Another analysis homework problem here:

Show that $A \subset M$ is sequentially compact $\iff$ every infinite subset of A has an accumulation point in $A$.

I have two questions:

1) For the $\Rightarrow$ implication, would it be incorrect to say that since $B$ is infinite, there exists a sequence $x_n$ in $B$. And since $B \subset A$, $x_n$ is a sequence in $A$ which means there exists a subsequence $x_{n_k}$ that converges in $A$ (call the limit point $x_0$). If this is true then I think I can show $x_0$ is an accumulation point of $B$.

2) For the $\Leftarrow$ implication, I want to say that for some sequence $x_n$ in $A$ since $x_n$ has infinitely many elements of $A$ it is technically an infinite subset of $A$, which means there exists an accumulation point $x_0$ in $A$. If this is true, I think I can show there is a subsequence $x_{n_k}$ that converges to $x_0$.


1 Answer 1


For $(\Rightarrow)$ you’re on the right track. The easiest way to handle $(\Leftarrow)$ is to prove the contrapositive: assume that $\langle x_n:n\in\Bbb N\rangle$ is a sequence in $A$ with no convergent subsequence, and deduce that $\{x_n:n\in\Bbb N\}$ is an infinite set with no accumulation point. (Don’t forget to prove that $\{x_n:n\in\Bbb N\}$ is infinite.)

  • $\begingroup$ And to show that $x_n$ is infinite could I simply make the remark, "since $\mathbb{N}$ is infinite, the sequence $x_n$ is also infinite"? $\endgroup$ Oct 16, 2014 at 1:19
  • $\begingroup$ @Kevin: No: if $\langle x_n:n\in\Bbb N\rangle$ is a constant sequence, for instance, $\{x_n:n\in\Bbb N\}$ has only one point. You need to use the hypothesis that the sequence has no convergent subsequence. $\endgroup$ Oct 16, 2014 at 1:21
  • $\begingroup$ @Kevin: You’re welcome! $\endgroup$ Oct 16, 2014 at 1:25
  • $\begingroup$ @KevinSheng wait a minute, $(\Leftarrow)$ is certainly not true without some extra assumptions of the space, at least if considering this definition of accumulation point. $\endgroup$
    – Anguepa
    Feb 7, 2018 at 15:42

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