Following some argument over a question on math.SE, I began to wonder:

We all know that in the standard topology there are no continuous bijections from $[0,1]$ to $(0,1)$ (for example by arguments of compactness).

However, if we consider the discrete topology on $\mathbb R$ then every function is continuous. In particular, every bijection between the $[0,1]$ and $(0,1)$ is continuous.

**Question:** Can we characterize all the topologies on $\mathbb R$ which refine the standard topology (i.e. open intervals are still open), and there exists a function $f$ which is continuous w.r.t to the topology, and is a bijection *from* $[0,1]$ to $(0,1)$?

Note that a typical bijection takes a monotonously decreasing sequence to $0$, places $0$ to the first term, $1$ to the second, and $x_n\mapsto x_{n+2}$, any other element in the interval is fixed.

This leads me to conjecture that such topology will have to have $0,1$ and some $\{x_n\mid n\in \mathbb N\}$ a monotonically decreasing sequence of isolated points, in turn this implies that $[0,1)$ is open, as well $(0,1]$ and we could probably cook some continuous bijections between half-open/half-closed intervals as well.

However, my topological toolbox is not very rich, though, and I could not prove that.

**A slightly generalized question:** What if we require $f$ to be a homeomorphism?

`$X \to X \setminus \{ x,y \}$`

. There are going to be tons of these. For example, take any topological space $Y$ of continuum cardinality and take its disjoint union with a countable number of isolated points. – David Speyer Jun 1 '11 at 20:32