Euclidean Neighbourhoods are always open sets

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  • #2
HallsofIvy
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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
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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
 
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  • #4
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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
This is not true. A subset N of a topological space M, with basis B, is a neighborhood of a point [itex]p\in M[/itex] iff [itex]N\in B[/itex] and [itex]p\in N[/itex]. A set O is open in M iff [itex]\forall x\in O, \exists N\in B[/itex] such that [itex]x\in N[/itex] and [itex]N\subseteq O[/itex]. 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.
 
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  • #5
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This is not true. A subset N of a topological space M, with basis B, is a neighborhood of a point [itex]p\in M[/itex] iff [itex]N\in B[/itex] and [itex]p\in N[/itex]. A set O is open in M iff [itex]\forall x\in O, \exists N\in B[/itex] such that [itex]x\in N[/itex] and [itex]N\subseteq O[/itex]. 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.
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:
munkres said:
[Mathematicians often] shorten the statement "U is an open set containing x" to the phrase "U is a neighborhood of x". (...) Some mathematicians use the term "neighborhood" differently. They say that A is a neighborhood of x if A merely contains an open set containing x.
thank you for your comment slider.
 
  • #6
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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
 
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  • #8
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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
The general Invariance of domain actually talks about n-dimensional manifolds, not Rn directly. This is also mentioned in the Wiki article.
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'?
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.
 

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