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Topology - How do they know this is open?

  1. Jun 24, 2007 #1

    quasar987

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    1. The problem statement, all variables and given/known data
    The lemma sets out to show that if A in R^n is compact and x_0 is in R^m, then A x {x_0} is compact in R^n x R^m.

    They say, "Let [itex]\mathcal{U}[/itex] be an open cover of A x {x_0} and

    [tex]\mathcal{V}=\{V\subset \mathbb{R}^n:V=\{y:(y,x_0)\in U\}, \ \mbox{for some} \ U\in \mathcal{U}\}[/tex]

    Then [itex]\mathcal{V}[/itex] is an open cover of A."

    How do they know that given some U in [itex]\mathcal{U}[/itex], the associated V is open???

    Edit: In fact, consider the following counter example: Let n=m=1, A=[-½,½], x_0=0. Then A x {x_0} is just the segment [-½,½] considered in the R² plane. Let [itex]\mathcal{U}=\{B_n(0,0)\}_{n\in\mathbb{N}}[/itex] (the collection of open balls centered on the origin of radius n). Then, let V_n be the set V associated with B_n(0,0) as described above, i.e. [itex]V_n=\{y\in\mathbb{R}:(y,0)\in B_n(0,0)\}[/itex]=[-½,½], a set that is not open in R. ah!
     
    Last edited: Jun 24, 2007
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  3. Jun 24, 2007 #2

    morphism

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    V_n = (-n, n)

    V_n [itex]\cap[/itex] A = [-1/2, 1/2] (for n>0)
     
  4. Jun 24, 2007 #3

    quasar987

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    Hm, yes. ":frown:" I wish I were smarter than the book.

    Meanwhile I think I found the answer. Given an U and it's associated V, for any point y in V, there is an open set O_y in R^n and another one O'_y in R^m such that (y,x_0) is in O_y x O'_y and O_y x O'_y [itex]\subset[/itex] U. And actually, we can write

    [tex]V=\bigcup_{y\in V}O_y[/tex]

    which is open as an arbitrary union of opens.
     
  5. Jun 25, 2007 #4

    matt grime

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    Why are you introducing O_y's? If curly(U) is some cover of A x {x}, then projecting into the first coordinate gives an open cover of A, take a finite subcover, and pull back.
     
  6. Jun 25, 2007 #5

    quasar987

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    I'm introducing the O_y because otherwise, I was not convinced that the projection was open.

    After all, as far as I understand, if U is open in a topological space X x Y, it does not mean that it can be written as the product of an open of X with an open of Y.
     
    Last edited: Jun 25, 2007
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