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.


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 $x \in I_n$.
  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.

Simplification in your proof.

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)

  • $\begingroup$ 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? $\endgroup$ – rodeo_flagellum Mar 31 at 3:31
  • 1
    $\begingroup$ 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. $\endgroup$ – Kaind Mar 31 at 3:36
  • $\begingroup$ @rodeo_flagellum Check the edit I made. $\endgroup$ – Kaind Mar 31 at 3:40
  • $\begingroup$ I have noted the edit. Again, thank you. I am going to wait to provide the check mark to prompt others to answer. Wish you the best and I appreciate help. $\endgroup$ – rodeo_flagellum Mar 31 at 3:45

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