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Show that no two of the intervals $[0,1]$ $(0,1)$ and $(0,1]$ are homeomorphic.

Instead of using connectedness. I simply use the openness and closedness of each interval. For example:

Suppose $(0,1]$ and $(0,1)$ are homeomorphic, with the homeomorphism $f: (0,1] \to (0,1)$. $f$ is a continuous bijection, thus $f^{-1}((0,1)) = (0,1]$. $(0,1)$ is open, but $(0,1]$ is not. Contradiction.

Is it approach correct?

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    $\begingroup$ $(0,1]$ is open, in the topology of $(0,1]$. $\endgroup$ – user228113 Jan 3 '16 at 13:22
  • $\begingroup$ Note that $(0,1]$ is open in the subspace topology on $(0,1]$. Thus (at the very least) you must be more precise as to what you mean by "$(0,1]$ isn't open". $\endgroup$ – Micapps Jan 3 '16 at 13:22
  • $\begingroup$ Actually $(0,1]$ is open when the topological space in question is $(0,1]$ itself. So you'd have to start by proving that you can extend $f$ to something $\mathbb R\to\mathbb R$, before you can speak about "open in $\mathbb R$". $\endgroup$ – Henning Makholm Jan 3 '16 at 13:22
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    $\begingroup$ I don't think so. For this proof to work, you need $f:\mathbb{R}\rightarrow\mathbb{R}$, not as you have defined it. Because being open/closed are properties of the subspace, so $(0,1]$ is open in the subspace topology. $\endgroup$ – Simon_Peterson Jan 3 '16 at 13:22
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This is not correct. You must be careful about the domain and range when speaking of a homeomorphism, and you have to be careful about using the term "open" when there may be several underlying spaces being discussed. $(0,1]$ is indeed open in $(0,1]$ since it is the entire space. You can't say a set is open without it being clear what the space is in which it is supposed to be open.

I suspect you mean that it is not open in $\mathbb R$, but the domain of the homeomorphism is not $\mathbb R$, it is $(0,1]$.

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