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?

  • 2
    $\begingroup$ Define "compact." With some definitions, this is trivial; with others, it's harder. $\endgroup$ – user296602 Mar 24 '18 at 19:20
  • 1
    $\begingroup$ 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! $\endgroup$ – user296602 Mar 24 '18 at 19:21
  • $\begingroup$ Open is not the same as not closed. $\endgroup$ – JKEG Mar 24 '18 at 19:26
  • $\begingroup$ The definition being used for compact is that it is closed and bounded $\endgroup$ – sandy Mar 24 '18 at 19:28
  • 1
    $\begingroup$ Are yourking in general metric spaces? Or in subspaces of $\mathbb{R}^n$? Are there statements about compact sets that you can use? $\endgroup$ – 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.


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.