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Question on cardinality of sequences.

  1. Jan 6, 2007 #1
    i need to show that there exists a class of sets A which is a subset of P(Q) such that it satisfies:
    1) |A|=c (c is the cardinality of the reals)
    2) for every A1,A2 which are different their intersection is finite (or empty).

    basically i think that i need to use something else iv'e proven: which for every a in R, let s(a) in Q^N is an increasing sequence which converges to a, then |{s(a)|a in R}|=c

    i think that such a class could be: A={P(A')||A'|<alephnull} where A' is a subset of Q, A is the union of all P(A') where A' is Q.
    im not sure if A's cardinality is c, but other than this example i dont see how to show it.
    i think it's related to what i typed in the second paragraph, perhaps i need to find a subset to {s(a)|a in R} which is still uncountable, but i dont see how.
     
  2. jcsd
  3. Jan 6, 2007 #2

    matt grime

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    The question states 'show there exists', right? That suggests to me that the first thing you shuold do is forget about constructing one explicitly.
     
  4. Jan 6, 2007 #3
    so how should i prove it?
    i mean do i have to assume that A is a subset of P(Q) and then prove that |A|=c and for every 2 sets in A, their intersection is finite or empty.
    then one way to do so is, bacause A is a subset of P(Q) then |A|<=c
    but we also have c<=|A| cause |A|=c... well im stuck.
    i dont think i can prove it without showing explicitly.
     
  5. Jan 6, 2007 #4

    matt grime

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    IMPORTANT: I haven't proved this myself, so do what you will with this advice.

    ARGUMENT: how do you demonstrate that there are transcendental numbers? Cardinality of real numbers=c. Cardinality of Algebraic numbers=/=c. Hence there are transcendental numbers.

    ANALOGY: take the set of all possible As with the finite intersection property. Count them. Count those where |A|=/=c. Are they the same? I suspect the answer is no.
     
  6. Jan 7, 2007 #5
    what i did so far is:
    A is a subset of P(Q), so i defined A as the union of the sets {Q_n} where Q_n are finite subsets of Q, which have cardinality |Q_n|=n, i think that bacuase there are c sets in P(Q) A equals alephnull*c=c so its cardinality is c.
    and we have that every two sets which are in A are either their intersection is finite or empty, is this good definition?
     
  7. Jan 7, 2007 #6

    matt grime

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    There are countalby many Q_n for each n. Thus A is a countable union of countable sets, hence countable.

    the fact that |P(Q)|=c doesn't have anything effect.
     
  8. Jan 7, 2007 #7

    AKG

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    There's a simple constructive solution. HINT: think about decimal expansions.
     
  9. Jan 7, 2007 #8

    Hurkyl

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    Incidentally, the "finite power set" of an infinite set is the same cardinality as that set:

    [tex]
    |S| = | \{ \, A \subseteq S \, | \, |A| < |\mathbb{N}| \, \} |
    [/tex]
     
  10. Jan 7, 2007 #9
    my mistake, perhaps this appraoch will do:
    iv'e defined Q_n to be finite subsets of Q, and A={Q_n|Q_n subset of Q} such that |Q_k|=k for every k nonnegative integer.
    |A|=|UQ_n|=c
    Q_n a subset of Q
    and every two elements of A their intersection is finite or empty.
    will that be enough?
     
  11. Jan 7, 2007 #10

    matt grime

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    In what way is that different from what you jsut did? Why is A uncountable? It is a countable union of countable sets.
     
  12. Jan 7, 2007 #11

    AKG

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    To add to my hint, think about decimal expansions of all real numbers. And think about how R is constructed as the completion of Q.
     
  13. Jan 7, 2007 #12

    HallsofIvy

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    Second this but note that there are several ways to construct R from Q. Since AKG uses the phrase "completion of Q", I assume he means defining R as equivalence classes of the set of Cauchy sequences of rational numbers.
    One could also use the definition in terms of equivalence classes of the set of non-decreasing sequences of rational numbers with upper bound but not the definition as Dedeking Cuts.
     
  14. Jan 8, 2007 #13
    the decimal expansion of all real numbers is infinite, but the decimal expansion of Q can also be finite.
    perhaps i can construct A to be a family of sets of the finite decimal expansions of a particular real number, for example pi would have the set {3,3.1,3.14,...} and so on.
    but i dont see how the intersection of two different sets in A WOULD BE FINITE OR EMPTY.
     
  15. Jan 8, 2007 #14

    Hurkyl

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    Well, what happens if two of these sequences agree infinitely often?
     
  16. Jan 8, 2007 #15
    then they are the same.
    anyway, i saw the solution in class, i tried to do something similar beore i handed the assignment, i hope i did well.
     
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