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I am reading the first volume of Spivak's Differential Geometry Series, A Comprehensive Introduction to Differential Geometry, and I don't understand his definition of a manifold.

He says that

A metric space $M$ is a manifold if for each $x\in M$, there is some neighborhood $U$ of $X$ and some integer $n\ge 0$ such that $U$ is homeomorphic to $\mathbb{R}^n$.

I understand how the (usual) Hausdorff condition of a manifold comes from the fact that $M$ is a metric space in this definition (of course, one can get rid of the metric space condition and just let $M$ be Hausdorff or just a topological space), but how does Spivak's definition account for the condition that $M$ must be second countable?

Thank you for your help.

Nicholas James
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  • I see another problem with this defintion: According to this one the disjoint union of $\mathbb{R}$ and $\mathbb{R}^2$ would be a manifold, which I have never ever seen. – Ulli Oct 20 '22 at 14:28
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    The fact that manifold are allowed not to be connected is fairly common (looking up the expression "connected manifold" seems to back this up). Maybe you are referring to the fact that it is not even equidimensional. I think that's fair game too. – Captain Lama Oct 20 '22 at 14:47
  • Yes, in my world a manifold has everywhere the same dimension. Do you have another reference for the unequality case? Of course, also in my opinion manifolds need not be connected. Sorry, I should have mentioned that, since my example could be misleading, indeed. – Ulli Oct 20 '22 at 15:25
  • I think if you want to fixed the dimension of a manifold, that is if you want to say a metric space "$M$ is said be a $n$ dimensional manifold " then you are going to assign some non negative integer corresponding the manifold M. In that stage since we are fixes the dimension that should be unique otherwise if taking two points on the manifold having different open ball one of is homeomorphic to $\mathbb{R}^n$ another one is homeomorphic to $\mathbb{R}^m$ for different $m, n$ then by Invarience of domain it will give a contradiction. – T ghosh Jan 29 '24 at 21:59
  • Another remark: I think if take a $M=$ is disk disjoint union of a line segment, then it is a manifold, here you can't define the dimension of then manifold in that case is manifold without defending the dimension. I never seen any use of it. – T ghosh Jan 29 '24 at 22:16

3 Answers3

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Short answer: the definition without "second countable" is not equivalent to the definition with "second countable".

For a longer answer, read the rest of that page, and follow the instructions to read Appendix A. After reading Appendix A, you will hopefully have a better understanding of how the definition of manifold is not written in stone, and that when one writes about manifolds one should be precise regarding which of the various topological properties one is assuming (e.g. metrizable; Hausdorff; second countable; paracompact; ...)

Lee Mosher
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It doesn't. The given definition could include non-second countable spaces, like $\mathbb{R}^2$ with the metric $$d((x_1,x_2),(y_1,y_2)) = \begin{cases}|x_2-y_2|& \text{ if } x_1=y_1,\\ 1 & \text{ otherwise}\end{cases},$$ which is homeomorphic to the disjoint union of uncountably many copies of $\mathbb{R}$.

Dan Rust
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In his appendix A, Spivak defines a manifold as a locally euclidean Hausdorff space (i.e. same as in chapter 1 but with "metric" replaced by "Hausdorff") and proves the following theorem:

The following are equivalent for any manifold M:

(a) Each component of M is σ-compact.

(b) Each component of M is second countable.

(c) M is metrizable.

(d) M is paracompact.

Anne Bauval
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