The real numbers in the interval $[0,1]$ can be put in one-to-one correspondence with the (uncountable) power set of the natural numbers, $2^{\mathbb{N}}$, and from this we know that the real numbers are uncountable since every subset of a countable set is also countable. I have seen this proof also used to show that the cardinality of the real numbers is that of $2^{\mathbb{N}}$, however, what troubles me is that this leaves out the integer component of real numbers, which have cardinality $\aleph_0$. It would seem to me that if we take into account all real numbers we must take into account the cardinality of those in the interval $[0,1]$ and the integers, which would be the product of both cardinalities, $\aleph_0 \cdot 2^{\aleph_0}$. What am I misunderstanding here?


You are right that $|\mathbb R|=\aleph_0\cdot 2^{\aleph_0}$.

What you are missing is that $\aleph_0\cdot 2^{\aleph_0}$ is the same cardinality as $2^{\aleph_0}$ itself.

(Multiplication of infinite cardinal numbers behaves weirdly: Under the axiom of choice $\kappa\cdot\lambda=\max(\kappa,\lambda)$ for all infinite cardinals $\kappa$ and $\lambda$, and it may be weirder yet -- though not in the particular case of $\aleph_0\cdot 2^{\aleph_0}$ -- if we don't assume the axiom of choice).

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    $\begingroup$ Regarding the parenthetical paragraph, it may be worth pointing out that the equality $\aleph_02^{\aleph_0}=2^{\aleph_0}$ does not require the axiom of choice. $\endgroup$ – Andrés E. Caicedo Jul 30 '17 at 13:23
  • $\begingroup$ @AndrésE.Caicedo: Good point; I've tried to make it clearer. $\endgroup$ – hmakholm left over Monica Jul 30 '17 at 13:26
  • $\begingroup$ Perfect answer, thank you! For reference, I am studying Szekeres' "Modern Mathematical Physics" and multiplication of infinite cardinals is not covered. Also, thanks to those who fixed up my MathJax. $\endgroup$ – Boon Jul 30 '17 at 14:05

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