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Euclidean Neighbourhoods are always open sets

  1. Jul 16, 2008 #1
  2. jcsd
  3. Jul 17, 2008 #2


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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.
  4. Jul 17, 2008 #3
    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
  5. Jul 17, 2008 #4
    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.
    Last edited: Jul 17, 2008
  6. Jul 17, 2008 #5

    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


    munkres topology book has the following definition/remarks:
    thank you for your comment slider.
  7. Jul 17, 2008 #6
    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: Apr 23, 2017 at 2:19 PM
  8. Jul 17, 2008 #7
  9. Jul 17, 2008 #8
    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.
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