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.


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.

  • $\begingroup$ 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 Feb 1 '14 at 12:51
  • $\begingroup$ Yes, that works. But it suffices to show that this is an injection. $\endgroup$ – Asaf Karagila Feb 1 '14 at 13:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.