Topological Embedding Which is Neither Open nor Closed I'm having trouble coming up with an example of an embedding which is neither open nor closed.
My attempts have included trying to find such a map from $\mathbb{R}$ (given the usual Euclidean topology, of course) to some subset of $\mathbb{R}$, which I now believe impossible, and trying to find one from some topology on $\{1, 2, 3\}$ to some other topology on $\{1, 2, 3, 4\}$. Both of these attempts seem to have failed me. So what do I do?
 A: Stefan has given a machine for generating counterexamples in the comments. But since this question seems to have gone unanswered for so long, I figured I'd point out to provide a standard, not-so-obvious, but fairly useful instance of Stefan's general result.
A somewhat stronger condition than what you've asked for is to construct an immersed submanifold which is neither open nor closed. The immersion map is not an embedding in general, but the inclusion map on its image will be.
The example is given by $f:\Bbb R\to \Bbb R^2/\Bbb Z^2$ (*) defined as
$$f(x)=(x,\alpha x) \mod 1$$
for some irrational $\alpha$. It's fairly straightforward to see that this map is continuous, but the topology induced by $f$ on its image is different than the topology induced from the ambient $\Bbb R^2$.
[To be clear, $f$ is the immersion of manifolds— not a topological embedding. The topological embedding is the inclusion map im$(f)\to\Bbb R^2$.]
(*) If this is not familiar, it is just a torus: take the unit square $[0,1]^2$ and identify the top with the bottom, and the right with the left.
