Image of an accumulation point under a continuous function Let $X$ be a compact Hausdorff topological space and let $f:X \to F$ ($F$ can be assumed metric space) be a continuous map. Let $\left(x_n\right)$ be a sequence in $X$. Now, there are two possibilities:

*

*$\left(x_n\right)$ is eventually constant, i.e., $x_n=x_0$ for all but finitely many $n$.


*We can think of $\left(x_n\right)$ as an infinite set in $X$, which has an accumulation point in $X$, say $x_0$.
Now, what can we say about $f(x_0)$?
Can we say that the sequence $(f(x_n))$ converges to $f(x_0)$?
A detailed answer will be of great help. Thanks in advance.
 A: I will assume that $X$ is $T_1$.
In the first case it is clear that $f(x_n)=f(x_0)$ for all but finitely many $n$ and hence that $\langle f(x_n):n\in\Bbb Z^+\rangle$ converges to $f(x_0)$.
In the second case it need not be true that $\langle x_n:n\in\Bbb Z^+\rangle$ even has a subsequence that converges to $x_0$. However, if $F$ is first countable (and hence in particular if it is metrizable), there is a local open base $\{B_n:n\in\Bbb Z^+\}$ at $f(x_0)$ such that $B_{n+1}\subseteq B_n$ for each $n\in\Bbb Z^+$. For each $n\in\Bbb Z^+$ let $U_n=f^{-1}[B_n]$; each $U_n$ is an open nbhd of $x_0$, so $U_n\cap\{x_k:k\in\Bbb Z^+\}\ne\varnothing$.
In fact, since $X$ is $T_1$, $U_n\cap\{x_k:k\in\Bbb Z^+\}$ is infinite for eack $n\in\Bbb Z^+$. This means that we can recursively construct a subsequence $\langle x_{n_k}:k\in\Bbb Z^+\rangle$ of $\langle x_n:n\in\Bbb Z^+\rangle$ such that $x_{n_k}\in U_k$ for each $k\in\Bbb Z^+$ and hence $f(x_{n_k})\in B_k$ for each $k\in\Bbb Z^+$. This construction ensures that $\langle f(x_{n_k}):k\in\Bbb Z^+\rangle$ converges to $f(x_0)$ in $F$. However, there is no guarantee that the sequence $\langle x_n:n\in\Bbb Z^+\rangle$ converges to $f(x_0)$.
A: A bijective continuous map from a compact space to a Hausdorff space is a homeomorphism.
Thus every open set $U\subset X$, $f(U)$ is open in F. So, if every open set containing $x_0$ interests X (limit point), the image of those open sets (containing $f(x_0)$ )will also intersect $f(X)$.
