# Limit point compactness implies sequential compactness

I am trying to go through the proof of: Suppose $X$ is metrizable space. If $X$ is limit point compact, then $X$ is sequentially compact.

Proof:

Let $(x_n)$ where $n\in \mathbb{N_0}$ be a sequence of points in $X$. We need to find a convergent subsequence.

Case: $A=\{x_n: n\in \mathbb{N_0}\}$ is infinite. By limit point compactness there is a limit point $x$ of $\{x_n, n\in \mathbb{N}\}$. We can find a subsequence that converges to $x$.

I don't understand/ have intuition when they say "By limit point compactness there is a limit point $x$ of $\{x_n, n\in \mathbb{N}\}$". I am well aware of the definition of limit points and limit point compactness. But in this case what if we choose a neighborhood of say $x_1$ which will contain only $x_2$. Now if we intersect that with $A$ how can we get $x$? Any help?

## 3 Answers

There’s no reason to choose a nbhd of $x_1$. When we say that $x$ is a limit point of $A$, we’re saying that every open nbhd of $x$ contains a point of $A$ different from $x$, so $x$ is the only point whose nbhds are of interest. It’s entirely possible that for each $n\in\Bbb N$ the point $x_n$ has an open nbhd that contains no other point of $A$ and does not contain the limit point $x$.

Suppose, for example, that the space is $\Bbb R$, and $A=\left\{\frac1n:n\in\Bbb Z^+\right\}$; the only limit point of $A$ is $0$. If $n\ge 2$, $\left(\frac1{n+1},\frac1{n-1}\right)$ is an open interval around $\frac1n$ whose intersection with $A$ is just $\left\{\frac1n\right\}$, and $\left(\frac12,2\right)$ is an open interval around $1$ whose intersection with $A$ is just $\{1\}$, so each point of $A$ has an open nbhd that contains no other point of $A$. Morover, none of these nbhds contains the limit point $0$. The thing that makes $0$ a limit point of $A$ is that every open nbhd of $0$ contains points of $A$ different from $0$.

Now go back to the general situation. If $A$ is infinite, limit point compactness says that it must have a limit point $x$. This means that for each $\epsilon>0$, $B(x,\epsilon)\cap(A\setminus\{x\})\ne\varnothing$: each $\epsilon$ ball centred at $x$ contains a point of $A$ other than $x$. (Of course it’s possible that $x\notin A$, as in the example above, but we have to cover the possibility that $x$ is in $A$ as well.) In particular, for each $k\in\Bbb Z^+$ there some $x_{n_k}\in B\left(x,\frac1k\right)\cap(A\setminus\{x\})$. Then $\langle x_{n_k}:k\in\Bbb Z^+\rangle$ is a sequence in $A$ converging to $x$. The only problem is that it might not be a subsequence of the original sequence $\langle x_k:k\in\Bbb N\rangle$, because the indices $n_k$ might not be strictly increasing. To complete the proof, you must show that the sequence $\langle n_k:k\in\Bbb Z^+\rangle$ of indices has a strictly increasing subsequence.

• How can we show that the sequence we got is a subsequence of the original sequence? – Cosmic Dec 4 '19 at 14:29

The definition of limit-point compactness simply says that every infinite set has a limit point. We don't know what the limit point of $\{ x_n : n \in \mathbb{N} \}$ is, but whatever it is we can construct a subsequence of $\langle x_n \rangle_n$ converging to it using a couple of simple facts.

1. By Hausdorffness (well, T$_1$-ness) of $X$, if $x$ is a limit point of $A$, then every neighbourhood of $x$ contains infinitely many points of $A$.
2. By first-countability of $X$ every $x \in X$ has a countable (open-)neighbourhood basis $\{ U_i : i \in \mathbb{N} \}$, and for simplicity we may take this to be descending ($U_{i+1} \subseteq U_i$).

Now we work as follows:

If $x$ is a limit point of $A$, choose a countable (descending) (open-)neighbourhood basis $\{ U_i : i \in \mathbb{N} \}$ for $x$. Recursively construct a sequence $\langle n_i \rangle_i$ of natural numbers so that

1. $x_{n_i} \in U_i$; and
2. $n_{i+1} > n_i$.

(Since $U_{i+1}$ is an open neighbourhood of $x$, then $U_{i+1} \cap A$ is infinite, and so there must be an $n_{i+1} > n_i$ with $x_{n_{i+1}} \in U_{i+1}$.)

Then it is (relatively) easy to show that $\langle x_{n_i} \rangle_i$ is a subsequence of $\langle x_n \rangle_n$ which converges to $x$.

Limit point compactness does not imply Sequential Compactness. However, Sequential Compactness does always imply Limit Point compactness. To understand it, you need to go through the definitions of both type of compactness a little more carefully.

While the limit point compactness does not require the sub-sequence of any infinite sequence to be convergent, the Sequential Compactness does require it, and hence Sequential Compactness is more restrictive.

• Any example to help explaining? – Salomo May 3 '15 at 10:05