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Definition 1

A surface of dimension k (or k-dimensional surface or k-dimensional manifold) in $\mathbb R^n$ is a subset $S\subset \mathbb R^n$ each point of which has a neighborhood in $S$ homeomorphic to $\mathbb R^k$.

Definition 2

A set $A(S) := \{\phi : I_k \to U_i, i \in\mathbb N\} $of local charts of a surface $S $ whose domains of action together cover the entire surface (that is, $S = \bigcup U_i$), is called an atlas of the surface $ S$.

Since $A(S)$ is countable, but I wanna know if every surface $S$ has an atlas? If not, could you give a counter example?

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  • $\begingroup$ This is not a right definition of an atlas. What book are you reading? $\endgroup$ Jun 30 at 4:25
  • $\begingroup$ Mathematical Analysis by Vladimir A.Zorich $\endgroup$
    – LEY
    Jun 30 at 4:26
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    $\begingroup$ What is $I_k$? What are the $U_i$? What definition of a surface do you have in mind? A 2-dimensional manifold admits an atlas by definition… $\endgroup$
    – PrudiiArca
    Jun 30 at 4:27
  • $\begingroup$ Then take another look at Zorich's definition. $\endgroup$ Jun 30 at 4:28
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The trick is that any subspace of $\mathbb{R}^n$ is second-countable and hence Lindelöf: any open cover has a countable subcover. So, since your $S$ according to Definition 1 has an open cover by sets homeomorphic to $\mathbb{R}^k$, there is a countable subcover which gives an atlas according to Definition 2.

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