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Every infinite set has a countable dense subset?

  1. Sep 14, 2008 #1
    1. The problem statement, all variables and given/known data

    Prove that every infinite set has a countable dense subset.

    2. Relevant equations



    3. The attempt at a solution

    I have almost no idea how to solve this problem of my analysis homework. I was thinking that i need to show that there is a countable subset that has all points of the initial set as limit points. So i was thinking of saying that that any open infinite set can be written as a countable union of subsets or like neighbourhoods, and using something like a pigeon hold principle to say that each of these neighbourhoods must contain some points in the countable subset. but we have countable nbhds and countable points, i dont think the pigeonhole principle works here... I'm probably way off the solution... ANY HELP?
     
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  3. Sep 14, 2008 #2

    Dick

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    Not every metric space has a countable dense subset. Google the "long line". Only separable spaces have one. If it's a subset of the reals then you are ok.
     
  4. Sep 14, 2008 #3
    ok, since this is real analysis, i think i am mostly interested in the real line. so how do i go about solving it for the reals?
     
  5. Sep 14, 2008 #4
    well we know that the rationals are countable and dense in the reals. but umm..i donno, at this point i am lost about how to prove this.
     
  6. Sep 15, 2008 #5

    Dick

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    That's a good start. Think of all of the intervals (q-1/n,q+1/n) for all q rational and fixed n. That's a countable set of intervals, right? Pick one x in your infinite set in each interval (if there is one). That's a countable set of x's, correct? Now union all of those countable sets over n. Isn't that dense? There's more economical ways to do this, but who cares?
     
  7. Sep 17, 2008 #6
    i think i get what u're saying.. so for every q rational for a fixed n we choose and x in the set that belongs to (q-1/n ; q+1/n). Giving us a set of countable x's, then we do this for all n, which is a countable number of countable sets, thus countable.

    But i am having a little trouble seein that this is dense. To show this we need that any point in the infinte set is either in the union of our new sets or a limit point.
    To show that any point will be a limit point we need that there is a point within 1/n of it for every n. Certainly there is a rational number 1/n close to it, but are there x's in that range too?

    Oh wait a minute...i think i missunderstood the phrase you said "(if there is one)" .. There is definitely one, since there are infinite points and only countable intervals (pigeonhole). (unless by infinite we mean countable, then we definitely have a countable dense subset, since the set is dense in itself). Unless (q - 1/n; q+1/n) is totally disjoint from the set. I just need some convincing that every point is now a limit point
     
    Last edited: Sep 17, 2008
  8. Sep 17, 2008 #7

    Dick

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    Any given interval may miss hitting your set. That's not a problem. Call the countable subset of X that we have defined A. Now pick any point x in X. x is either isolated or it's a limit point of X. If it's isolated then you can find one of the rational intervals that contains only x. By the way we have constructed A, that means x must be in A. Right? Now suppose it's a limit point. Pick any epsilon. Then there is a y in X at a distance less than epsilon from x. Find a rational interval that contains y and fits inside of the epsilon interval. By the construction of A, either y is in A or there is a z in A in the rational interval. In either case there is a point from A within epsilon of x.
     
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