The question I am trying to answer reads as follows.
Let $(X,d)$ be a metric space and $A \subseteq X$ a countably compact subset. We suppose $(x_{n})_{n=1}^{\infty}$ is a sequence in $A$ and that the corresponding set $\{x_{n}\}_{n=1}^{\infty}$ has an accumulation point $a \in A$. Prove there is some subsequence of $(x_{n})_{n=1}^{\infty}$ which converges to $a$.
This seemed initially straightforward to me. We can construct the sequence $(y_{n})_{n}$ where $y_{k} \in (B(a, \frac{1}{k}) \setminus \{a\}) \cap (x_{n})_{n=1}^{\infty}$, after which showing the convergence of $(y_{n})_{n}$ seems to follow naturally.
I'm pretty sure I must have the wrong idea or completely be missing something here however, since as far as I can tell I didn't have to rely upon the fact that $A$ is countably compact. I know that countably compact means that every infinite subset $Q$ of your considered set $R$ has an accumulation point within $R$, but I'm struggling to see both:
- What is wrong with my proof?
- How I am going to use countable compactness of $A$ to construct a correct proof.
Any hints or pointers would be appreciated.
You’re right that countable compactness isn’t needed here, but you have to work a little harder than that in order to make $\langle y_k:k\in\Bbb Z^+\rangle$ a subsequence of $\langle x_n:n\in\Bbb Z^+\rangle$. What you get from your argument is a sequence $\langle n_k:k\in\Bbb Z^+\rangle$ of positive integers such that $y_k=x_{n_k}$ for each $k\in\Bbb Z^+$, but you don’t know that the sequence $\langle n_k:k\in\Bbb Z^+\rangle$ is strictly increasing. To complete the argument you must show that it has a strictly increasing subsequence; that will give you a subsequence of $\langle x_n:n\in\Bbb Z^+\rangle$ converging to $a$. It’s not hard to do: the key step is to show that $\{n_k:k\in\Bbb Z^+\}$ must be infinite.
The reason that countable compactness isn’t needed is that we assumed that $\{x_n:n\in\Bbb Z^+\}$ has an accumulation point $a$. If $A$ is countably compact that assumption is unnecessary: either $\{x_n:n\in\Bbb Z^+\}$ is finite, in which case the sequence has a constant subsequence, or $\{x_n:n\in\Bbb Z^+\}$ is infinite and therefore has an accumulation point.