Are topological spaces $X,Y$ homeomorphic if each is homeomorphic to a subspace of the other? The Schroeder-Bernstein theorem states that

If $A$ and $B$ are two sets and there exists injective maps $f:A \to B$ and $g:B \to A$, then there exists a bijection $h:A \to B$.

The statement above is equivalent to the following

Suppose $A_1 \subset A$ and $B_1 \subset B$ are sets such that there exists a bijective map $f:A \to B_1$ and $g:B \to A_1$, then there exists a bijection $h:A \to B$.

So I was thinking about the problem below. I couldn't prove it, but not getting a counterexample either. Any hints will be appreciated.

Now suppose $A \subset X$, $B \subset Y$ are topological spaces, and suppose that there exists a homeomorphism $\alpha:A \to Y$ and $\beta: B \to X$, then can we conclude that $X$ and $Y$ are homeomorphic?

 A: The topological version of Schroeder-Bernstein fails badly. For example, consider $(0,1)$ and $[0,1]$ with their usual topologies, which are of course non-homeomorphic:

*

*$(0,1)$ is homeomorphic to itself, which is a subspace of $[0,1]$.


*$[0,1]$ is homeomorphic to $[{1\over 3}, {1\over 2}]$, which is a subspace of $(0,1)$.
In general, the question of when Schroeder-Bernstein holds for a given category of mathematical objects is a quite interesting one; see e.g. this MathOverflow question. For example, while the counterexample above shows that Schroeder-Bernstein fails for metric spaces, it in fact holds for compact metric spaces (and while $[0,1]$ is compact, $(0,1)$ isn't).
A: No, Suppose $A=[-1,1], X=\mathbb{R}, B=(-2,2)$ and $Y=[-3,3]$. Then this gives a counter example
A: No, consider $X = [0,1]$ and $Y = [0,1] \cup [2,3]$, with their usual subspace topologies. You may let $A = [0,1/3] \cup [1/3,2/3]$ and $B = [0,1]$. $X$ is connected while $Y$ is not.
A: For example, consider $X=[0,1]$ and $Y=(0,1)$. $X$ is not homeomorphic to $Y.$ but $X$ is homeomorphic to $[1/4,3/4] \subset Y=(0,1).$ and Y is homeomorphic to $(0,1)\subset X.$
