Continuous map from topological space to real line and $\Omega_f(x)=0$

Question

I'm trying to prove that $Ω_f (x) = 0$ iff f is continuous at x.

Given the follow definitions:

Let $(X, T )$ be a topological space. For $x ∈ X$ let $N (x)$ denote the collection of all neighborhoods of $x$. For any function $f : X → R$ we define the function $Ω_f : X → R$ by $Ω_f (x) := \inf_{U∈N(x)}diam(f(U))$, where we have used the notation $diam(A) := sup_{x,y∈A}|x − y|$, for subsets $A ⊂ R$.

I was thinking that by the continuity of the metric in real space, you can make a neighborhood as small as you'd like. But does that imply by the definition of continuity in topological spaces that it's preimage in $X$ can also be made arbitrarily small?

I.e. can I say that for $V = (y-e, y+e)$, as $e \to 0 $,
$diamf^{-1}(V) \to 0$

Similarly, can I say the same going in the other direction? Or is there more to the proof than that?

Answer 1

This is indeed true.

Supposing $f$ is continuous at $x$, fix $\epsilon>0$; by the definition of continuity, the $f$-preimage of $(f(x)-\epsilon, f(x)+\epsilon)$ is an open neighborhood of $x$, so $\Omega_f(x)\le\epsilon$. Thus, $\Omega_f(x)=0$.

Conversely, suppose $\Omega_f(x)=0$ and $\epsilon>0$. Since $\Omega_f(x)=0<\epsilon$, we can find some open neighborhood $U$ of $x$ such that $Diam(f(U))<\epsilon$. But then $f(U)\subseteq (f(x)-\epsilon, f(x)+\epsilon)$. So for every $\epsilon>0$ there is a neighborhood $U$ of $x$ such that $f(U)\subseteq (f(x)-\epsilon, f(x)+\epsilon)$, which means $f$ is continuous at $x$.