a simple example of a non-continuous function whose graph is homeomorphic to the domain There is a theorem that says that if $f$ is a continuous map, then its graph is homeomorphic to the domain. The converse is false. I'd like to find the simplest counterexample possible. I've found this 
$f:\mathbb{R}\longrightarrow \{0,1\}$, 
where $\mathbb R$ is endowed with the topology generated by the sets $\{1\},\{2\},\{3\},...$ and $f(x)=1$ for $x=0,1,2,...$ and $f(x)=0$ otherwise. 
Is there a simpler example?
edit: In my example, $\{0,1\}$ has discrete topology.

Here's one that seems easy to visualize.
Take $X = [0,1) \times \mathbb{N}$ with its usual topology; i.e. $X$ is a countable disjoint union of half-open "sticks".  Define $f : X \to \mathbb{N}$ by
$$f(x,n) = \begin{cases} 1, & n=1, x < 1/2 \\ 2, & n=1, x \ge 1/2 \\ n+1, & n > 1.\end{cases}$$
Then a homeomorphism from the graph of $f$ to $X$ is given by
$$F((x,n),k) = \begin{cases} (2x,1), & k=1 \\ (2x-1, 2), & k=2 \\ (x,k), & k > 2. \end{cases}$$
(Hopefully I have written it correctly.)
The idea is that the graph looks just the same, except that the first stick has been broken in half.  By stretching the half-pieces we get two full-size sticks.  But there were already infinitely many sticks, so adding one more doesn't produce an "extra" if we re-index.
