# Preimage of a closed set is a closed implies f is continuous. Some concerns about the proof

Ok. I have managed thru:

1. If f is continuous then the preimage of open set is a open set
2. If the preimage of the open set is a open set then f is continuous
3. If f is continuous then the preimage of a closed set is closed.

Silly me, I'm struggling with:

1. If the preimage of a closed set is closed then f is continuous.

The proof I'm working with is:

Let $f: X \to Y$ and let $U \subset Y$ a open set therefore we have $f^{-1}(U)$ a open set (by statement 1)

Lets take $Y \setminus U$, this set is closed (Ok here)

Therefore $f^{-1}(Y\setminus U)$ is closed, by hypothesis (Ok here)

Now the part that I'm struggling. (not funny)

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

If I accept this affirmation, ok, the proof finishes because I can tatke the complement . But I can't stand this affirmation can be true from nowhere. Let me explain my reasoning.

I now that I can map $U$ open, to a $f^{-1}(U)$ an open set (already proved). But if I take the complement of $U$ I will take everything else from he image, how can I garantee that it will map to everything else from the domain (besides the $f^{-1}{U}$), I didn't said anything if the mapping is onto or bijective. Therefore I think that $f^{-1}(Y \setminus U) \subset X$ but I can't garantee $f^{-1}(Y \setminus U) \cup f^{-1}(U) = X$ and $f^{-1}(Y \setminus U) \cap f^{-1}(U) = \emptyset$.

Can someone enlighten me?

• Just look at the definition of the preimage: $f^{-1}(Y\setminus U)$ is the set of points $x$ in $X$ such that $f(x)\not\in U$, while $f^{-1}(U)$ is the set of point $x$ such that $f(x)$ is in $U$. Can you see why these sets are disjoint and why their union is all of $X$? Jun 13, 2015 at 12:40

By definition $$f^{-1}(A):=\{x\in X\mid f(x)\in A\}$$In general: $$x\in f^{-1}(A)\iff f(x)\in A$$
Statement $x\in f^{-1}(Y\setminus U)\vee x\in f^{-1}(U)$ is equivalent with statement $f(x)\in Y\setminus U\vee f(x)\in U$, wich is evidently true for every $x\in X$. So: $$f^{-1}(Y\setminus U)\cup f^{-1}(U)=X$$
Statement $x\in f^{-1}(Y\setminus U)\wedge x\in f^{-1}(U)$ is equivalent with statement $f(x)\in Y\setminus U\wedge f(x)\in U$, which is evidently not true for any $x\in X$. So: $$f^{-1}(Y\setminus U)\cap f^{-1}(U)=\varnothing$$