# Prove $C$ is compact $\iff$ every sequence in $C$ has a subsequence that converges to a point in $C$

I am able to prove the direction from Compact to subsequence but I can't seem to get all the other direction quite right.

I can prove that $C$ must be bounded, but then my professor says that I am not correct about why it must be closed. I said if $C$ is open then there must exist a convergent sequence in $C$ such that the limit is not in $C$ which is a contradiction to the assumption which would imply that $C$ is closed.

Can someone explain what I am doing wrong?

• Define "compact." With some definitions, this is trivial; with others, it's harder. – user296602 Mar 24 '18 at 19:20
• Also, the argument that if $C$ is open then there is a convergent sequence whose limit is not in $C$ isn't correct: $C = \mathbb{C}$ is a counterexample! – user296602 Mar 24 '18 at 19:21
• Open is not the same as not closed. – JKEG Mar 24 '18 at 19:26
• The definition being used for compact is that it is closed and bounded – sandy Mar 24 '18 at 19:28
• Are yourking in general metric spaces? Or in subspaces of $\mathbb{R}^n$? Are there statements about compact sets that you can use? – José Carlos Santos Mar 24 '18 at 19:29

It seems that $C \subseteq \mathbb{R}^n$ and you wish to prove that $C$ is bounded and closed, assuming that every sequence in $C$ has a convergent subsequence with the limit in $C$.

If $C$ is not bounded, then there exists a sequence $(x_n)_n$ in $C$ such that $\|x_n\| \ge n, \forall n \in \mathbb{N}$. However, every subsequence of $(x_n)_n$ is unbounded, and hence cannot converge to an element of $C$.

To show that $C$ is closed, assume $(x_n)_n$ is a sequence such that $x_n \xrightarrow{n\to\infty} x$, where $x \in \mathbb{R}^n$. We wish to prove $x\in C$.

There exists a subsequence $(x_{p(n)})_n$ such that $x_{p(n)} \xrightarrow{n\to\infty} y \in C$. Since every subsequence of a convergent sequence also converges to the same limit, we conclude $x = y \in C$.

Hence, $C$ is closed.

If $C$ is not compact, then it is not both closed and bounded.

If $C$ is not bounded, then, for each natural $n$, take $x_n\in C$ such that $|x_n|>n$. Then the sequence of al $x_n$'s has no convergent subsequence.

If $C$ is not closed, let $x\in\overline C\setminus C$. Take a sequence $(x_n)_{n\in\mathbb N}$ of elements of $C$ which converges to $x$. Then the sequence $(x_n)_{n\in\mathbb N}$ has no convergent subsequence.