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Why not closed?

  1. Jun 12, 2007 #1
    Consider the set {(x,y)|x in Q, y in R}

    This is just a bunch of vertical lines in the R^2 plane. For every point there exists a ball no matter what size which contains another point in this set, namely another point on the same line, dictated by the x value.

    So this set contains all its adherent points. However the answers suggested it is not closed (nor open for that matter which is obvious).

    Why isn't it closed? Haven't I just shown it is closed?
     
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  3. Jun 12, 2007 #2

    Office_Shredder

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    Uhh... what's the definition of closed? I don't think you have the right one (actually,you've got it backwards... a closed set contains all its adherent points, it isn't a set where all its points are adherent points. Look at an (x,y) where x is irrational and see what you can do)
     
  4. Jun 12, 2007 #3

    Hurkyl

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    No it doesn't; it fails for the same reason that the subset Q of R does.
     
  5. Jun 12, 2007 #4
    Closed set is where all its adherent points are in the set. Or A=A closure. Or a set that contains all its adherent points as I said in the OP.

    I never specified 'a set where all its points are adherent points' is a closed set. In fact this statement is true for open sets as well.

    In {(x,y)|x in Q, y in R}, x can't be irrational.


    The set A={Q in R} is not closed because (A closure) is the empty set. But A is not empty. Not open because the interior of A is empty whereas A isn't.

    But with B={(x,y)|x in Q, y in R}, its different because every adherent point of B is in B.
     
  6. Jun 12, 2007 #5

    Office_Shredder

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    Pixova, you just verified that if you take a point in the set, it's an adherent point. I asked, what if you take a point not in the set? For example, (x,y) such that x is irrational. Is that an adherent point?

    And A closure of Q in R isn't empty, it's just R (as for any point in R, you can find a point in Q infinitely close). That may be the problem you're having
     
  7. Jun 12, 2007 #6
    I was trying to say that if any adherent of a set is in the set than that set is closed.

    If for any (x,y) such that x is irrational, I can't see how it is an adherent point of {(x,y)|x in Q, y in R}. If a ball centered around x were to contain an x point that is in Q then I can always shrink that radius of the ball so that it contains no points in Q.
     
  8. Jun 12, 2007 #7

    morphism

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    No you cannot! Between any two reals you can always find a rational.

    So suppose you start with (x,y) and take the ball centered at that point of radius e. Then the horizontal line through (x,y) will contain a copy of the interval (x-e,x+e). This interval contains infinitely many rationals. If you let the radius be e' < e, then the same can be said about (x-e', x+e'), because there is always a rational q such that x-e' < q < x+e'.

    The rationals are dense in the reals, after all, i.e. [itex]\bar{\mathbb{Q}} = \mathbb{R}[/itex].

    By the way, you previously said that the closure of Q was empty. That should have immediately struck you as absurd, because Q is a subset of its closure. So Q's closure at least contains Q.
     
  9. Jun 12, 2007 #8
    Or is the point that between any two irrational points there exist rational pints (in fact infinitely many rational ones). Between two rational points there are infinitely many irrational ones. So the adherent points of the set is R^2.

    So {(x,y)|x in Q, y in R} is actually a dense set. So is Q in R.

    It is neither open nor closed. Not closed because {(x,y)|x in Q, y in R} is not R^2.
     
  10. Jun 12, 2007 #9
    I made my above post before I read Morphisms which pretty much summed up what I had to say. My mistake was in thinking about the natural numbers when I really should have been thinking about the rationals.
     
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