Continuity between topological spaces Let $X,Y$ be topological spaces, and $f:X\to Y$. I am trying to show that, if "$A$ closed in $Y\Rightarrow$ $f^{-1}(A)$ closed in $X$", then we have sequential continuity:
$$x_n\to x\;\text{in }X\Rightarrow f(x_n)\to f(x)\;\text{in }Y$$
I am stuck because this is "in the wrong direction": if I talk about things in $Y$ to then get to things in $X$ (from the closed-set definition) then I can't see how to prove the sequential statement.
I'd like to prove this using only the closed-sets statement, without mentioning continuity explicitly. The reason for this is that I want to show several definitions $p,q,r$ are equivalent to the usual open-set definiton of continuity. So proving "continuity" $\Rightarrow p\Rightarrow q\Rightarrow r\Rightarrow$ "continuity" seems the most efficient way to do it.
 A: It is maybe easier to think about proving the contrapositive rather than proving it directly.  Suppose $x_n\to x$ but $f(x_n)\not\to f(x)$ for some sequence $(x_n)$ in $X$.  You want to somehow get from this a closed set $A\subseteq Y$ such that $f^{-1}(A)$ is not closed.  To do this, look at what it means for $f(x_n)$ to not converge to $f(x)$.  It means that there is some open set $U$ containing $f(x)$ such that there are infinitely many $n$ such that $f(x_n)\not\in U$.  You can now take $A=Y\setminus U$ and show that it is exactly the $A$ you are looking for; I'll leave the details to you.
A: Let $U \subseteq Y$ be n open neighborhood of $f(x)$. Then $f^{-1}(U)$ is an open neighborhood of $x$, so there exists $N$ such that for all $k \geq N$, we have that $x_{k} \in f^{-1}(U)$. Thus for all $k \geq N$, we have that $f(x_{k}) \in U$. So for any open neighborhood of $f(x)$, we can choose $k$ large enough that $f(x_{k}) \in U$. Thus $\lim_{k \to \infty} f(x_{k}) = f(x)$.
(This follows because continuity is equivalent to pre-images of open sets being open.)
