Let $A, B \subseteq \mathbb{R^n}$ and $f:A \rightarrow B$ be an homeomorphism with the subspace topologies. Is it the case that $A$ is regular closed if and only if $B$ is regular closed? By regular closed, I mean $A = \text{cl}_{\mathbb{R}^n}(\text{int}_{\mathbb{R}^n}(A))$, where $\text{cl}_{\mathbb{R}^n}$ and $\text{int}_{\mathbb{R}^n}$ are the closure and interior in $\mathbb{R}^n$, respectively.
I know that if $g:\mathbb{R}^n \rightarrow \mathbb{R^n}$ is an homeomorphism and that $B = g(A)$, the equivalence I'm looking for holds. But my question is slightly different.
I also know that if $A \subseteq \mathbb{R}^n$, $B \subseteq \mathbb{R}^m$ with $n \not= m$, and $h:A \rightarrow B$ is an homeomorphism (with the subspace topologies), then it is not necessarily true that $A = \text{cl}_{\mathbb{R}^n}(\text{int}_{\mathbb{R}^n}(A)) \iff B = \text{cl}_{\mathbb{R}^m}(\text{int}_{\mathbb{R}^m}(B))$. A counterexample would be $f:[0,1] \rightarrow [0,1] \times \{0\}$ where $f(x) = (x,0)$. So I suspect a proof of the result I'm looking for would use the fact that $m=n$.
Are there additional simple conditions on $f$ that would make the statement true? For example, what if $f$ is a diffeomorphism from $A$ to $B$?
I hope I'm not thoroughly confused! Thanks!
Theorem 1. Suppose $A, B\subseteq\mathbb{R}^n$ are homeomorphic. If $A$ is regular closed in $\mathbb{R}^n$, then $\overline{\text{int}(B)} = \overline{B}$. In particular, $\overline{B}$ is regular closed.
Proof: If $f:A\to B$ is a homeomorphism, then from invariance of domain $f[\text{int}(A)] = \text{int}(B)$ (the interior is in $\mathbb{R}^n$). This is since it implies that $f[\text{int}(A)]$ and $f^{-1}[\text{int}(B)]$ are both open in $\mathbb{R}^n$, so that $f[\text{int}(A)]\subseteq \text{int}(B)$ and $f^{-1}[\text{int}(B)]\subseteq \text{int}(A)$, thus $f[\text{int}(A)] = \text{int}(B)$.
We have $B = f[A] = f[\overline{\text{int}(A)}] \subseteq \overline{f[\text{int}(A)]} = \overline{\text{int}(B)}$. Thus $\overline{B}\subseteq \overline{\text{int}(B)}$. Since the other inclusion is clear, we see that $\overline{B} = \overline{\text{int}(B)}$. Of course, $\overline{\text{int}(B)}\subseteq \overline{\text{int}(\overline{B})}\subseteq \overline{B}$ so this implies that $\overline{B}$ is regular closed. $\square$
Corollary. If $A, B\subseteq \mathbb{R}^n$ are homeomorphic, $A$ is a compact regular closed set, then so is $B$.
Various modifications of theorem 1 fail:
Example 1. If $A, B\subseteq \mathbb{R}$ are homeomorphic and $A$ is regular closed, then $B$ doesn't have to be closed: $[0, \infty), [0, 1)\subseteq\mathbb{R}^n$ are homeomorphic, $[0, \infty)$ is regular closed, but $[0, 1)$ isn't.
Example 2. If $A, B\subseteq \mathbb{R}$ are homeomorphic and $\overline{A}$ is regular closed, then $\overline{B}$ doesn't have to be regular closed: Let $A = \mathbb{Q}\cap (0, 1)$ and $B$ be the endpoints of intervals removed in the construction of the Cantor set. Then $A, B$ are homeomorphic, $\overline{A} = [0, 1]$ is compact and regular closed, but $\overline{B} = C$ is the Cantor set, for which $\overline{\text{int}(C)} = \emptyset$, so that $\overline{B}$ is not regular closed.
Converse of theorem 1 fails:
Example 3. If $A\subseteq \mathbb{R}$ and $\overline{\text{int}(A)} = \overline{A}$ then $A$ doesn't have to be homeomorphic to a closed subset of $\mathbb{R}$: Let $A = (0, 1)\cup (1, 2)$, then if $B\subseteq \mathbb{R}$ is homeomorphic to $A$, then from invariance of domain, $B$ is a disconnected open subset of $\mathbb{R}$, thus can't be closed (for the only clopen subsets of $\mathbb{R}$ are $\mathbb{R}$ and $\emptyset$, which are connected).
Converse of theorem 1 for convex sets:
Remark. All convex subsets $A\subseteq\mathbb{R}^n$ with $\text{int}(A)\neq \emptyset$ satisfy $\overline{\text{int}(A)} = \overline{A}$.
Example 4. Note that if $A\subseteq \mathbb{R}$ is connected and $\text{int}(A)\neq \emptyset$, then $A$ is homeomorphic to a regular closed subset of $\mathbb{R}$: $A$ is homeomorphic to either $[0,\infty), [0, 1]$ or $\mathbb{R}$.
Example 5. If $A\subseteq \mathbb{R}^2$ is convex and $\text{int}(A)\neq \emptyset$, then $A$ doesn't have to be homeomorphic to a closed subset of $\mathbb{R}^2$: Let $A = \{(x_1, x_2)\in\mathbb{R}^2 : x_1^2+x_2^2 < 1 \text{ or } x_1^2+x_2^2 = 1\text{ and }x_2\leq 0\}$, $K = \{x\in A : x_1^2+x_2^2 = 1\}$ and $f:A\to B\subseteq\mathbb{R}^2$ a homeomorphism, where $B$ is closed. Then $K$ has no compact neighbourhood in $A$, but $f[K]$ has a compact neighbourhood in $B$. This is a contradiction, so such homeomorphism $f$ cannot exist.
Note that in above counter-examples weren't homeomorphic to a closed subset of $\mathbb{R}^n$. With that assumption we have the converse of theorem 1:
Theorem 2. $A\subseteq \mathbb{R}^n$ is homeomorphic to a regular closed subset of $\mathbb{R}^n$ iff $A$ is homeomorphic to a closed subset of $\mathbb{R}^n$ and $\overline{\text{int}(A)} = \overline{A}$.
Proof: If $A$ is homeomorphic to a regular closed subset, then its homeomorphic to a closed subset, and theorem 1 implies that $\overline{\text{int}(A)} = \overline{A}$. Conversely, if $\overline{\text{int}(A)} = \overline{A}$, $f:A\to B$ is a homeomorphism and $B\subseteq\mathbb{R}^n$ be closed, then from invariance of domain $f[\text{int}(A)] = \text{int}(B)$ like before. We have $B = f[A] = f[\text{cl}_A(\text{int}(A))]\subseteq \overline{f[\text{int}(A)]} = \overline{\text{int}(B)}$, and since $B$ is closed, $\overline{\text{int}(B)}\subseteq B$ holds as well. Thus $B$ is actually regular closed. $\square$