Continuous bijection from $\mathbb R-\{0\}$ to $\mathbb R$ I know such a function exists if continuity is not required.
How do you prove it doesn't exist when continuity is required? 
For the sake of completeness,  note that a continuous bijection from $\mathbb R$ to $\mathbb R-{0}$ cannot exist either (the IVT applies trivially). 
 A: Here is a slight variation on Dan's excellent answer.
Note that a continuous bijection between two connected sets whose topology is an order topology are either order preserving or order reversing. But it has to be either an order isomorphism or an anti-isomorphism (order reversing).
Assume by contradiction that there is such $f$. As Dan points out, the image of $\Bbb R_{>0}$ and $\Bbb R_{<0}$ are both intervals, which are disjoint and whose union is $\Bbb R$. Therefore one has the form $[x,\infty)$ or $(-\infty,x]$.
In either case this means that the restriction of $f$ to one of the two connected parts is not an order isomorphism (or anti-isomorphism), since neither parts has endpoints but one of the parts is mapped to an interval with an endpoint.
A: Assume $f: \mathbb{R}\setminus \{0\} \to \mathbb{R}$ is a continuous bijection. Denote $\mathbb{R}_{>0} = \{x \in \mathbb{R} \mid x > 0\}$ and $\mathbb{R}_{<0} = \{x \in \mathbb{R} \mid x < 0\}$.


*

*Spaces $\mathbb{R}_{>0}$ and $\mathbb{R}_{<0}$ are connected and $f$ is continuous, therefore $f(\mathbb{R}_{>0})$ and $f(\mathbb{R}_{>0})$ are also connected. Since $f$ is a bijection, we see that $\mathbb{R}$ is a union of two disjoint non-empty connected subsets: $\mathbb{R} = f(\mathbb{R}_{<0}) \cup f(\mathbb{R}_{>0})$.

*Connected subsets in $\mathbb{R}$ are precisely the same thing as intervals. It's easy to see then that one of the two sets $f(\mathbb{R}_{<0})$ and $f(\mathbb{R}_{>0})$ must be unbounded on one side and closed on the other side. For instance, let's assume that $f(\mathbb{R}_{>0}) = [y, +\infty)$, where $y \in \mathbb{R}$. (The other possible cases are analogous).

*Let $x = f^{-1}(y)$. The restricted function $f|_{\mathbb{R}_{>0}}$ reaches its minimum at point $x > 0$. It follows that there are $x_1$ and $x_2$ somewhere close to $x$ such that $0 < x_1 < x < x_2$, $f(x_1) > f(x)$ and $f(x_2) > f(x)$. Now it's easy to reach a contradiction with the injectivity of $f$ using the intermediate value theorem. Done.

A: The topology is different. $\mathbb{R}=(-\infty,\infty)$ has two open ends and $\mathbb{R}-\{0\}=(-\infty,0)\cup(0,\infty)$ has four. Continuous functions map boundaries to boundaries, which cannot happen bijectively. For instance, $\log |x|$ is an example that maps two-to-one.
Continuous bijection only exists between spaces with equal topology.
