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This question already has an answer here:

Is it true that $|\mathbb{R}|=2^\omega=\omega_1$?

Note that $\omega_1$ is the successor of $\omega$ and $2^\omega$ is |all functions from $\omega$ to 2|.

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marked as duplicate by Asaf Karagila, copper.hat, Shuchang, Antonio Vargas, ncmathsadist Dec 5 '13 at 1:51

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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This is known as the Continuum hypothesis.

http://en.wikipedia.org/wiki/Continuum_hypothesis.

And this cannot be proved either way using the standard Zermelo–Fraenkel set theory (ZF). Even when the axiom of choice is assumed, it still cannot be proved!

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  • $\begingroup$ I find your last remark strange. Why would the axiom of choice have anything to do with proving the continuum hypothesis? If anything, it proves that all the ways of expressing the continuum hypothesis are in fact equivalent. $\endgroup$ – Asaf Karagila Dec 5 '13 at 0:13
  • $\begingroup$ It is probably naivety on my behalf. I personally find it highly non-obvious to whether or not it has an affect. $\endgroup$ – user93826 Dec 5 '13 at 0:29
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The first equality is true. The second is the continuum hypotheses and is independent of the axioms of ZFC. You could also see this question

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