I want to prove

A manifold $M$ can be covered by a countable collection of neighbourhoods each diffeomorphic to an open subset of $\mathbb R^m$.

By definition we have for each $x \in M$ an open neighbourhood diffeomorphic to an open subset of $\mathbb R^m$. But I don't understand why they have to be countable. Here a manifold is subset of Euclidean space.
Manifold is a subset $M$ of $\mathbb R^n$ is a $k$ - dimensional manifold if $\forall x\in M $ there is open $U,V \subset \mathbb R^n$ $x \in U$ and a diffeomorphism $f:U\rightarrow V$ such that $f(U \cap M)=V \cap(\mathbb R^k \times 0) $.

  • 2
    $\begingroup$ What is your definition of a manifold? $\endgroup$
    – D. Brogan
    Jul 5 '18 at 16:01
  • 1
    $\begingroup$ Presumably, you have to use second-countability of the Euclidean space your manifold is embedded into. $\endgroup$ Jul 5 '18 at 16:13
  • $\begingroup$ If we assume that then there is nothing left to prove. I just learned that manifold is also defined such that it is second countable. But the definitions (the one I provided and this) should be equivalent and hence I am now concerned with proving the equivalence at least one way. $\endgroup$ Jul 5 '18 at 16:28

As you state that manifolds are a subspace of some $\mathbb{R}^n$, this means they are second countable, and in particular Lindelöf. This implies that the cover by "open neighbourhoods of $M$ that are diffeomorphic to an open subset of $\mathbb{R}^k$" of $M$ has a countable subcover. Done.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.