# Euclidean Neighbourhoods are always open sets

1. Jul 16, 2008

### lantern

2. Jul 17, 2008

### HallsofIvy

Staff Emeritus
What exactly is your difficulty? A "Euclidean neighborhood" of a manifold is defined, in the first site you give, as a subset of the mainifold that has a continuous one-to-one map to a open set in Euclidean space. The "invariance of domain", the second site you reference, says that such a set must be open. I don't see anything left to prove, unless you want to prove "invariance of domain" itself, which is very hard.

3. Jul 17, 2008

### lantern

Sorry this is a new subject to me so please bear with me.

I'm having difficulty with two things

First, we are talking about neighborhoods of arbitrary topological spaces. 'Invariance of domain' is a theorem about R^n. How does this theorem, which is restricted to R^n say anything about sets in topological spaces other than R^n? In particular, I don't see how it says that our original Euclidean neighborhood must be open, since it is a set in an arbitrary topological space

Second, I'm having some trouble with the statement 'By definition, every point of a locally Euclidean space has a neighborhood homeomorphic to an open subset of R^n'

By definition, a neighborhood of a point p in a topological space is a set (not necessarily open) containing an open set that contains p

By definition, a topological space X is called locally Euclidean if there is a non-negative integer n such that every point in X has a neighborhood which is homeomorphic to Euclidean space R^n (not a subset of R^n)

So how does it follow that 'By definition, every point of a locally Euclidean space has a neighborhood homeomorphic to an open subset of R^n'? Is the word 'neighbourhood' used differently in this instance, or am I being stupid as usual?

I should say that this claim is 'left to the reader' in a text book I am reading and that Euclidean neighbourhoods are defined in the text as neighbourhoods homeomorphic to R^n for some non-negative integer n

Last edited: Jul 17, 2008
4. Jul 17, 2008

### slider142

This is not true. A subset N of a topological space M, with basis B, is a neighborhood of a point $p\in M$ iff $N\in B$ and $p\in N$. A set O is open in M iff $\forall x\in O, \exists N\in B$ such that $x\in N$ and $N\subseteq O$. The neighborhoods of a topological space determine the open sets; not the other way around. N need not contain any open sets other than itself, and as a corollary, N is necessarily open.

Last edited: Jul 17, 2008
5. Jul 17, 2008

### lantern

hmmmm

well what i wrote down is the definition of neighbourhood given in the book im trying to read. and it is also the definition given on this page

http://en.wikipedia.org/wiki/Neighbourhood_(mathematics)

munkres topology book has the following definition/remarks:
thank you for your comment slider.

6. Jul 17, 2008

### slider142

Interesting. The books I have start at neighborhood and define open sets from that. Yours starts at open sets and defines neighborhoods. The definition I know is also given at http://planetmath.org/encyclopedia/Neighborhood.html [Broken]. I didn't know it was this ambiguous! :D

Last edited by a moderator: May 3, 2017
7. Jul 17, 2008

### lantern

8. Jul 17, 2008

### slider142

The general Invariance of domain actually talks about n-dimensional manifolds, not Rn directly. This is also mentioned in the Wiki article.
Open subsets of Rn are topological n-manifolds, so if an n-manifold is homeomorphic to Rn, then since homeomorphism yields an equivalence relation between manifolds, we have the statement above.