Let $X$ be the cofinite topology and let $f\colon X\to X$ a non-constant function. Show that $f\colon X\to X$ is continuous iff for every inifinite subset $A\subseteq X$, $f(A)$ is infinite.
I am lost with this problem and don't even know if it's true. For the implication from left to right, I supposed that $f(A)$ is finite, so there is a infinite subset of $A$, $B$, such that there are not elements in $X$ that their image lies in $B$. But from here, how does the continuity help us? $A$ is not necessarily an open set, so don't know how to proceed.
Best Answer
Here’s an approach using closed sets.
$(\rightarrow)$
Assume by contradiction that $A$ is infinite but $f(A)$ is finite. Then $f(A)$ is closed; since $f$ is continuous, $f^{-1}(f(A))$ is closed. Therefore, $f^{-1}(f(A))$ is either finite or all of $X$. If it is finite, then $A\subseteq f^{-1}(f(A))$ implies $A$ is finite, a contradiction.
Now suppose $f^{-1}(f(A))=X$. Suppose $V\subseteq f(A)$. Since $V$ is finite, $V$ is closed. But $f^{-1}(f(A)-V) = f^{-1}(f(A))-f^{-1}(V)$, so $f^{-1}(V)$ must either be $\varnothing$ or $X$. Therefore, there exists a singleton $\{y\}$ such that $f^{-1}(y)=X$, contradicting the assumption that $f$ is not constant. As such, $f^{-1}$ cannot be continuous, as desired.
$(\leftarrow)$
Let $V\ne X$ be a closed set of $X$. Then $V$ is finite, so if $f^{-1}(V)$ was infinite, then $f(f^{-1}(V))\subseteq V$ would be infinite by assumption, a contradiction. Clearly $f^{-1}(X)$ is closed, so the proof is complete.
EDIT: Here is some more justification for the $f^{-1}(f(A)) = X$ case.
Recall we assumed the existence of an infinite set $A\subseteq X$, so $X$ is infinite. Because $V$ is closed, $f^{-1}(V)$ is closed. $f(A)-V$ is also closed because $f(A)$ is finite, so
$$f^{-1}(f(A)-V) = f^{-1}(f(A))-f^{-1}(V) = X - f^{-1}(V)$$
is also closed. $f^{-1}(V)$ can either be $\varnothing$, a finite set, or $X$ itself. If $f^{-1}(V)$ is finite and not empty, then $X-f^{-1}(V)\ne X$ is closed, impossible because it is an infinite set.
Hence for all singletons $y\in f(A)$, $f^{-1}(y) = X$ or $f^{-1}(y)=\varnothing$. If all of the fibers $f^{-1}(y)=\varnothing$, then
$$f^{-1}(f(A)) = f^{-1}\left(\bigcup_{y\in f(A)} y\right) =\bigcup_{y\in f(A)} f^{-1}(y)= \varnothing,$$
a contradiction.