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The manifold of $\Bbb R^n$ with its usual topology certainly has this property, every open set has the same size as every other open set, so there exists a bijection between any two open sets. But, does the fact that a smooth manifold is locally homeomorphic to some euclidean space imply that every open set has same cardinality?

I'm not able to prove it because the definition says that it is locally homeomorphic, but I think it is because every open set has a representation, in one chart or another so it must be the same as some open set in $\Bbb R^n$. Is this right?

Garmekain
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    Isn't this immediate from the fact that for each $n$ and for each nonempty open set $U$ in ${\mathbb R}^{n},$ the cardinality of $U$ is $c$ (and any homeomorphism is a bijection)? Or am I missing something? – Dave L. Renfro Dec 01 '18 at 19:31

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Assuming you require your manifolds to be second-countable, then it is true that every nonempty open subset of a manifold of positive dimension has cardinality $2^{\aleph_0}$. Indeed, suppose $M$ is a manifold of dimension $n>0$. Since $M$ is a manifold, it has a basis of open sets that are homeomorphic to $\mathbb{R}^n$ (and thus have cardinality $2^{\aleph_0}$. By second-countability, this basis can be chosen to be countable. Any open subset of $M$ is a union of countably many basis sets. So, any open subset has cardinality at most $\aleph_0\cdot 2^{\aleph_0}=2^{\aleph_0}$. Conversely, any nonempty open subset contains at least one basis set and so has cardinality at least $2^{\aleph_0}$.

(Of course, this is false for $0$-dimensional manifolds since you could take a discrete space of any countable cardinality which has open subsets of all smaller cardinalities. It also fails if you do not require second-countability since then you could take a disjoint union of more than $2^{\aleph_0}$ copies of $\mathbb{R}^n$.)

Eric Wofsey
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