# Interior of two homeomorphic spaces

Suppose X and Y are two homeomorphic subspaces. I'm trying since a few days to construct an example that their interiors need not be homeomorphic.

X and Y can be subspaces of different topological spaces.

Any help would be appreciated.

• The interior of any topological space is the whole space. – Brian M. Scott Jun 21 '16 at 19:36
• I edited the question. – Sahiba Arora Jun 21 '16 at 20:04
• Consider $[0,1]$ as a subspace or $\Bbb R$ and of itself. The interior of $X$ in $\Bbb R$ is $(0,1)$, and its interior in itself is $[0,1]$. – Brian M. Scott Jun 21 '16 at 20:09

## 1 Answer

As Brian M. Scott has pointed out, the way that the question is worded is confusing. I assume you mean that you want $X\subseteq A$ and $Y\subseteq B$ such that $X$ and $Y$ are homeomorphic, but the interior of $X$ (when $X$ is considered as a subset of $A$) is not homeomorphic to the interior of $Y$ (when $Y$ is considered as a subset of $B$). I'll give an example in which $A = B$.

Let $S = \{x,y\}$ be the Sierpiński space, whose open sets are $\{\varnothing,\{x\},\{x,y\}\}$. Let $X = \{x\}$ and $Y = \{y\}$. When equipped with the subspace topologies, $X$ and $Y$ are homeomorphic (there is a unique topology on a one point space). But when viewed as subsets of $S$, the interior of $X$ is $X$, but the interior of $Y$ is $\varnothing$, since $\{x\}$ is open, but $\{y\}$ is not.

Edit: Thanks to Prism for the interesting follow-up question in the comments.

For an example in which $A = B$ is a metric space, take $X = [0,1]$ and $Y = [1/4,3/4]$ as subspaces of $[0,1]$. They are homeomorphic (by the map $x\mapsto \frac{1}{2}x + \frac{1}{4}$), but their interiors $[0,1]$ and $(1/4,3/4)$ are not (one is compact, and the other isn't).

There is no example when $A = B = \mathbb{R}^n$, thanks to the Invariance of domain theorem. Indeed, suppose $X$ and $Y$ are subsets of $\mathbb{R}^n$ and $f\colon X\to Y$ is a homeomorphism. When I say "open set" and "interior", I mean with respect to the topology on $\mathbb{R}^n$, not the subspace topologies. Let $U = \operatorname{int}(X)$, and let $V = f(U)$. Now $U$ is an open set, and the restriction of $f$ to $U$ is certainly injective and continuous, so the invariance of domain theorem tells us that $V$ is open and $f$ restricts to a homeomorphism $U\cong V$. It remains to show that $V = \operatorname{int}(Y)$. $V$ is an open set contained in $Y$, so $V\subseteq \operatorname{int}(Y)$. For the converse, apply the invariance of domain theorem again, this time with $f^{-1}$ and $\operatorname{int}(Y)$, to see that $f^{-1}(\operatorname{int}(Y))$ is an open subset of $X$, so $f^{-1}(\operatorname{int}(Y))\subseteq U = \operatorname{int}(X)$. Hence $\operatorname{int}(Y)\subseteq f(U) = V$ (since $f$ is a bijection).

• Do you know any examples when $A=B=\mathbb{R}^{n}$ with the standard topology (or in general a metric space)? – Prism Jun 22 '16 at 3:23
• @Prism See my edit. – Alex Kruckman Jun 25 '16 at 21:17