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 $x_0$.