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I want to show that no two of the spaces $(0,1), \ (0, 1], \ [0,1]$ are homeomorphic, with the hint: what happens if you remove a point from each one of these spaces.

No idea where to begin, except that it concludes a chapter on connectedness.

All these spaces are all in the subspace topology of the usual topology on $\mathbb{R}$.

Edit

(Thanks @Chris Culter)

So far what I've got is: Let $A = (0,1] - \{1\}$. If $f:(0,1] \rightarrow (0,1)$ is a homeomorphism, then $f$ restricted to $A$ is a homeorphism onto a subset of $(0,1)$ with one and only one point removed. But any removal of a point in $(0,1)$ results in a disconnected space, while at the same time $f_A$ ensures that the space is connected, it being a continuous image of a connected space $A$. Therefore there is no continuous bijective map between the two spaces let alone a homeomorphism. A similar agument shows that $[0,1]$ and $(0,1)$ are not homeomorphic.

I can't seem to show that $(0,1]$ and $[0,1]$ are not homeomorphic using this method though.

Daniel Donnelly
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2 Answers2

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All of them have a different number of non-cut-points, i.e. points that you can remove without disconnecting the set.

Stefan Hamcke
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Well, following the hint: if you remove a point from $(0,1)$, is the remaining space connected?

Chris Culter
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  • Okay so if I remove a point from $(0,1]$, namely $1$ then the remaining space is connected but no point can be removed from $(0,1)$ and make the remaining space connected. But how do I prove that that's a necessary homeomorphism-invariant property? – Daniel Donnelly Sep 24 '13 at 22:24
  • Connectedness is defined in terms of the topology of a space, so you could just say that it's "manifestly homeomorphism-invariant". Or, go ahead and prove that any space homeomorphic to a connected space is also connected. – Chris Culter Sep 24 '13 at 22:28
  • Please see my edit above. Thanks for your help! – Daniel Donnelly Sep 24 '13 at 22:37