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Prove the boundary of rationals is real

  1. Aug 13, 2007 #1
    1. The problem statement, all variables and given/known data

    Let [tex]Q[/tex] be the set of all rational numbers
    Prove [tex]bd(Q)=R[/tex]

    2. Relevant equations

    3. The attempt at a solution
    Let x be a real number, then since the interval |x-r| contains both rationals and irrationals for arbitrary small r, so R is the boundary of Q. Is that right?
  2. jcsd
  3. Aug 13, 2007 #2


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    Slow down a bit. What, exactly, is the definition of "boundary" and "boundary point". If you can answer that you should be able to answer your own question. (I might point out that, strictly speaking, |x-r| is not an interval, it is a number! You mean either the interval (x-r, x+r) or the set {y| |x-y|< r}.)
  4. Aug 13, 2007 #3


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    I would assume that
    • The boundary is the closure minus the interior
    • A point is in the closure if every epsilon-ball around it has non-empty intersection with the set
    • A point is in the interior if there is an epsilon-ball which fits entirely in the set

    From these definitions it is quite straightforward to prove what the boundary of |R is, given (or proven) that the rationals are dense.

    So kekido, your idea is right, but if you want to do it rigorously you should put a little more work into it (show that it satisfies the definitions yadayada andsoforth).
  5. Aug 13, 2007 #4
    Ok, here "boundary" is the set of all boundary points, i.e., [tex]\partial(Q)=R[/tex]

    You're right, the interval should be like what you said. I was being sloppy here.

    Since the boundary point is defined as for every neighbourhood of the point, it contains both points in S and [tex]S^c[/tex], so here every small interval of an arbitrary real number contains both rationals and irrationals, so [tex]\partial(Q)=R[/tex] and also [tex]\partial(Q^c)=R[/tex]
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