Let $Y$ be a topological space, and $M$ be second countable Hausdorff topological space such that for any $p\in M$ it holds that there is an open neighborhood $U(p)$ in $M$ such that $U(p)\approx Y$ where with $\approx$ the homeomorphism relation is denoted. I wonder what $M$ can look like if $Y$ is different from $\mathbb R^n$.

First of all I was thinking about $Y = \mathbb S^1$ but then realized that taking $Y$ compact is quite restricting. Indeed, if $Y$ is Hausdorff second countable and compact then for any $p\in M$ there is a compact neighborhood $U(p)$ and since $M$ is Hausdorff, $U(p)$ is closed and open and hence is a union of connected components of $M$ which is homeomorphic to $Y$. If I am not wrong, it means that $M$ is a disjoint union of $Y$, i.e. $$ M\approx \coprod\limits_{i\in I}Y_i $$ for some countable (i.e. finite of countably infinite) index set $I$. Please correct proof if it's wrong.

On the other hand, if $Y$ is Hausdorff and second countable but not compact, the space $M$ may be more interesting, i.e. for $Y$ being $C\cap (0,1)$ with $C$ - Hausdorff set or $Y$ being torus without one point. I wonder if there are any spaces of interest which are Hausdorff second countable but not locally Euclidean, being rather locally homeomorphic to some more complicated spaces.

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    $\begingroup$ Your compact neighbourhood $U(p)$ need not be open in $M$. Indeed, you could require that manifolds be locally homeomorphic to a closed disc and get the same spaces as with the classic definition. Also, a space is not necessarily homeomorphic to the disjoint union of its connected components (just look at the rationals). $\endgroup$ – Miha Habič Nov 21 '11 at 14:50
  • $\begingroup$ Could you tell me please, why $U(p)$ need not to be open in $M$ if it is defined to be open? Thanks for the example with rationals - but I guess that in my case the space is still homeomorphic to the disjoint union, isn't it? $\endgroup$ – Ilya Nov 21 '11 at 14:58
  • $\begingroup$ Sorry, I missed the fact that you defined $U(p)$ to be open. In this case I'm not sure what happens, even if you require $Y$ to be connected, but my feeling is $M$ still won't be a disjoint union. As a remark on the general case, I believe you get nothing new if $Y$ is a manifold itself. $\endgroup$ – Miha Habič Nov 21 '11 at 15:48

Here’s a complete answer for the case in which $Y$ is compact, the space $\mathbb{P}$ of irrational numbers, or $\mathbb{Q}$.

If you require $Y$ to be compact, it is true that $M$ is homeomorphic to a disjoint union of at most countably many copies of $Y$. To prove this, note first that each point of $Y$ must have a compact open nbhd homeomorphic to $Y$, and in particular $Y$ must be zero-dimensional. If $Y$ is a singleton, $M$ is discrete and therefore is the disjoint union of copies of $Y$ $-$ at most $\omega$ copies, since $M$ is second countable.

If $Y$ is infinite, it can have no isolated points; a second countable compact Hausdorff space is metrizable, so it must be a zero-dimensional compact metrizable space without isolated points, i.e., a Cantor set. $M$ is Lindelöf, so it has a countable open cover $\{U_n:n\in\omega\}$ by Cantor sets. For $n\in\omega$ set $$V_n=U_n\setminus\bigcup_{k<n}U_k\;;$$ then $\{V_n:n\in\omega\land V_n\ne\varnothing\}$ is a partition of $M$ into Cantor sets, and hence either $M\approx Y$, or $M\approx\omega\times Y$ (which is also homeomorphic to $Y\setminus\{p\}$ for any $p\in Y\;$).

For a non-compact example we can take $Y=\mathbb{P}$, the space of irrational numbers with the usual topology. Since $Y$ is zero-dimensional, we can argue as before that there is a partition of $M$ into at most $\omega$ copies of the irrationals. But it’s well known that $\mathbb{P}\approx\omega^\omega$ with the product topology, so $\omega\times\mathbb{P}\approx\mathbb{P}$, and $M$ must be homeomorphic to $Y$.

If $Y=\mathbb{Q}$, second countability of $M$ ensures that $M$ is countable. Thus, $M$ is a countable, metrizable space with no isolated points and as such is homeomorphic to $\mathbb{Q}$.


To get a more interesting answer you should modify the notion of a neighborhood that you are using. Following Bourbaki, a neighborhood of a point $x$ in a topological space $X$ is a subset $N\subset X$ such that there exists an open subset $U\subset X$ such that $x\in U\subset N$. With this in mind, an $n$-dimensional manifold (possibly with boundary) is a (2nd countable, Hausdorff) topological space such that every point in $X$ admits a basis consisting of neighborhoods homeomorphic to the closed $n$-ball in $R^n$.

Similarly, an $n$-dimensional Menger manifold is a (2nd countable, Hausdorff) topological space such that every point in $X$ admits a basis consisting of neighborhoods homeomorphic to the $n$-dimensional Menger space $\mu^n$. The theory of Menger manifolds is quite rich and interesting.

See for instance

Kazuhiro Kawamura, "A survey on Menger manifold theory—Update", Topology and its Applications 101 (2000) 83–91.

Similarly, one defines $Q$-manifolds (locally homeomorphic to the Hilbert cube). This theory is again quite interesting and, in some ways, similar but simpler than the theory of topological manifolds.


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