Why $[0,1]\times (0,1)$ and $[0,1)\times (0,1)$ are not homeomorphic

Question

This question has been asked here Homeomorphisms in $\mathbb{R}^2$., but I don't understand in the answer said that they are not homeomorphic since boundary of $[0,1]\times (0,1)$ has two connected components, while $[0,1)\times (0,1)$ only has one.

I'm confused about the "boundary" in that answer, aren't boundaries of $[0,1]\times (0,1)$ and $[0,1)\times (0,1)$ same in $\Bbb{R}^2$? And why this shows the two subspaces are not homeomorphic? Thanks!

Read the answer you’re referring to more carefully. The two sets you’re looking at are not the ones that are proved to be non homeomorphic — marwalix, Dec 28 '19 at 23:18
Where must a homeomorphism send a connected boundary component? — Eric Towers, Dec 28 '19 at 23:25
@EricTowers I'm confused about the "boundary" in that answer, aren't $[0,1]\times (0,1)$ and $[0,1)\times (0,1)$ have same boundary in $R^2$? — , Dec 28 '19 at 23:43
He is not talking about the topological boundary, but the boundary as a manifold. — Severin Schraven, Dec 28 '19 at 23:49

score 2 · Accepted Answer · answered Dec 28 '19 at 23:47

Let $A_1 = [0,1] \times (0,1)$ and $A_2 = [0,1)\times(0,1)$.

We have $$\partial A_1 = \partial A_2 = [0,1]^2 \setminus (0,1)^2$$

and hence

$$A_1 \cap \partial A_1 = \{0,1\} \times (0,1)$$

$$A_2 \cap \partial A_2 = \{0\} \times (0,1)$$

If $f : A_2 \to A_1$ is a homeomorphism, we have $f(\partial A_2) = \partial A_1$ (why?) so $$f(A_2 \cap \partial A_2) = f(A_2) \cap f(\partial A_2) = A_1 \cap \partial A_1$$

Since $A_2 \cap \partial A_2$ is a connected set and $f$ is continuous, it follows that $A_1 \cap \partial A_1$ is a connected set as well. This is a contradiction.

Why $[0,1]\times (0,1)$ and $[0,1)\times (0,1)$ are not homeomorphic

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