty}|I_n| = 0$ factor into proving that the $x$ is unique? Does this involve the Nested Interval Theorem? – rodeo_flagellum Mar 31 at 3:31
• Not quite. Let $I_n = [a_n, b_n] \rightarrow I = [a,b]$. This implies that $a_n \rightarrow a$ and $b_n \rightarrow b$, and as it is given that $|I| = 0 \Rightarrow a = b$. Hence $x \in I$ is unique. (Here I have obviously assumed that a sequence of nested intervals will converge to an interval (i.e. a closed, bounded and connected set) - this can be proved trivially by method of contradiction. – Kaind Mar 31 at 3:36