I have looked through some of the other answers to very similar questions, however, it seems that the book I am using defines a continuous function on a topology differently than in the other questions. In my book, a function $f\colon X\to Y$ from the topological space $X$ to $Y$ is continuous iff for every $x$ for every neighborhood $U(f(x))$, the preimage $ f^{-1}(U(f(x)) $ is also a neighborhood of $x$. According to my book, this implies (and even is equivalent to) that of every closed set $A$, the preimage $ f^{-1}(A) $ is also closed. I have absolutely no idea how to do this (tried to first prove that continuity implies that preimage of any open set is also open, but failed), so help would be appreciated.
Topology: Preimage of open set is open if function is continuous
continuitygeneral-topology
Best Answer
So we start with this definition:
Now we prove the following.
Proof. "$\Rightarrow$": Consider any open set $U$ in $Y$. To show that $f^{-1}(U)$ is open, consider any $x\in f^{-1}(U)$. Since $U$ is a neighborhood of $f(x)$ it follows that $f^{-1}(U)$ is a neighborhood of $x$. Hence $f^{-1}(U)$ is a neighborhood of every of its elements and thus open.
"$\Leftarrow$": Consider any $x\in X$ and a neighborhood $U$ of $f(x)$. Pick an open set $f(x)\in V\subseteq U$. Then $f^{-1}(V)$ is open by assumption and satisfies $x\in f^{-1}(V) \subseteq f^{-1}(U)$. Hence, $f^{-1}(U)$ is a neighborhood of $x$. We conclude that $f$ is continuous at $x$. Since $x$ was arbitrary in $X$, the map $f$ is indeed continuous. $\square$
The statement about preimages of closed sets follows by taking complements, noting that $$ f^{-1}(Y\setminus U) = X\setminus f^{-1}(U). $$