Ordinal-indexed exponentation I have read from somewhere that $2^{\aleph_0}$ is uncountable.
However, I have also read from somewhere else that $2^\omega=\omega$.
Do the above statements contradict with each other?
In particular, I would like to see a proof of the latter statement.
 A: The statements do not contradict with each other. Although the set $\aleph_0$ is identical to the set $\omega$, the operation "exponentiation" is different in the two statements, one being cardinal exponentiation, one being ordinal exponentiation.
Intuitively, $2^{\aleph_0}$ is the cartesian product $\displaystyle\prod_{n \in \aleph_0}2$, where $2 = \{0,1\}$, and $2^\omega$ is the order-type of the cartesian product $\displaystyle\prod_{n \in \omega}2$ with finite support under the lexicographical order. It is the finite support which makes the cardinality of $2^\omega$ the same as $\omega$.
Formally, $2^{\aleph_0}$ is the set of functions from $\aleph_0$ to $2$, and can easily biject with the power set of $\aleph_0$, which, by Cantor's theorem
 is greater than $\aleph_0$. $2^\omega$ is defined as $\sup \{2^n \mid n < \omega\}$. All elements in $\{2^n \mid n < \omega\}$ is finite and strictly increasing, and the set is countably infinite, so the supremum is $\omega$, so $2^\omega = \omega$.
