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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?

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    $\begingroup$ Do you want to impose some sort of compatibility between this topology and the standard one? Otherwise, you are just asking: What all are the topological spaces $X$, with cardinality the continuum, such that there is a continuous bijection $X \to X \setminus \{ x,y \}$. 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. $\endgroup$
    – David E Speyer
    Commented Jun 1, 2011 at 20:32
  • $\begingroup$ @David Speyer: This is a good point, I'll restate my question. $\endgroup$
    – Asaf Karagila ♦
    Commented Jun 1, 2011 at 20:35
  • $\begingroup$ In reply to the ``Note that ...'': there is no `typical' bijection between $[0,1]$ and $(0,1)$. All that happens is that $0$ and $1$ disappear along the sequences of iterates $f^n(0)$ and $f^(1)$; these can be as wild as you want (dense, somewhere dense but not everywhere, dense in the Cantor set, ...) and outside those sequences you can mess up the structure of the interval even more. $\endgroup$
    – KP Hart
    Commented Jun 4, 2011 at 7:19

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