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Let $S = \{ s\mid s\text{ is an infinite binary sequence}\}$ be the set of all infinite binary sequences.

By Cantor's Diagonal Argument, $S$ is uncountable.

But does $|S \times S| = |S|$ also hold?

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    $\begingroup$ Your definition of $S$ is just $\{0,1\}$. How about you write in exact and explicit words what is $S$? $\endgroup$ – Asaf Karagila Aug 13 '17 at 10:28
  • $\begingroup$ @Asaf Karagila Done. $\endgroup$ – Henry Durham Aug 13 '17 at 11:27
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    $\begingroup$ If it is an uncountable set, why do you write it as $\{s_1, s_2,\ldots\}$ which seems to imply that it is countable? Also, again this is not how set-builder notation is used, and you got $S=\{0,1\}$ again. $\endgroup$ – Asaf Karagila Aug 13 '17 at 11:28
  • $\begingroup$ @Asaf Karagila Can you please edit the question ro the correct form? Thanks. $\endgroup$ – Henry Durham Aug 13 '17 at 12:29
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    $\begingroup$ Also, $S$ is often denoted by $2^\Bbb N$, or sometimes $2^\omega$. $\endgroup$ – Asaf Karagila Aug 13 '17 at 12:41
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Yes. Simply separate each sequence into two sequences. If $s$ is a sequence, then we can decompose it to $u$ and $t$ such that $u(n)=s(2n)$ and $t(n)=s(2n+1)$. Namely, taking the even coordinates as the sequence $u$ and the odd coordinates as the sequence $t$.

It is not hard to check that the map $s\mapsto(u,t)$ as defined above is indeed a bijection between $S$ and $S\times S$.

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