# Topology: Preimage of open set is open if function is continuous

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.

A map $$f\colon X\to Y$$ is continuous at $$x$$ if for every neighborhood $$U$$ of $$f(x)$$ the preimage $$f^{-1}(U)$$ is a neighborhood of $$x$$. The map is continuous if it is continuous at every point in $$X$$.

Now we prove the following.

$$f$$ is continuous iff for every open set $$U\subseteq Y$$ the preimage $$f^{-1}(U)$$ is open in $$X$$.

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 $$f^{-1}(Y\setminus U) = X\setminus f^{-1}(U).$$

Let $$f: X \to Y$$ be a continuous map of topological spaces. Suppose $$C \subseteq Y$$ is a closed set. Now $$X \setminus f^{-1}(C)= f^{-1}(Y \setminus C)$$. But by your definition, $$f^{-1}(Y \setminus C)$$ is open because $$Y \setminus C$$ is an open set, but then $$f^{-1}(C)$$ is the complement of an open set, hence closed.

• @DeapSoup A neighborhood is the same thing as an open. Depending on the definition choice, this may be something which requires proof, but is never more than a line line proof. – mathematics2x2life Feb 25 '20 at 18:28
• @mathematics2x2life The usual definition is that $U$ is a neighborhood of $x$ if there exists an open $V$ with $x\in V\subseteq U$. In particular, $U$ need not be open itself. – Christoph Feb 26 '20 at 7:34