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!
Best Answer
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.