Question about a definition in "Topology from the differentiable viewpoint" (Milnor)

Hello everyone,

currently I'm reading the book "Topology from the differentiable viewpoint" by John Milnor.

On page one, he defines what a *smooth manifold of dimenson m* is. I already know some other definitions of this, but they are (slightly) different. Also, I'm not sure if I understand his definition.

He writes:

"A subset M $\displaystyle \subset$ IR^k is called a smooth manifold of dimensoin m if each x$\displaystyle \in$M has a neighborhood W$\displaystyle \cap$M that is diffeomorphic to an open subset U of the euclidian space IR^m."

What does he mean when he writes "neighborhood W$\displaystyle \cap$M"?

I know what a neighborhood is, but I don't understand what he means here. Is W a neighborhood of x in IR^k?

If W$\displaystyle \cap$M (as a subset of M) is diffeomorphic to U and U is open, then W$\displaystyle \cap$M is open (in M) and therefore W is an open set in IR^k which contains x. So W would be an *open *neighborhood of x in IR^k.

Can someone give me an (accurate) explanation?

Thanks in advance!

engmaths

Re: Question about a definition in "Topology from the differentiable viewpoint" (Miln

You're right that W is a open neighborhood of x in R^k

Re: Question about a definition in "Topology from the differentiable viewpoint" (Miln

I made a mistake in my first post (seems that I can't edit). If W$\displaystyle \cap$M is open in M, it doesn't mean that W is open in R^k.

Choose R^k = R, M=(-1,1), W=(0,2], then W$\displaystyle \cap$M=(0,1) and therefore open in M, but W isn't open in R.

Re: Question about a definition in "Topology from the differentiable viewpoint" (Miln

right W is a neighborhood of x in R^k, not necessarily open, but that doesn't matter.