# Continuous Functions - Topology

I'd like to prove the following.

A function $f:X \to Y$ is continuous if whenever $A$ is closed in $Y$, $f^{-1}(A)$ is closed in $X$.

Proof. By definition, a function is continuous if the inverse image of every open set is open. Suppose that $A\in Y$ is closed. Then, $Y-A$ is open, so $f^{-1}(Y-A)$ is open.

$f^{-1}(Y-A) = X - f^{-1}(A)$ is open. So $f^{-1}(A)$ is closed.

Is this correct?

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Looks like your proof is in the wrong direction. You have proved that if $f$ is continuous, then whenever $A$ is closed $f^{-1}(A)$ is also closed. – Dan Shved Jun 29 '13 at 8:39
The last phrase in your proof is "thus $f^{-1}(A)$ is closed". Looks like "closeness" is on the right hand side of the arrow. – Dan Shved Jun 29 '13 at 8:50
@saadtaame, what you wrote " Then, $Y-A$ is open, so $f^{-1}(Y-A)$ is open" wrongly assumes that $\,f\,$ is already continuous, which is what you want to prove. – DonAntonio Jun 29 '13 at 8:53
But you want to prove that $\,f\,$ is continuous if the inverse image of every closed set is closed, @saadtaame! It is continuity of $\,f\,$ that you want to prove, so you can not assume it. – DonAntonio Jun 29 '13 at 9:26
Because you don't know that $f$ is continuous. – Cortizol Jun 29 '13 at 9:30

Hints:

You cannot assume what you want to prove: suppose that whenever $\,A\subset Y\,$ is closed then also $\,f^{-1}(A)\subset X\,$ is closed.

Let $\,U\subset Y\,$ be open $\;\implies Y\setminus U\;$ is closed, so by asumption $\,f^{-1}\left(Y\setminus U\right)\;$ is closed in $\;X\;$ and thus $\;X\setminus f^{-1}(Y\setminus U)\;$ is open.

But $\,X\setminus f^{-1}(Y\setminus U)\subset f^{-1}(U)\;$ since:

$$z\in X\setminus f^{-1}(Y\setminus U)\implies z\notin f^{-1}(Y\setminus U)\implies f(z)\notin Y\setminus U\implies$$

$$f(z)\in U$$

Deduce now that in fact $\,f^{-1}(U)\;$ is open and thus $\,f\,$ fulfills the usual definition of continuity, i.e. $\,f\,$ is continuous.

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I know that a set is closed is it's complement is open. It works in the other direction as well? – saadtaame Jun 29 '13 at 9:44
Yes, it is an iff thing: a set is open iff its complement is closed. – DonAntonio Jun 29 '13 at 9:49
Okay, I'll use your proof (excluding the inclusion part). So we know that $X-f^{-1}(Y-U)$ is open. But $$X-f^{-1}(Y-U)=X-(X-f^{-1}(U))=f^{-1}(U)$$. Thus $f^{-1}(U)$ is open proving that $f$ is continuous by the usual definition. Right? – saadtaame Jun 29 '13 at 9:53
Hmmm...can you prove your first equality? If so then yes: you're done. – DonAntonio Jun 29 '13 at 9:55