# Generalizing Cantor's Nested Intervals Theorem

I am trying to prove the following: If $$\{A_n\}_{n=1}^{\infty}$$ is a nested sequence of sets in $$\mathbb{R}$$ each of which is closed, bounded, and non-empty, then $$\cap_{n=1}^{\infty} A_n$$ is non-empty.

I know the nested interval theorem, so I tried to make progress as follows. Let $$a_n = \inf A_n$$ and $$b_n = \sup A_n$$, which are well-defined since each $$A_n$$ is non-empty and bounded. The intervals $$[a_n, b_n]$$ are a nested sequence of non-empty, closed intervals and as such $$\cap_{n=1} [a_n, b_n]$$ is not empty. Also, $$\cap_{n=1}^{\infty} A_n \subset \cap_{n=1} [a_n, b_n]$$.

However, it is unclear if I can make further progress in this direction. Simple counterexamples show that $$\cap_{n=1}^{\infty} A_n \neq \cap_{n=1} [a_n, b_n]$$, e.g., $$A_n = [0, 1/n] \cup [1, (n+1)/n]$$ whence $$\cap_{n=1}^{\infty} A_n = \{0, 1\}$$, while $$\cap_{n=1} [a_n, b_n] = [0, 1]$$.

I have noticed some other things while trying to prove this statement. For example, if $$A_n$$ is a finite set for any $$n$$, say $$A_n = \{a_1, ..., a_r\}$$, then there must be some $$i = 1,...,r$$ such that $$a_i \in A_m$$ for all $$m \geq n$$ (because the sequence is nested and each set in the sequence is non-empty). Thus $$\cap_{n=1}^{\infty} A_n$$ contains $$a_i$$ and is not empty as desired.

Also, if $$a$$ is a limit point of $$A_n$$, then for $$m \geq n$$ either $$a$$ is in $$A_m$$ or $$a$$ is not a limit point of $$A_m$$ (since $$A_m$$ is closed).

You can make progress. You don't need to consider endpoints. What I would do is choose $$x_n\in A_n$$ for all $$n$$. Since the $$A_1$$ is bounded and $$x_n\in A_n\subseteq A_1$$ for all $$n$$ it follows that $$\{x_n\}$$ is bounded. It now follows from the Bolzano-Weierstrass property that $$\{x_n\}$$ has at least one convergent subsequence, let $$x$$ be a limit of one such subsequence. Since $$x$$ is a limit of a subsequence of$$\{x_n:n\in\mathbb{N}\}$$ it follows that it is a limit of a subsequence of $$\{x_m:m\in\mathbb{N} \ \text{and } m\geq n\}\subseteq A_n$$ for all $$n$$. Well then as $$A_n$$ is closed and $$x$$ is a limit of a sequence in $$A_n$$ we have that $$x\in A_n$$. Since $$n$$ was arbitrary $$x\in \bigcap_{n\in\mathbb{N}} A_n$$. Thus the set is not-empty.
• thanks for your answer! One thing that I noticed is that $x$ need not be a limit point of $\cap_{n \in \mathbb{N}} A_n$, e.g., if $A_n = [0, 1/n]$, when $\cap_{n \in \mathbb{N}} A_n$ is a singleton set and hence has no limit points. Sep 9 '21 at 20:10
• @neophyte_autodidact It appears I was using a slightly different version of the definition of a limit point. I was considering a point a limit point if it was a limit of any sequence in the set. I changed the proof to accomodate your definition. You'll find the argument the same, the language just changed slightly. With the new language you'll see that the argument holds for finite or infinite sets $A_n$. Sep 9 '21 at 23:37