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Boundary and homeomorphism

  1. Aug 4, 2011 #1

    aleazk

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    Hi, I need to know if the following statement is false or true. Given two topological spaces, X and Y, and an homeomorphism, F, between them, if bA is the boundary of the subset A of X, this implies that F(bA) is the boundary of the subset F(A) of Y?
     
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  3. Aug 4, 2011 #2

    micromass

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    What did you try already?? What is the definition of the boundary??
     
  4. Aug 4, 2011 #3

    aleazk

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    sorry if this is a silly question, I'm new in topology. bA=CA-IA, where CA is the closure and IA is the interior.
     
  5. Aug 4, 2011 #4

    micromass

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    Indeed, so what happens if you take

    [tex]f(CA-IA)[/tex]

    can you show that this equals [itex]Cf(A)-If(A)[/itex]??

    Simplified, can you show that

    [tex]Cf(A)=f(CA)~\text{and}~If(A)=f(IA)[/tex]
     
  6. Aug 5, 2011 #5

    HallsofIvy

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    That is equivalent to saying that point p is in the boundary of A if and only if any open set containing p contains points in A and points not in A. And, of course, homeomorphisms map open sets to open sets.
     
  7. Aug 8, 2011 #6

    aleazk

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    So, using the property AcB then F[A]cF of maps, the open sets that contain F(p) will also contain points inside and outside of F[A], and then F(p) is in the boundary of F[A] if p is in the boundary of A?
     
  8. Aug 8, 2011 #7

    micromass

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    Indeed!! You may want to rigorize that however. For example, why will open sets containing F(p) also contain points in and out F(A)?
     
  9. Aug 8, 2011 #8

    aleazk

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    Hi, I suppose yes. int A is the largest open set contained in A, so int A c A and then F[int A] c F[A]. If F is an homeomorphism, then F^-1 is continuous, which implies that F[int A] is open. Then, because the last relation, F[int A] c F[A], F[int A] c int F[A]. Now, int F[A] c F[A], so F^-1[int F[A]] c A. Because F is continuous, then F^-1[int F[A]] is open. Using the last relation, F^-1[int F[A]] c A, it follows that F^-1[int F[A]] c int A, which is equivalent to int F[A] c F[int A]. Thus, using the previous result, F[int A] c int F[A], it follows that F[int A] = int F[A]. CA is the smallest closed set containing A, so F[A] c CF[A] and then A c F^-1[CF[A]]. F^-1[CF[A]] is closed because CF[A] is closed and F is continuous. Then CA c F^-1[CF[A]], which is equivalent to F[CA] c CF[A]. Using similar arguments, but invoking the continuity of F^-1, it follows that CF[A] c F[CA]. So, F[CA] = CF[A].
     
  10. Aug 8, 2011 #9

    micromass

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    Seems ok! :smile: Nicely done!!
     
  11. Aug 8, 2011 #10

    aleazk

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    Thanks. I needed the result to convince myself of a claim that I read below proposition 4.5.1 of Hawking and Ellis. The proposition says that in a convex normal neighborhood Np of point p, the points in Np that can be reached by timelike curves diverging from p are those of the form Q=exp\p(V), where V is timelike. But then, below the proposition, he says: in other words, the null geodesics that diverge from p form the boundary of the region in Np that can be reached by timelike curves diverging from p. :confused: why?
    The zone exp\p(V), where V is timelike, is generated by the timelike geodesics that diverge from p, and the zone exp\p(V), where V is null, is generated by the null geodesics that diverge from p, all this by definition of the exponential map at p. In the tangent space to p, Tp, the null vectors form the boundary of the zone where the timelike vectors lie. Thus, because exp\p is a diffeomorphism (and then an homeomorphism) at Np, using the result we discused in this tread, the null geodesics that diverge from p really form the boundary of the zone generated by the timelike geodesics that diverge from p. At least I think so :redface:
     
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