If $\pmb A$ is not compact, then every convergent subsequence and $(x_j)_{j=1}^\infty$ itself do not have to converge to the same limit?

Question

Is it true that if a set $\pmb A$ is not compact, then every convergent subsequence of a sequence of points $(x_j)_{j=1}^\infty$ and $(x_j)_{j=1}^\infty$ itself do not have to converge to the same limit?

Can anyone show me some examples?

Answer 1

Let $A$ be a subset of a metric space. Then $A$ is not compact, iff and only if there is a sequence $(x_n)$ in $A$ such that each subsequence of $(x_n)$ does not have a limit in $A$.

Answer 2

If the series $x_n$ is convergent, then every subsequence converges to the same limit. This statement holds regardless whether $A$ is compact or not. However if the series $x_n$ is not convergent, it is true that some subsequences can converge to different limits in a non-compact space $A$.

Take $A$ = $\Bbb R$ and $x_n = (-1)^n$

$x_{2n}$ converges to $1$ and $x_{2n+1}$ converges to $-1$.

Hope it helps.