I was trying to prove the equivalence between the epsilon delta definition and open ball definition of continuity, and there have already been quite a few discussions such as here already. I realised that in proving that "epsilon delta $\Rightarrow$ open ball" we try to show that every point in the preimage of an open ball in the domain is an interior point, which requires the global $\epsilon-\delta$ continuity of a function. This seems to suggest to me that open ball continuity is stronger, as in these two examples:
1) $f(x) :=\begin{cases}
0 & x \text{ is rational}\\
x & x\text{ is irrational}
\end{cases}$
Here f(x) is continuous only at $x = 0$. However, the preimage of any open ball containing zero is never open.
2) $ f(x):=\begin{cases}
1 & x<-1 \\
0 & -1\leq x \leq 1 \\
1 & x>1
\end{cases}
$
Here $f(x)$ is continuous at zero, yet there exist an open ball $(-1/2,1/2)$ whose preimage is not open.
If continuity on functions only 'makes sense' for global continuity, why do we then still talk about continuity at a point in a topological space (i.e. a function is continuous at $x$ if every neighbourhood of $x$ pulls back to open sets) ?
Best Answer
As you find in this post, $f(x)$ being continuous in a certain point $x\in X$ is topologically defined in the following way: If $V$ is an open neighborhood of $f(x)$, then the preimage of $V$ contains an open neighborhood $U$ of $x$.
When thinking about $V$ being of size $\varepsilon$ and $U$ being of size $\delta$, this resembles the $\varepsilon$-$\delta$-definition in an intuitive way.