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I only know that the each point has some neighborhood with cardinality same as $\mathbb R$, but I have no idea about the cardinality of the whole manifold, must it have cardinality as $\mathbb R$? Please give some example to illustrate.

JSCB
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Manifolds are typically assumed to be second countable and thus they have a countable base which can be taken to consist of open subsets homeomorphic to open balls in $\mathbb{R}^n$. If $n > 1$, then open balls have the same cardinality as $\mathbb{R}$ and so an $n$-dimensional second countable manifold has the same cardinality as $\mathbb{R}$.

A zero dimensional second countable manifold is finite or countable.

If you don't assume your manifold is second countable, any discrete space can be considered as a zero-dimensional manifold and thus manifolds can be of arbitrary cardinality.

levap
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