# $f : X\to Y$ is continuous if and only if $f$ is continuous at $x$ for all $x\in X$

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.

• What is the definition of $f$ continuous at $x$? $f^{-1}(V)$ open for all $V$ containing $f(x)$? – user99914 Nov 10 '15 at 12:51
• isn't this a definition? How else do you explain what $f$ continuous over a set $X$ means? – gt6989b Nov 10 '15 at 12:51
• Yes, I've proven that f−1(V) is open for all open sets V containing f(x)? – Emily Nov 10 '15 at 12:54
• 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 – Emily Nov 10 '15 at 12:55
• I am asking for the definition, what is the definition of "$f$ continuous at $x$?" @Emily – user99914 Nov 10 '15 at 12:55

## 2 Answers

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.)

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

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.