Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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
up vote 4 down vote accepted


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.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.