# Finishing a Supposedly Simple Proof

1. Sep 29, 2013

### Mandelbroth

I started to reread Spivak's A Comprehensive Introduction to Differential Geometry last night, for the sake of attempting to improve my ability to do differential geometry. I noticed that I had skipped what Spivak calls an "easy exercise" after he introduces Invariance of Domain.

Embarrassingly, I did not find it quite as easy as I'd have hoped. Flustered, I tried looking on the internet. I found various proofs of the statement that "the neighborhood U in our definition [of a manifold] must be open." I found one that was particularly easy to follow (All the proofs I saw certainly weren't what I'd call "easy exercises."), but I didn't understand the last step. The proof-writer explains his or her last step as "We have proved that any point x in U has an open neighborhood W contained in U, therefore U is open."

Can someone explain why this is true? I'm probably just missing something really simple here.

Additionally, would you consider the proof that "the neighborhood U in our definition [of a manifold] must be open" to be trivial? I certainly don't see the "easy exercise" Spivak must have envisioned.

Thank you.

Last edited: Sep 29, 2013
2. Sep 29, 2013

### UltrafastPED

If the set is closed there will be points (on the boundary) which are not in open sets ... because they don't have any neighbors beyond the boundary.

Think in terms of topology here.

3. Sep 29, 2013

### Mandelbroth

I...don't follow how this shows $U$ is open. :uhh:

4. Sep 29, 2013

### UltrafastPED

Perhaps I did not understand your specific question. Please quote the actual exercise, including the definitions which have been provided.

5. Sep 29, 2013

### Mandelbroth

The quoted part of the text is attached.

I consider a manifold to be a second countable Hausdorff space such that every point in the space has a neighborhood homeomorphic to $\mathbb{R}^n$.

I'm trying to understand a proof by someone else, who finishes their proof with the phrase "We have proved that any point x in U has an open neighborhood W contained in U, therefore U is open."

Edit: Never mind. I figured it out. I read it as "a point x" rather than "any point x."

#### Attached Files:

• ###### spivak1.JPG
File size:
80.1 KB
Views:
119
6. Sep 29, 2013

### UltrafastPED

Glad you were able to figure it out!