TITLE: Continuous bijection from $(0,1)$ to $[0,1]$
QUESTION [102 upvotes]: Does there exist a continuous bijection from $(0,1)$ to $[0,1]$? Of course the map should not be a proper map.

REPLY [3 votes]: Recall the following result:
Proposition 1: Let $f: (a,b) \to (c,d)$ be an injective continuous mapping from one open interval to another. Then the image of $f$ is an open interval.
Now assume we have a continuous bijection $f: (0,1) \to [0,1]$. The function $f$ maps some point in $(0,1)$ to $0$ and another point to $1$. If we remove these points we can restrict $f$ and define an injective and surjective continuous mapping
$\tag 1 f: (0,a) \sqcup (a,b) \sqcup (b,1) \to (0,1)$
on three non-overlapping intervals.
By proposition 1 the image of each of these three intervals is an open interval of $(0,1)$. Since $f$ is also surjective, we have a partition of $(0,1)$ into three open subsets. But it is not possible to express a connected topological space in such a manner.