# Does every manifold in $\mathbb{R}^n$ have a countable altas?

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?

• This is not a right definition of an atlas. What book are you reading? Jun 30 at 4:25
• Mathematical Analysis by Vladimir A.Zorich
– LEY
Jun 30 at 4:26
• 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… Jun 30 at 4:27
• Then take another look at Zorich's definition. Jun 30 at 4:28

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.