# Why is cardinality of reals not $\aleph_0 \cdot 2^{\aleph_0}$?

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).
• 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. – Andrés E. Caicedo Jul 30 '17 at 13:23