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Homework Help: Is this set really not closed ? ?

  1. Feb 7, 2006 #1

    quasar987

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    Is this set really not closed ?!!?!

    Ok, consider the metric space [itex]\mathbb{R}^2[/itex] armed with the pythagorean metric [itex]d(x,y) = ||x-y||[/itex] and B, the closed ball of radius 1 centered on the origin: [itex]B = B_1(0) \cup \partial B_1(0)[/itex]. Now construct the metric space composed of this closed ball and the appropriate restriction on the pythagorean metric.

    Now consider D, a closed ball centered on the origin but of radius r<1, and ask the question: is D closed in B? D is closed iff [itex]D^c = B \slash D[/itex] is open in B. But it is not, for consider a point on the border of B ([itex]x_0 \in \partial B_1(0)[/itex]). No ball with x_0 as center is contained entirely in B. Hence D is not closed in B.

    Is that correct?!
     
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  3. Feb 7, 2006 #2

    AKG

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    No, it's not correct; D is closed in B. If B is your metric space, then the ball of radius s centered at x0 is defined as:

    [tex]\{x\ \mathbf{\in B}\, :\, ||x - x_0|| < s\}[/tex]

    They key part is the "[itex]\in B[/itex]" part. So a ball in B is the intersection of the ball in R² with the space B itself.
     
  4. Feb 8, 2006 #3

    quasar987

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    Thx AKG.

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  5. Feb 9, 2006 #4

    quasar987

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    But then how would any set NOT be open?

    G is open if for all x in G, there exists r>0 s.t. B(r,x) is in G.

    But the definition of B(r,x) is all g in G such that d(x,g)<r. So for any r, B contains only elements of G, so B(r,x) is necessarily in G, making G open.

    what the heck?!

    I would appreciate help quickly. I leave for exam in 13 minutes exactly. *sweat*
     
    Last edited: Feb 9, 2006
  6. Feb 9, 2006 #5

    quasar987

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    At least there were no questions on open sets. phew.
     
  7. Feb 9, 2006 #6

    Hurkyl

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    No: if X is your metric space, then B(r,x) is all g in X such that...
     
  8. Feb 9, 2006 #7

    matt grime

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    very simply, in a metric topology. The set of things distance less than or equal to 1 from some given point would not be open in general.
     
  9. Feb 9, 2006 #8

    quasar987

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    Oh, of course!
     
  10. Feb 10, 2006 #9

    HallsofIvy

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    If G is the entire set, (as you say "the definition of B(r,x) is all g in G such that d(x,g)<r") then yes, it is open. The entire base set in a topological space is always both open and closed.

    If you mean that G is a proper subset of some set X on which the topology is defined then "the definition of B(r,x) is all g in G such that d(x,g)<r." is not true. B(r,x) is all g in X such that d(x,g)<r.
     
  11. Feb 10, 2006 #10

    quasar987

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    Exactly.

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