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Thread: Need a bit of help with indexed sets

  1. #1
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    Need a bit of help with indexed sets

    Hi,

    I'm a real dunce when it comes to indexed sets/Collections etc. I can't seem to understand this question at all, so I would really appreciate some help with the following question.

    Let Ak is a set for each positive integer k. Define another collection of sets Bk:
    B1 := A1, B2 := A2 \ A1, , Bk = Ak \ Ut<kAt
    , .
    Prove that
    (a) The sets Bk are pairwise disjoint;
    (b) Ut≤kAt = Ut≤kBt for each positive integer k;
    (c) Ut≥1At = Ut≥1Bt
    .
    Last edited by BlackZeppelin; Feb 12th 2018 at 10:04 AM.
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  2. #2
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    Re: Need a bit of help with indexed sets

    I'm sorry this notation is just too cryptic.

    what are Ut, kAt, and kBt?
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  3. #3
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    Re: Need a bit of help with indexed sets

    I'm sorry, I should have made it more clear.

    Here's the image of the question:

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    Re: Need a bit of help with indexed sets

    Quote Originally Posted by BlackZeppelin View Post
    Hi,I'm a real dunce when it comes to indexed sets/Collections etc. I can't seem to understand this question at all, so I would really appreciate some help with the following question.
    Let Ak is a set for each positive integer k. Define another collection of sets Bk:
    B1 := A1, B2 := A2 \ A1, , Bk = Ak \ Ut<kAt
    , .
    Prove that
    (a) The sets Bk are pairwise disjoint;
    (b) Ut≤kAt = Ut≤kBt for each positive integer k;
    (c) Ut≥1At = Ut≥1Bt.
    Suppose that $\{j,k\}\subset\mathbb{Z}^+~\&~j<k$
    By definition ${B_k} = {A_k}\backslash \bigcup\limits_{t = 1}^{k - 1} {{A_t}}$
    Suppose that $y\in B_j\cap B_k$ so that means $y\in A_j~\&~y\in A_k$.
    Study the definition of $A_k$ carefully. Has not $Aj\cap A_k$ been removed from $A_k$(recall $j<k$) to get $B_k~?$
    Is that a contradiction?
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  5. #5
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    Re: Need a bit of help with indexed sets

    Quote Originally Posted by Plato View Post
    Suppose that $\{j,k\}\subset\mathbb{Z}^+~\&~j<k$
    By definition ${B_k} = {A_k}\backslash \bigcup\limits_{t = 1}^{k - 1} {{A_t}}$
    Suppose that $y\in B_j\cap B_k$ so that means $y\in A_j~\&~y\in A_k$.
    Study the definition of $A_k$ carefully. Has not $Aj\cap A_k$ been removed from $A_k$(recall $j<k$) to get $B_k~?$
    Is that a contradiction?
    If I'm correct, then yes it is a contradiction. Since Bk is supposed to have no element from any set denoted by an integer less than k. That proves that the sets are pairwise disjoint. So, that's a bit clear now, thanks!

    But what do parts b) and c) even imply. I don't even understand their meaning...
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  6. #6
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    Re: Need a bit of help with indexed sets

    Quote Originally Posted by BlackZeppelin View Post
    But what do parts b) and c) even imply. I don't even understand their meaning...
    Here are parts b) & c).

    b) $\left( {\forall k} \right)\left[ {\bigcup\limits_{t = 1}^k {{A_t}} = \bigcup\limits_{t = 1}^k {{B_t}} } \right]$

    c) $\bigcup\limits_{t = 1}^\infty {{A_t}} = \bigcup\limits_{t = 1}^\infty {{B_t}} $

    See what you can do with those.
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  7. #7
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    Re: Need a bit of help with indexed sets

    I know it is late in Lahore. Here are some clues.
    Can you see that $(\forall k)[B_k\subset A_k]$ If so that is the trick for b).

    Say $x\in\bigcup\limits_{t = 1}^\infty {{A_t}}$ then $(\exists j)[x\in A_j]$
    Let $n$ be the least such $j$. If $n=1$ then $A_n=B_1$ and we are done.

    But if $n>1$ then $x\notin A_k$ for $k=1,\cdots n-1$ which means that $x \in {B_n} = {A_n}\backslash \bigcup\limits_{t = 1}^{n - 1} {{A_t}} $

    The other direction. Suppose that $y\in\bigcup\limits_{t = 1}^\infty {{B_t}}$ Can you finish?
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