Let $X,Y$ be (compact connected) topological manifolds of dimensions $n,m$, respectively. Assume that a dense $G_{\delta}$ subspace $A$ of $X$ is homeomorphic to a dense $G_{\delta}$ subspace $B$ of $Y$. Can we claim that $n=m$?
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2$\begingroup$ Every manifold with dimension greater than 0 is going to have a dense $G_\delta$ subspace homeomorphic to Baire space, so any two such manifolds satisfy this condition. $\endgroup$– James E HansonCommented Oct 30 at 16:08
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$\begingroup$ That is very nice, James, I would appreciate a reference for this construction. Would things change if we assume, in addition, that there are homeomorphisms $f\colon A\to B$ and $g\colon B\to A$ extending to continuous maps $F\colon X\to Y$ and $G\colon Y\to X$, respectively? $\endgroup$– William of BaskervilleCommented Oct 30 at 17:18
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$\begingroup$ I don't actually know a reference off the top of my head, but I can write out an argument for it later today if nobody else answers. If the maps extend to continuous maps, is that enough for the extension to be a homeomorphism since $A$ and $B$ are dense? $\endgroup$– James E HansonCommented Oct 30 at 17:42
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$\begingroup$ Extensions $F,G$ may fail to be injective. Such an example can be constructed on $X=Y=\mathbb T^2$ and with $A=B$. $\endgroup$– William of BaskervilleCommented Oct 30 at 17:54
1 Answer
Recall that a metric space is perfect if it has no isolated points. Also recall that a Polish space is a separable completely metrizable space. Finally recall that Baire space is the space of sequences of natural numbers, $\omega^\omega$ with some product metric. Baire space is homeomorphic to the subspace of $\mathbb{R}$ of irrational numbers. Baire space has a lot of nice characterizations among Polish spaces. Up to homeomorphism, it is the unique non-empty zero-dimensional nowhere locally compact Polish space (where zero-dimensionality is the property of admitting a basis of clopen sets).
Proposition. Every non-empty perfect Polish space has a dense $G_\delta$ subset that is homeomorphic to Baire space.
Proof. Fix a non-empty perfect Polish space $X$ and fix a metric $d$ on $X$ inducing its topology satisfying $d(x,y) \leq 1$ for all $x$ and $y$. Let $\omega^{<\omega}$ be the set of finite sequences of natural numbers. I'm going to define a family $(U_{\sigma})_{\sigma \in \omega^{<\omega}}$ of non-empty open subsets of $X$ satisfying the following conditions:
- $U_\varnothing = X$.
- For every $\sigma$, the diameter of $U_\sigma$ is at most $2^{-|\sigma|}$ (where $|\sigma|$ is the length of the sequence $\sigma$).
- For any $\sigma$ and $n$, $\overline{U_{\sigma\frown n}} \subseteq U_\sigma$.
- For any $\sigma$, the sequence $(U_{\sigma\frown n})_{n \in \omega}$ is pairwise disjoint and has dense union in $U_\sigma$.
Note that having such a family gives us fairly immediately what we want. For any $\alpha \in \omega^\omega$, we get that $\bigcap_{n \in \omega} U_{\alpha | n}$ (where $\alpha | n$ is the initial segment of $\alpha$ of length $n$) is a singleton $\{x_\sigma\}$ by the condition on the diameters of the $U_\sigma$'s and the fact that the closure of each $U_{\alpha | n+1}$ is contained in $U_{\alpha | n}$. The map $\sigma \mapsto x_\sigma$ is then a uniformly continuous bijection with continuous inverse. Moreover, the image of this map is $\bigcap_{n \in \omega} \bigcup_{\sigma \in \omega^{<\omega},|\sigma|=n} U_\sigma$, which is clearly a countable intersection of dense open sets, giving us that the image of the map is a comeager $G_\delta$ set.
Building the family $(U_\sigma)_{\sigma \in \omega^{<\omega}}$ is a fairly straightforward induction argument. Suppose that we're given a non-empty open set $U_\sigma$. Since $X$ is a non-empty perfect Polish space, $U_\sigma$ must be infinite and so has a countable dense sequence $(x_i)_{i \in \omega}$. For each $n$, given $U_{\sigma\frown k}$ for all $k < n$, find the smallest $i$ such that $x_i$ is not in $\bigcup_{k < n} \overline{U_{\sigma \frown k}}$. Since the complement of this set is open, we can find some open ball $B_r(x_i)$ with $r \leq 2^{-|\sigma|-2}$ such that $\overline{B}_r(x_i) \subseteq U_\sigma \setminus \bigcup_{k < n} \overline{U_{\sigma \frown k}}$. Let $U_{\sigma \frown n} = B_r(x_i)$. Note that we have maintained the induction hypothesis in that $U_\sigma \setminus \bigcup_{k \leq n} \overline{U_{\sigma \frown k}}$ is non-empty.
Proceeding by induction then builds the entire family $(U_\sigma)_{\sigma \in \omega^{<\omega}}$ satisfying the bullet points above, so we get that this is the required dense $G_\delta$ subset homeomorphic to Baire space. $\square$
More generally, the same argument basically shows that every Polish space has a dense $G_\delta$ subset that is homeomorphic to either a finite or countably infinite discrete space, Baire space, or the coproduct of a finite or countably infinite discrete space and Baire space.
So in particular, we get that the only properties of (metrizable) manifolds that are preserved by having homeomorphic dense $G_\delta$ subsets are the number of isolated points and whether there is a connected component of positive dimension.