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Metrics and open balls

  1. Feb 19, 2008 #1
    1. The problem statement, all variables and given/known data

    Consider [tex]\Re[/tex] with metric [tex]\rho[/tex] (x,y) = |x-y|. Verify for all x [tex]\in[/tex] [tex]\Re[/tex] and for any [tex]\epsilon[/tex] > 0, (x-[tex]\epsilon[/tex], x+[tex]\epsilon[/tex]) is an open neighborhood for x.

    2. Relevant equations

    Neighborhood/Ball of p is a set Nr(p) consisting of all q s.t. d(p,q)<r for some r>0.

    3. The attempt at a solution

    Take [tex]\alpha[/tex] > 0, [tex]\alpha[/tex] < [tex]\epsilon[/tex]. Take [tex]\rho[/tex](x, x-[tex]\alpha[/tex]) = |x-(x- [tex]\alpha[/tex] )| = [tex]\alpha[/tex] < [tex]\epsilon[/tex].
    Take [tex]\rho[/tex](x, x+[tex]\alpha[/tex]) = |x-(x+[tex]\alpha[/tex])| = [tex]\alpha[/tex] < [tex]\epsilon[/tex].
    Therefore, any positive [tex]\alpha[/tex] < [tex]\epsilon[/tex] is in N[tex]\epsilon[/tex](x).

    I initially misplaced this thread, and was told that this shows that [itex](x-\epsilon, x+\epsilon)[/itex] is a neighborhood but it was not an "open neighborhood." But there is a theorem (2.19 in Rudin) that says: "Every neighborhood is an open set." I must be misinterpreting this theorem then?

    (x - [tex]\epsilon[/tex]) is a limit point of the set, but (x - [tex]\epsilon[/tex]) [tex]\notin[/tex] N[tex]\epsilon[/tex] (x), and (x + [tex]\epsilon[/tex]) is a limit point of the set, but (x + [tex]\epsilon[/tex]) [tex]\notin[/tex] N[tex]\epsilon[/tex] (x). Therefore the set is open.

    Will this complete the proof?
  2. jcsd
  3. Feb 19, 2008 #2
    don't see what there is to prove here.
    one definition of open set in R is that it contains an open ball, well obviously this interval is an open ball is it not?
  4. Feb 19, 2008 #3


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    Homework Helper

    Different authors will sometimes use different definitions for particular terms. That said, loop quantum gravity has pointed out that the set you're dealing with is a basis element, thus clearly open.
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