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I need to check whether the open interval $(0,1)$ is homeomorphic to the semi-closed interval $[0,1)$ or not? (Usual topology is defined on both the intervals).

I am not sure how to approach. I just have a vague idea. I can consider a homeomorphic map between these two intervals and then show that it's actually not one-one or onto by removing a point (perhaps $0$) from $[0,1)$.

Silvia Ghinassi
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user262860
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    That's the right idea. If there were a homeomorphism $[0,1)\to (0,1)$, then deleting the image of $0$ disconnects the range space; however, removing $0$ from $[0,1)$ does not have such an effect. Thus, ... – BrianO Nov 22 '15 at 22:29
  • @MoebiusCorzer Take this to mean: subspace topologies of the usual topology. – BrianO Nov 22 '15 at 22:30
  • @BrianO Yes, I understood it after. Thanks. To the OP: my bad, I misunderstood what you meant (and it is why I deleted my comment). – MoebiusCorzer Nov 22 '15 at 22:31
  • @MoebiusCorzer Actually it's a pretty heavyweight theorem (Brouwer's theorem on invariance of domain) that openness in $\mathbb{R}^n$ is invariant under homeomorphism, and it isn't true for general topological spaces (there can be subsets homeomorphic to open subsets that are not open). – Matt Samuel Nov 22 '15 at 22:33
  • @MattSamuel I misunderstood what he meant and mixed up some notions accordingly, so that I deleted my comment when I realized I was completely mistaken. Thanks anyway :) – MoebiusCorzer Nov 22 '15 at 22:34

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Only one of those sets remains connected when removing a point (not any point, mind you).

[An entirely different approach would be to show that any homeomorphism between connected and complete ordered topological spaces is in fact an order isomorphism or an anti-isomorphism (order reversing). Then use the fact that only one of these ordered sets has any endpoints.]

Asaf Karagila
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