Nested Sequence Theorem

metric-spacesreal-analysis

I am trying to solve the following problem in Kolmogorov's real analysis textbook.

By the diameter of a subset $A$ of a metric space $R$ is meant the number
\begin{equation*}
d(A) = \sup\limits_{x, y \in A} \rho(x,y).
\end{equation*}
Suppose $R$ is complete, and let $\{A_n\}$ be a sequence of closed subsets of $R$ nested in the sense that
$$
A_1 \supset A_2 \supset \cdots A_n \supset \cdots.
$$
Suppose further that
$$
\lim\limits_{n \to \infty} d(A_n) = 0.
$$
Prove that the intersection $\bigcap\limits_{n =1}^{\infty} A_n$ is nonempty.

Here is my best attempt at a proof, trying to incorporate the assumption that $R$ is complete.

Let $\{A_n\}$ be a sequence of closed, nested subsets of $R$ such that $A_1 \supset A_2 \supset A_3 \cdots$ and $\lim\limits_{n \to \infty} d(A_n) = 0$.

For each $n \in \mathbb{N}$, pick $x_n \in A_n$, constructing the sequence $\{x_n\}$. Since $A_n \subset R$ for all $n$, $x_n \subset R$ for any $n$.

Observe that $A_m \subset A_j$ implies that $\sup A_m \leq \sup A_j$. Similarly, if $A_m \subset A_j$, we have $\sup_{x, y \in A_m} \rho(x,y) \leq \sup_{x,y \in A_j} \rho(x,y)$, so $d(A_m) \leq d(A_j)$. Furthermore, since $d(A_n) \to 0$, we have that $\forall \epsilon > 0$, $\exists N$ such that for all $n > N$, $|d(A_n)| < \epsilon$.

We will prove that $\{x_n\}$ is Cauchy. Let $\epsilon > 0$. By drawing on convergence of $d(A_n)$ to $0$, choose $N$ such that for all $n > N$, $|d(A_n)| < \epsilon$. For any $m, n > N$, in which case $x \in A_m$ and $x \in A_n$, have $A_m, A_n \subset A_N$. Without loss of generality, we can take $m > n$. (If $m = n$, then $|x_m – x_n| = 0 < \epsilon$ for all $\epsilon > 0$, and the result is proved.) This implies that $A_m \subset A_n$, so $x_m \in A_m$ implies that $x_m \in A_n$. We have:
\begin{align*}
\rho(x_m, x_n) & \leq \sup_{x, y \in A_n} \rho(x, y) & & \text{Definition of supremum} \\
& \leq \sup_{x,y \in A_N} \rho(x,y) & & \text{Since $A_n \subset A_N$} \\
& = d(A_N) & & \text{Definition} \\
& \leq |d(A_N)| & & \text{Definition of absolute value} \\
& < \epsilon & & \text{Choice of $N$ since $d(A_N) \to 0$}
\end{align*}
Thus, $\{x_n\}$ is a Cauchy sequence in $R$. Since $R$ is a complete space, $\{x_n\} \to x \in R$.

Since this holds for an arbitrary $n$, it holds for all $n$. Thus, any such sequence $\{x_n\}$ is Cauchy. But $\{x_n\} \in A_n$, and $A_n$ is closed, so $x \in A_n$, as closed sets contain all of their limit points. Thus, $x \in \bigcap\limits_{n=1}^{\infty} A_n$.

I have some additional questions on this proof:

  1. I am unsure of whether I should be using the "absolute value" metric in my proof that $x_n$ is Cauchy, particularly because I do not use this when writing the proof that $\{x_n\}$ is Cauchy. Have I sacrificed generality in doing this?

  2. The last paragraph of the proof is not completely clear to me or, at the very least, I am unsure completely on how to formalize it. Some help with this would be very much appreciated.

Best Answer

Your proof in general is absolutely fine, I have juts a couple of comments for the streamlining of the proof.

  • First of all, the diameter of a set is always non-negative, so the absolute value signs are not necessary. (Positive side effect, no generality was sacrificed, just efficiency. :-))
  • Second, the sentence "Observe that $A_m \subset A_j$ implies that $\sup A_m \leq \sup A_j$." is nonsensical as $\sup A_{m}$ means nothing in a general metric space. It is enough to simply obeserve that if $A_{m}\subset A_{n}$, then $$d(A_{m})=\sup_{x,y\in A_{m}}\rho(x,y)\leq\sup_{x,y\in A_{n}}\rho(x,y)=d(A_{n}).$$
  • Third, you only need that $x_{m},x_{n}\in A_{N}$ to show that $$\rho(x_{m},x_{n})\leq d(A_{N}).$$ Assuming wlog that $A_{m}\subset A_{n}$ is not required. (On that note, I assume you meant $\rho(x_{m},x_{n})=0$ instead of $|x_{m}-x_{n}|=0$.
  • Finally for the last step of your proof I would suggest the following: So we find for all $N\in\mathbb{N}$ that $\{x_{n}\}_{n\geq N}$ is a sequence in $A_{N}$ converging to $x$. Since $A_{N}$ is closed it follows that $x\in A_{N}$ for all $N\in\mathbb{N}$. Thus $x\in\bigcap_{n=1}^{\infty}A_{n}\neq\emptyset$.
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