Let $\{x_n\}$ be a sequence in a compact metric space X such that:

1) For all $n \in \mathbb{N}$ the set $\{ x_n : n \ge N\}$ is closed.
2) The sequence $\{x_n\}$ is Cauchy.

Prove that there exists $x \in X$ and a subsequence $\{x_{n_k}\}$ such that $x_{n_k} =x$ for all $k \in \mathbb{N}$.

Since we are in a compact metric space, the Cauchy sequence is convergent to some $x \in X$ and as $\{x_n\}$ is closed that would mean that $x= x_M$ for some $m\in \mathbb{N}$. But I'm not sure how to extract a subsequence which is eventually constant.


As you note, we can take $x$ to be the limit of the sequence $x_n$, which we know exists since compact spaces are complete.

In order to prove that the desired subsequence exist, it suffices to show that infinitely many elements of the sequence satisfy $x_n = x$. Suppose (for the purpose of contradiction) that only finitely many elements of the sequence satisfy $x_n = x$. Then there exists an $N$ such that the set $\{x_n : n \geq N\}$ does not contain the element $x$. The sequence $\{x_n\}_{n \geq N}$ converges to $x$, so $x$ is a limit point of the set $\{x_n : n \geq N\}$. The fact that $x \notin \{x_n : n \geq N\}$ contradicts the premise that $\{x_n : n \geq N\}$ is closed.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.