# Cardinality

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• Jan 3rd 2010, 02:18 PM
Drexel28
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.
• Jan 3rd 2010, 03:22 PM
Plato
Quote:

Originally Posted by Drexel28
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?
• Jan 3rd 2010, 03:24 PM
Drexel28
Quote:

Originally Posted by Plato
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.
• Jan 3rd 2010, 03:44 PM
Plato
Quote:

Originally Posted by Drexel28
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.
Let’s change the example.
$X_i$ be the set of infinite bit-strings in which only the first $i^{th}$ terms are possibly non-zero.
• Jan 3rd 2010, 06:15 PM
NonCommAlg
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 $\text{card} \ \bigcup_{i \in I} Y_i \leq \text{card} \ J. \ \Box$
• Jan 4th 2010, 02:45 AM
Drexel28
Quote:

Originally Posted by NonCommAlg
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.