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Math Help - Cardinality

  1. #1
    MHF Contributor Drexel28's Avatar
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    Cardinality

    Hello friends, I have an idea of how to do this, but a nudge in the right direction wouldn't hurt!


    Let \left\{X_i\right\},\text{ }i\in\mathcal{I} be a class of countably infinite sets where \mathcal{I} is infinite (countably or uncountably). Prove that \text{card }\bigcup_{i\in\mathcal{I}}X_i\le\text{card }\mathcal{I}

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    If \mathcal{I} is countably infinite this follows easily, otherwise I feel as though I can use Zorn's lemma.
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    Quote Originally Posted by Drexel28 View Post
    Let \left\{X_i\right\},\text{ }i\in\mathcal{I} be a class of countably infinite sets where \mathcal{I} is infinite (countably or uncountably). Prove that \text{card }\bigcup_{i\in\mathcal{I}}X_i\le\text{card }\mathcal{I}
    I freely admit that I am not at all sure if I understand your question.
    But suppose that \mathcal{I}=\mathbb{Z}^+ the let X_i be the set of infinite bit-strings in which the i^{th} entry is not 0.

    Is this a counter-example?
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  3. #3
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Plato View Post
    I freely admit that I am not at all sure if I understand your question.
    But suppose that \mathcal{I}=\mathbb{Z}^+ the let X_i be the set of infinite bit-strings in which the i^{th} entry is not 0.

    Is this a counter-example?
    Thanks Plato. I'm not sure, but it looks as though your X_i is uncountable though.
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    Quote Originally Posted by Drexel28 View Post
    Thanks Plato. I'm not sure, but it looks as though your X_i is uncountable though.
    Well, I did say that I might not fully understand the question.
    Letís change the example.
    X_i be the set of infinite bit-strings in which only the first i^{th} terms are possibly non-zero.
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    for every i \in I put Y_i=X_i \times \{i \}. see that \text{card} \ \bigcup_{i \in I}X_i \leq \text{card} \ \bigcup_{i \in I} Y_i. clearly \text{card} \ X_i = \text{card} \ Y_i and Y_i \cap Y_j = \emptyset, for all i \neq j. now let J=I \times \mathbb{N}. then \text{card} \ I = \text{card} \ J. we also have an

    one-to-one map f_i : Y_i \longrightarrow \mathbb{N}, for every i \in I. finally define g : \bigcup_{i \in I} Y_i \longrightarrow J by g(y)=(i,f_i(y)), for all i \in I, \ y \in Y_i. see that g is one-to-one and thus \text{card} \ \bigcup_{i \in I} Y_i \leq \text{card} \ J. \ \Box
    Last edited by NonCommAlg; January 3rd 2010 at 08:40 PM.
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  6. #6
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by NonCommAlg View Post
    for every i \in I put Y_i=X_i \times \{i \}. see that \text{card} \ \bigcup_{i \in I}X_i \leq \text{card} \ \bigcup_{i \in I} Y_i. clearly \text{card} \ X_i = \text{card} \ Y_i and Y_i \cap Y_j = \emptyset, for all i \neq j. now let J=I \times \mathbb{N}. then \text{card} \ I = \text{card} \ J. we also have an

    one-to-one map f_i : Y_i \longrightarrow \mathbb{N}, for every i \in I. finally define g : \bigcup_{i \in I} Y_i \longrightarrow J by g(y)=(i,f_i(y)), for all i \in I, \ y \in Y_i. see that g is one-to-one and thus \text{card} \ \bigcup_{i \in I} Y_i \leq \text{card} \ J. \ \Box
    That looks great, except I don't see why it follows so easily that \text{card }\mathcal{I}\times\mathbb{N}=\text{card }\mathcal{I}? I think what we could do to prove it is to take \mathfrak{F} to be a sufficiently large family (maybe \mathfrak{F}=\left\{\mathcal{I}\times\{1,\cdots,n\  },n\in\mathbb{N}\right\})and transform it into a poset by letting \left(\mathfrak{F},\le\right) be a poset with A\le B\Leftrightarrow A\subseteq B\text{ and }\text{card }A\le\text{card }B. Then if we can prove that \text{card }\mathcal{I}\times\left\{1,\cdots,n\right\}=\text{  card }\mathcal{I} for all n\in\mathbb{N} I think we can apply Zorn's lemma to show that \mathfrak{F} has a maximal element \mathfrak{M} and this \mathfrak{M} is forced to be \mathfrak{M}=\mathcal{I}\times\mathbb{N}. This may very wrong, I just can't think straight right now.
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