Let $f$ be a function from $\mathbb{R}$ to $\mathbb{R}$, and suppose that $f(x) = 0$, $\forall x \in \mathbb{R}$, except when $x=c$, for a fixed $c \in \mathbb{R}$. Now, $f$ is clearly discontinuous at $c$, and I can easily show this using the $\delta$-$\epsilon$ definition, but how do I show discontinuity using the definition of continuity that the preimage of every open set is open? I just don't see any nontrivial open sets in the image of the function, as $f(\mathbb{R}) = \{0,f(c)\}$.
Any ideas?
EDIT: Note that $f(c) \neq 0$.
Best Answer
You know that $f(c)\neq 0$. So take some open neighborhood $U$ such that $f(c)\in U$ but $0\notin U$. So $f^{-1}(U)=\{c\}$. However, singletons are not open in the usual topology.
To see this, note that $c\in\{c\}$. But $\{c\}$ does not contain any open interval about $c$, hence it cannot possibly be open, since any open set contains an open neighborhood of any of its points.