Confused about proving things using compactness

Question

Suppose $\{P_n\}$ is a sequence of points in $S$ and $S$ is a compact subset a subset of $\mathbb R^2$ .

Prove that some sequence $\{P_{n_{i}}\}$ converges to a point in S.

So far, What I have come up with is as follows:

1. Since $\{P_n\}$ is contained in a compact set $S$, then any open cover of of $S$ has a finite subcover and that finite sub-cover will also cover $\{P_n\}$

2. Let us choose the cover $\{B(0,N)\}_{N\in \mathbb N}$ , this covers $S$ and has a finite sub-cover . Then we have $\{P_n\}\subset S \subset B(0,M) $ ( for some $M>0$) and for the subcover is finite,the sequence it is bounded by some $M$.

3. Since $\{P_n\}$ is bounded by some $M$ , then by the Bolzano- Weierstrass theorem, a subsequence of $\{P_n\}$ converges to some point in $S$. Since every bounded sequence has a convergent subsequence.

Could you point out my mistakes here or give me an idea on how to prove this? I am just confused as where should I start and how I can start proving things..

P.s. pardon my english as my first language is Dutch.

Answer 1

You did a little bit of mistake in showing the sequence is bounded.(rather I would say your idea was right) Exactly this was the proof which I did during my undergraduate. observe that since $S$ is compact it is bounded. Hence ${p_n:n\in \mathbb N}$ is a bounded sequence. After this Point number 3 is on point !

But I would like to suggest you try to prove it by some other methods as well.

Try to prove the following fact: Any infinite subset of $S\subset \mathbb R^2$ has a limit point in $S$

Answer 2

Here is a proof directly from the open cover definition of compactness, one that works in any metric space.

Suppose that $\sigma=\langle x_n:n\in\Bbb Z^+\rangle$ is a sequence in $S$, and let $A=\{x_n:n\in\Bbb Z^+\}$. Suppose that $p$ is a limit point of the set $A$; I’ll construct a subsequence of $\sigma$ that converges to $p$.

Let $n_1=\min\{k\in\Bbb Z^+:x_k\in B(p,1)\}$. Now suppose that for some $m\in\Bbb Z^+$ we’ve already chosen $n_1,\ldots,n_m\in\Bbb Z^+$ such that $n_1<n_2<\ldots<n_m$ and $x_{n_k}\in B\left(p,\frac1k\right)$ for $k=1,\ldots,m$. $A\cap B\left(p,\frac1{m+1}\right)$ is infinite, so let

$$n_{m+1}=\min\left\{k\in\Bbb Z^+:k>n_m\text{ and }x_k\in B\left(p,\frac1{m+1}\right)\right\}\;;$$

then $n_1<\ldots<n_{m+1}$, and $x_{n_k}\in B\left(p,\frac1k\right)$ for $k=1,\ldots,m+1$, and the recursive construction goes through to produce a subsequence $\langle n_k:k\in\Bbb Z^+\rangle$ of $\sigma$ that converges to $p$.

Thus, we’re done unless $A$ has no limit point. In that case for each $x\in S$ there is an $\epsilon_x>0$ such that $B(x,\epsilon_x)\cap A\subseteq\{x\}$. Clearly $\{B(x,\epsilon_x):x\in S\}$ is an open cover of $S$, which is compact, so it has a finite subcover, $\{B(x_1,\epsilon_{x_1}),\ldots,B(x_n,\epsilon_{x_n})\}$, say. $A\subseteq S$, so

$$A=\bigcup_{k=1}^n\big(A\cap B(x_k,\epsilon_{x_k})\big)\;,$$

and therefore $|A|\le n$, since each of the sets $B(x_k,\epsilon_{x_k})$ contains at most one point of $A$. In particular, $A$ is finite, so there must be some $p\in A$ such that $\{n\in\Bbb Z^+:x_n=p\}$ is infinite, and in that case $\sigma$ clearly has a constant subsequence that converges to $p$.

Thus, in every case $\sigma$ has a convergent subsequence.