1
$\begingroup$

I'm working in topological spaces and I have proved the direction $(\Rightarrow)$, that $f$ is continuous at $x$.

I think it's quite intuitive that the function is continuous but I've been asked to prove this. Essentially I want to prove this statement in the 'only if' direction.

The definition I am using is:

$f : X\to Y$ is continuous at $x$, if for all open set $V$ in $Y$ containing $f(x)$, $f^{-1}(V)$ is open.

$\endgroup$
  • $\begingroup$ What is the definition of $f$ continuous at $x$? $f^{-1}(V)$ open for all $V$ containing $f(x)$? $\endgroup$ – user99914 Nov 10 '15 at 12:51
  • $\begingroup$ isn't this a definition? How else do you explain what $f$ continuous over a set $X$ means? $\endgroup$ – gt6989b Nov 10 '15 at 12:51
  • $\begingroup$ Yes, I've proven that f−1(V) is open for all open sets V containing f(x)? $\endgroup$ – Emily Nov 10 '15 at 12:54
  • $\begingroup$ Yes I originally thought it would it be sufficient to say that clearly the function f is continuous, but there is a point in my notes that asks me to prove that this is true $\endgroup$ – Emily Nov 10 '15 at 12:55
  • $\begingroup$ I am asking for the definition, what is the definition of "$f$ continuous at $x$?" @Emily $\endgroup$ – user99914 Nov 10 '15 at 12:55
3
$\begingroup$

This is a classic definition chase. The setup of the game is that you have two different definitions, "$f$ is continuous provided ..." and "$f$ is continuous at $x$ provided ...".

You need to take the first definition and prove $\forall x$ "$f$ is continuous at $x$."

(In many courses, e.g. analysis courses, it is common to define "$f$ is continuous" as meaning $f$ is continuous at $x$ for all $x$. This is presumably not the case here, so we can't really give you more specific advice without knowing precisely what your definitions are.)

$\endgroup$
  • 1
    $\begingroup$ Everything in math is a "definition chase"... $\endgroup$ – Najib Idrissi Nov 10 '15 at 13:07
  • 1
    $\begingroup$ @NajibIdrissi That very much depends on your definition of "definition chase"! ;) $\endgroup$ – Neal Nov 10 '15 at 13:12
0
$\begingroup$

We prove that $f$ is continuous if $f$ is continuous at $x$ for all $x\in X$. So take an open set $V$ in $Y$ And consider $f^{-1}(V)$. If $f^{-1}(V)$ is empty, then it is open. If $f^{-1}(V)$ is nonempty, let $x\in f^{-1}(V)$. Then $f(x) \in V$. Since $f$ is continuous at $x$, $f^{-1}(V)$ is open.

To sum up, $f^{-1}(V)$ is open whenever $V$ is open. Thus $f$ is continuous.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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