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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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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). –  Miha Habič Nov 21 '11 at 14:50
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? –  Ilya Nov 21 '11 at 14:58
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. –  Miha Habič Nov 21 '11 at 15:48

1 Answer 1

up vote 2 down vote accepted

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 $\mathbb{Q}$.

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