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Question on homeomorfism

  1. Sep 7, 2006 #1
    Hello guys. I need to prove the following:

    Let [itex]X=(\mathbb{R}\times \{0\})\cup(\mathbb{R}\times\{1\})[/itex] and [itex](x,0)\sim (x,1)[/itex] when [itex]x \neq 0[/itex]. Prove that [itex]L:=X/\sim[/itex] is a topological space locally homeomorphic to [itex]\mathbb{R}[/itex], but is not Hausdorff.

    In order to prove that [itex]L[/itex] is homeomorphic to [itex]\mathbb{R}[/itex], all I need to do is show a continuous function [itex]f:L\longrightarrow \mathbb{R}[/itex] such that [itex]f[/itex] is invertible and [itex]f^{-1}[/itex] is also continuous, right?

    I am new at this, so I am a bit confused on the Hausdorff part. A topological space is not Hausdorff if there is a pair of distinct points [itex]x,\,y[/itex] such that there are open sets [itex]U[/itex] and [itex]V[/itex] so that [itex]x\in U[/itex] and [itex]y\in V[/itex], but [itex]U\cap V \neq \emptyset[/itex], right?

    If what I stated above is true, then I need to find two open sets, one containing the point [itex](0,0)[/itex] and the other containing [itex](0,1)[/itex], such that their intersection is not empty? Will that be sufficient?

    Thx for the help and sorry for my english.
    Last edited: Sep 7, 2006
  2. jcsd
  3. Sep 7, 2006 #2


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    You don't need to prove L homeomorphic to R, indeed you can't. You need to prove that it is locally homeomorphic to R. What is the definition of "locally homeomorphic"?
    No, and the underlined part is wrong. You need to find distinct points x and y such that for every pair of open sets U and V containing x and y respectively, U and V have non-empty intersection.
  4. Sep 13, 2006 #3
    Sorry for the late reply AKG... thx anyway for the help...

    On the Hausdorff part, I think I got it.

    Any point in the upper line is related to any point in the lower line, except the origins. So, every open neigborhood of [itex](0,1)[/itex] will intersect with an open neighborhood containing the point [itex](0,0)[/itex]. Is this correct?

    On the locally homeomorphic part... If I topologize my space with a basis of open subsets [itex]V \subset X[/itex] consisting of the following:

    If [itex](0,1) \notin V[/itex], then [itex]V[/itex] is open as a subset of [itex]\mathbb{R}\times\{0\}[/itex].
    If [itex](0,1) \in V[/itex], then there is an open neighborhood [itex]W \subset (\mathbb{R}\times\{0\})[/itex] of [itex](0,0)[/itex] such that [itex]V=(W-\{(0,0)\})\cup\{(0,1)\}[/itex].

    Then, for every point [itex]p\in X[/itex], there is an open neighborhood [itex]N \subseteq V[/itex], such that there exist a homeomorphism [itex]f:N\longrightarrow U \subset \mathbb{R}[/itex].

    Such homeomorphism could be the projection function?

    What dou you think? Is this correct?
    Last edited: Sep 13, 2006
  5. Sep 14, 2006 #4


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    What is the topology for L? You can't say a space is Hausdorff unless you know its topology. How do you know that {(0,0)} isn't an open set? It isn't, don't worry, but how do you know?

    What is the definition of locally homeomorphic? Prior to this, I myself had never even seen anyone talk about a local homeomorphism. I just looked it up in my textbook, and if you're using the same definition as my book's, then whatever it is you're doing doesn't show local homeomorphism.
    Last edited: Sep 14, 2006
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