Homeomorphism between topological subspaces

Question

Let $X$ be a topological space. Take $A,B\subseteq X$ and assume that there is homeomorphism $\phi:A\rightarrow B$.

Is it true that $\phi(int(A))=int(B)$?

I know that homeomorphism preserve all topological structure so it seems true. But $int(A)$ can't be defined by the topology of $A$ alone, it must use the topology of $X$. Hence it sound reasonable to say that the homemorphism dosn't surely take the interior to the interior.

This being said, it sound strange to me that we have, for example, a homemorphism between some cube and some ball but in some way it doesn't map interior to interior. Do you have an exmplae for that?

Answer 1

Here is a counterexample:

$X = [0,1]$, $A = [0,1]$, $B = [0,1/2]$, $h(a) = a/2$.

Answer 2

Many counterexamples exist in abstract metric spaces, one of them given by Paul Frost. But given your particular objection:

It sound strange to me that we have, for example, a homemorphism between some cube and some ball but in some way it doesn't map interior to interior.

it could be interesting for you that the theorem holds when $X = \mathbb{R}^n$ with the standard topology. That is,

Assume $A, B \subseteq \mathbb{R}^n$ and $\phi : A \to B$ is a homeomorphism. Then $\phi[ \operatorname{int} A ] = \operatorname{int} B$.

To prove it, we can use the domain invariance theorem:

If $U \subseteq \mathbb{R}^n$ is open and $f : U \to \mathbb{R}^n$ is continuous and injective, then $f[U]$ is open.

From the theorem we quickly get that $\phi[\operatorname{int} A] \subseteq B$ is open in $\mathbb{R}^n$, so $\phi[\operatorname{int} A] \subseteq \operatorname{int} B$. The other inclusion follows by symmetry. $\square$

So indeed, a homeomorphism between a cube and a ball must preserve the interior.