# Are the topological spaces $[0,1)\times [0,1)$ and $(0,1)\times (0,1)$ homeomorphic

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.

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

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 $$\mathbb{R}^{+*} \times [0,\pi/2]$$.

• Now how do you show $\mathbb{R}^{+*} \times [0,\pi/2]$ is not homeomorphic to $S^1 \times \mathbb{R}^{+*}$? – David C. Ullrich May 17 '19 at 14:07
• I was trying to think of something else, but it all boils down to homotopy anyway, I would guess. – Mindlack May 17 '19 at 16:21

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.

• @Darman I guess he is referring to Manifold boundary – YuiTo Cheng May 17 '19 at 13:13
• 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... – David C. Ullrich May 17 '19 at 14:02
• 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. – David C. Ullrich May 17 '19 at 14:04