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Regard Q, the set of all rational numbers, as a metric space,

  1. Oct 25, 2011 #1
    1. The problem statement, all variables and given/known data

    Regard Q, the set of all rational numbers, as a metric space, with d(p, q) = |p − q|. Let E
    be the set of all p ∈ Q such that 2 < p2 < 3. Show that E is closed and bounded in Q, but that E is not compact. Is E open in Q?

    2. Relevant equations


    Definition of interior point, limit point, open set, closed set, and compact set.


    3. The attempt at a solution

    Obviously E is bounded, but I'm not convinced that E is closed. Take a point in Ec. Supposedly there is a neighborhood, about that point, containing only members of E. But that can't be true, since any neighborhood about that point will also contain irrational numbers (Right???).
     
  2. jcsd
  3. Oct 25, 2011 #2

    HallsofIvy

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    NO. This is in the space of rational numbers. There are NO irrational numbers in that space. For the purposes of this problem, irrational numbers do not exist.

    Notice that since there is no rational number whose square is 2 or 3, "for the purposes of this problem", again, the set of all x such that "[itex]2< x^2< 3[/itex]" is exactly the same the set of all x such that "[itex]2\le x^2\le 3[/itex]".
     
  4. Oct 26, 2011 #3
    I suspected that might be the case. Now take a look, if you don't mind, at the last paragraph of the solution to this problem, found here: http://www.math.ucla.edu/~elewis/Math230PDFs/Math%20230a%20HW%203%20Extra%20Credit.pdf [Broken]. It seems flawed to me, since they are putting using the given metric |p - q| and plugging irrational numbers into it, and it's only defined for rational numbers. nomsayin'?
     
    Last edited by a moderator: May 5, 2017
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