What is the cardinality of $\Bbb Q ^4$?
I understand that $\vert \Bbb Q \vert=\aleph _0$, and $\vert \Bbb Q ^4 \vert$ supposed to be $\aleph _0$, is there any formal proof to this?
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2Look here: https://proofwiki.org/wiki/Cartesian_Product_of_Countable_Sets_is_Countable. – Aniruddha Deshmukh Jul 18 '19 at 04:09
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Before tackling $\mathbb Q^4$, do you know the cardinality of $\mathbb Q^2$? – Chris Culter Jul 18 '19 at 04:23
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Basically, you just need to show that $\aleph_0\cdot\aleph_0=\aleph_0$. You can check the post The cartesian product $\mathbb{N} \times \mathbb{N}$ is countable and other questions linked there. – Martin Sleziak Jul 18 '19 at 07:06
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Well, suppose the set $A$ is denumerable, i.e., $A=\{a_0,a_1,a_2,\ldots\}$. Then the set $A^2$ is denumerable. This can be shown by writing the elements of $A^2$ as an infinite matrix $$\begin{array}{cccc} (a_0,a_0) & (a_0,a_1) & (a_0,a_2) & \ldots\\ (a_1,a_0) & (a_1,a_1) & (a_1,a_2) & \ldots\\ (a_2,a_0) & (a_2,a_1) & (a_2,a_2) & \ldots\\ \vdots & \vdots & \vdots \end{array}$$ These elements can be enumerated by walking through the array in zig-zag just as in Cantor's first diagonalization procedure. Now $A^2$ is denumerable and so $A^4=(A^2)^2$ is denumerable.
Wuestenfux
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Just in the same way you can prove $\mathbb{Q}$ is countable (Cantor's diagonal argument), you can also prove $\mathbb{N}^2$. Now. If $|\mathbb{N}| = |\mathbb{N}^2|$, $|\mathbb{N} \times \mathbb{N}| = |\mathbb{N}^2 \times \mathbb{N}^2| = |\mathbb{N}^4| = |\mathbb{N} \times \mathbb{N} \times \mathbb{N} \times \mathbb{N}|$
But $|\mathbb{N}| = |\mathbb{Q}|$, therefore $|\mathbb{N} \times \mathbb{N} \times \mathbb{N} \times \mathbb{N}| = |\mathbb{Q} \times \mathbb{Q} \times \mathbb{Q} \times \mathbb{Q}|$
David
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