No subsequence converges when sequence has limit point in topological space In general topological space, suppose $x_0$ is a limit point of a sequence $\{x_n\}$, is it possible that there is no subsequence that converges to it？
 A: Yes. Let $X=\Bbb N$, with the topology generated by the base
$$\big\{\{n\}:n\ge 2\big\}\cup\big\{\{0,1\}\big\}\;,$$
and let $x_n=n$ for $n\in\Bbb N$. Then $\{x_n:n\in\Bbb N\}=X$, and every nbhd of $0$ contains $1$, a point of $X$ different from $0$, so $0$ is a limit point of the set $X$, but the sequence clearly has no subsequence converging to $0$ (or to anything else!).

Added: This can easily be improved to a $T_0$ example. Again let $X=\Bbb N$, but for each $n\in\Bbb N$ let $V_n=\{k\in\Bbb N:k<n\}$, and let the topology be
$$\{\Bbb N\}\cup\{V_n:n\in\Bbb N\}\;.$$
The resulting space is easily seen to be $T_0$. Again consider the sequence $\sigma=\langle n:n\in\Bbb N\rangle$. Every point of $X\setminus\{0\}$ is a limit point of the set $X$, since every nbhd of every point contains $0$, but no $V_n$ contains a tail of any subsequence of $\sigma$.

For a less trivial example, let $\mathscr{U}$ be a free ultrafilter on $\Bbb N$, and let $X=\{p\}\cup\Bbb N$. Points of $\Bbb N$ are isolated, and $U\subseteq X$ is an open nbhd of $p$ if and only if $p\in U$ and $U\cap\Bbb N\in\mathscr{U}$. Then $p$ is a limit point of the set $\Bbb N$, but no subsequence of $\langle n:n\in\Bbb N\rangle$ converges to $p$. To prove this, let $A\subseteq\Bbb N$, and consider the subsequence $\langle n:n\in A\rangle$. If $A\notin\mathscr{U}$, then $X\setminus A$ is a nbhd of $p$ disjoint from $A$. If $A\in\mathscr{U}$, split $A$ into complementary infinite subsets $A_0$ and $A_1$. Exactly one of them is in $\mathscr{U}$; without loss of generality let it be $A_0$. Then $\{p\}\cup A_1$ is a nbhd of $p$ that excludes infinitely many terms of $\langle n:n\in A\rangle$, so $\langle n:n\in A\rangle$ does not converge to $p$.
