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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

- Showing any countable, dense, linear ordering is isomorphic to a subset of $\mathbb{Q}$
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- Is there a known well ordering of the reals?
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$$f(\langle x,y\rangle)=\begin{cases}y+1 & x=1\\ 1 &\text{otherwise}\end{cases}$$

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?

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- The p-adic numbers as an ordered group
- Is there a known well ordering of the reals?

$2^\omega$ means two different things, depending on whether you’re doing cardinal exponentiation or ordinal exponentiation. You’re thinking of cardinal exponentiation: for that it’s perfectly true that $2^\omega>\omega$. For ordinal exponentiation, however, $2^\omega$ is simply $\bigcup_{n\in\omega}2^n=\omega$.

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

Note that $2^\omega$ is not the powerset of $\omega$, and $|2^\omega|\ne2^{|\omega|}$, where the alatter is *cardinal* arithmetic.

If you read on in that Wikipedia article, you’ll see that $2^\omega$ is obtained by taking the maps $\omega\to 2$ with *finite* support, not arbitrary such maps.

That corresponds to the finite subsets rather than arbitrary subsets, thus to terminating binary sequences (esp. ratioanl numbers) instead of arbitrary binary sequences (real numbers) if you like.

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