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Open set

  1. Oct 18, 2011 #1
    " Let X be a metric space and p be a point in X and be a positive real number. Use the definition of openness to show that the set U(subset of X) given by:
    U = {x∈X|d(x,p)>r} is open. "

    I have tried:
    U is open if every point of U be an interior point of U. x is an interior point of U if there is an open ball B(x, r) is a subset of U. Let y belongs to B(x, r) then d(x, y)< r, and p doesn’t belong to B(x, r) since d(p, x) > r. If we can show that d(y, p) > r (y is in X) then since y is a arbitrary point of B(x, r) it means that the open ball is a subset of U.
    But I don’t know how to show d(y, p) > r ?
    Please help me.
  2. jcsd
  3. Oct 18, 2011 #2
    What you need to do is take an arbitrary x in U. This x will have as property that d(x,p)>r.

    Now you need to find an [itex]\varepsilon>0[/itex] suitably such that

    [tex]B(x,\varepsilon)\subseteq U[/tex]

    So for each y with [itex]d(x,y)<\varepsilon[/itex] must hold that d(y,p)>r.

    Now, draw a picture first. What must [itex]\varepsilon[/itex] be??
  4. Oct 18, 2011 #3
    d(x, p) + d(p, y)> d(x, y)
    d(x, y)< 2r < ε
    ε = 2r
    Is it correct?
  5. Oct 19, 2011 #4
    No, and I don't see what you did actually...
  6. Oct 19, 2011 #5
    I use triangle inequality, so:

    d(x, p) + d(p, y)> d(x, y)
  7. Oct 19, 2011 #6
    What you know is that [itex]d(x,y)<\varepsilon[/itex]. What you must prove is that d(y,p)>r. I don't see how your argument solves that.
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