Ordinal exponentiation - $2^{\omega}=\omega$ This is my understanding of ordinal arithmetic - two ordinals are the same as one another if there is an order-preserving bijection between them. So for instance 
$$1+\omega = \omega$$
because if 
$$f(\langle x,y\rangle)=\begin{cases}y+1 & x=1\\ 1 &\text{otherwise}\end{cases}$$
Then $f$ is an order-preserving bijection between $\{ 0 \} \times 1 \cup \{ 1 \} \times \omega$ and $\omega$, where $\{ 0 \} \times 1 \cup \{ 1 \} \times \omega$ is endowed with the addition order.
Likwise if
$$g(\langle x,y \rangle)=2 \times x+y$$
Then $g$ is an order-preserving bijection between $2 \times \omega$ and $\omega$, where $2 \times \omega$ is endowed with the multiplication order, and so $2 \cdot \omega =\omega$  , whereas $\lnot 2 \cdot \omega =\omega \cdot 2$ because $< 0,1 >$ is a limit of $\omega \times 2$ under the multiplication order whereas $2 \cdot \omega$ has no limit ordinals.
On Wikipedia's page, Exponentiation is described for ordinals, where in particular, it says that $2^{\omega} = \omega$. How can this be when $\omega$ does not even have the same cardinality as $2^{\omega}$ - to wit, isn't $2^{\omega}$ uncountable, with the same cardinality as the reals?

Ordinal exponentiation is not cardinal exponentiation.
The cardinal exponentiation $2^\omega$ is indeed uncountable and has the cardinality of the continuum.
The ordinal exponentiation $2^\omega$ is the supremum of $\{2^n\mid n\in\omega\}$ which in turn is exactly $\omega$ again.
Also related:


*

*How is $\epsilon_0$ countable?

*Do $\omega^\omega=2^{\aleph_0}=\aleph_1$?
