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Label the set of all binary series with an infinite amount of 0's and 1's as $C$.

It's easy to prove that the set (labeled $A$) of all binary series with a finite number of 1's is countable. I can then prove in an almost identical way that the set (labeled $B$) of binary series with a finite number of 0's is countable.

The set (labeled $D$) of all binary series is a direct sum: $A+B+C=D$.

$D$'s cardinality is a known $\mathfrak c$ (continum), so we get:

$$|A|+|B|+|C| = |D|\\ \aleph_0+\aleph_0 + |C|=\mathfrak c\implies |C| =\mathfrak c$$

What do you think of the proof? Thanks for your time.

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Your proof works out just fine. But you can also write a direct proof. Just find an injection from the set of all binary sequences, to $C$.

HINT: Fix one sequence which has infinitely $1$'s and $0$'s and call it $s$. Now given a binary sequence $t$, consider the new sequence where the $s$ is exactly the subsequence of even indices, and $t$ is the subsequence of odd indices.

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  • $\begingroup$ I'm not sure I understand what you're implying. Given a binary sequence $a_1a_2a_3...$, define $F(a_1a_2a_3...) = a_{1}1a_{2}0a_3{1}...$ . Then F is a bijection from C to D. Could that work? $\endgroup$
    – Mark Emacr
    Commented Feb 1, 2014 at 12:51
  • $\begingroup$ Yes, that works. But it suffices to show that this is an injection. $\endgroup$
    – Asaf Karagila
    Commented Feb 1, 2014 at 13:12

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