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Two topological spaces are homeomorphic if there is a bijective continuous map between them that the inverse of this map is continuous too. I think the spaces $[0,1)\times [0,1)$ and $(0,1)\times (0,1)$(as subsets of $\mathbb{R}^2$) are not homeomorphic. But how we can prove that.

The properties of compactness or connectivity cannot help me.

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Hint: If you remove $(0,0)$ from the first space, it is still simply connected

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Hint: when you remove any point of $(0,1)^2$, you get a space homeomorphic to $\mathbb{R}^2 \backslash \{0\}$, thus homeomorphic to $S^1 \times \mathbb{R}^{+*}$.

When you remove $(0,0)$ from $[0,1)^2$, you get a space homeomorphic to $\{x,y \geq 0,\,x+y >0\}$, which is (polar coordinates) homeomorphic to $\mathbb{R}^{+*} \times [0,\pi/2]$.

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  • $\begingroup$ Now how do you show $\mathbb{R}^{+*} \times [0,\pi/2]$ is not homeomorphic to $S^1 \times \mathbb{R}^{+*}$? $\endgroup$ – David C. Ullrich May 17 at 14:07
  • $\begingroup$ I was trying to think of something else, but it all boils down to homotopy anyway, I would guess. $\endgroup$ – Mindlack May 17 at 16:21
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In $[0,1)\times [0,1)$ you have boundary points which belong to the set.

On the other hand there is no boundary points of $(0,1)\times (0,1)$ which belong to it.

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  • $\begingroup$ @Darman I guess he is referring to Manifold boundary $\endgroup$ – YuiTo Cheng May 17 at 13:13
  • $\begingroup$ There's still a lot of work to do to show this implies the sets are not homeomorphic. All that's obvious from this is that there is no self-homeomorphism of $\Bbb R^2$ that takes one set to the other... $\endgroup$ – David C. Ullrich May 17 at 14:02
  • $\begingroup$ The problem is that "$p$ is a boundary point of $A$" does not depend just on the topology of$A$; "boundary point" applies to subsets of topological spaces, not to topological spaces per se. $\endgroup$ – David C. Ullrich May 17 at 14:04

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