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Closures on ordered sets

  1. Sep 13, 2010 #1

    radou

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    1. The problem statement, all variables and given/known data

    OK, this question may sound somewhat trivial, but I'd still like to check my answer. It's from Munkres.

    Let X be an ordered set with the order topology. One needs to prove that Cl(<a, b>) is a subset of [a, b] and investigate the conditions necessary for equality.

    3. The attempt at a solution

    So, if x is any element in <a, b> it is obviously in [a, b], too. If it happens that either a or b (or both) are in Cl(<a, b>), they are in [a, b] too, so Cl(<a, b>) is a subset of [a, b].

    Now, if x is any element of [a, b] different from a and b, obviously x is in <a, b>, too, which is a subset of Cl(<a, b>), and hence x is in Cl(<a, b>). Now, the interesting cases are a and b. Is a in Cl(<a, b>)? Well, if so, then every neighborhood of a intersects <a, b>. Let X be a finite ordered set. Is a is the least element of X, then every interval of the form [a, b> intersects <a, b>, and it's obviously the only type of neighborhood that contains a. But if a is not the least element of X, then <a1, a, a2> (where a1 < a and a2 is the first element such that a2 > a) is a neighborhood containing a which doesn't intersect <a, b>. So, a can't be in Cl(<a, b>) if X is finite, unless a is the least element of X. An analogous proof holds for b.

    Now, if my reasoning above is correct, it's natural to ask the question: are a and/or b in Cl(<a, b>) if X is infinite? Clearly, motivated by the arguments above, one can reason that between every element which is the first one greater than a and a itself there should be another element, and so on, so we could chose an open neighborhood of a which intersects <a, b>. So X should be infinite. But does every infinite ordered set satisfy this condition?

    I'm a bit confused on this one, thanks for any help.
     
  2. jcsd
  3. Sep 13, 2010 #2

    radou

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    Oops, it just occured to me that, if X has a smallest element a, {a} is a neighborhood of a which doesn't intersect <a, b> (since the smallest half-interval which contains a is {a}). So, actually, if X is finite, a and b can not belong to Cl(<a, b>). Still, my last question remains open.
     
  4. Sep 13, 2010 #3
    The inclusion can be strict - consider the least uncountable well-ordered set.
     
  5. Sep 13, 2010 #4

    radou

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    Of course, if we consider N with the order topology, then every interval <a, b> contains at least one element. Clearly, if it contains one element, i.e. if it is of the form <n-1, n+1>, where n is a natural number greater than 2, then <n-2, n> and <n, n+2> are neighborhoods of n-2 and n+2 which contain n-1 and n+1, respectively, and don't intersect <n-1, n+1>. Hence, neither n-1 nor n+1 are contained in the clusure of <n-1, n+1>.

    But still, the question about the other inclusion remains. Is there a general rule for an ordered set X so that [a, b] = Cl(<a, b>) for any a, b in X?
     
  6. Sep 13, 2010 #5

    radou

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    Here's an example where [a, b] is a subset of Cl(<a, b>): if <a, b> is a dense subset of X, then every open set in X intersects <a, b>. Hence, for every x in [a, b], and for any neighborhood Ux of x, Ux intersects <a, b>, so [a, b] is contained in Cl(<a, b>).
     
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