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I have read that $2^{\omega_0}=2^{\omega}=\omega$. (in the sense that they have the same order type). On the other hand, I know that $\omega=\aleph_0$, since it is the least infinite countable ordinal. I also know that $2^{\aleph_0} = \aleph_1 = \omega_1$ which is the least uncountable ordinal. Doesn't this gives us $2^{\omega_0}=\omega_1$? So, which of the above facts that i have stated is not correct?

Thank you,

Also. if you find my question duplicated, I will be greatfull if you could mention where can i find the answer.

A student
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1 Answers1

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Firstly, $2^{\aleph_0} = \aleph_1$ is not always true, this is independent of the ZFC axioms of set theory, and is known as the Continuum Hypothesis. What is true is that this is an uncountable cardinal, so $|\mathbb{R}| = 2^{\aleph_0} \ge \aleph_1$.

As a set, $\omega = \aleph_0$ (in the usual set theoretic set-up), but just like addition is different for ordinals, so is exponentiation. When we write $\omega$, we mean to look at the set as an ordinal (i.e. having a specific well-order), and when we write $\aleph_0$ we mean to use it as a "yard-stick" for infinite sets. So $\omega+1 \neq \omega$ as ordered sets, but $\aleph_0 + 1 = \aleph_0$ as cardinals (they have the same size). Also, exponentiation is different, so the meaning of $2^\omega$ is some specific order type, and $2^{\aleph_0}$ is the cardinality of the powerset of $\aleph_0$.

The operations $+$ and powers depend on the context. Also wikipedia warns for the confusion and the different symbol uses.

Henno Brandsma
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  • I think I am begining to get it. Is the following statement correct? $2^\omega = lim{2^n | n \in \omega}$ and $2^{\aleph_0} = { f | f $ is a function from $\omega$ to ${ 0,1 }$ $ }$. – A student Sep 28 '14 at 14:13
  • Indeed, the definition of $2^\omega$ is the $\sup$ or limit of the set of ordinals ${2^n: n \in \omega}$, while $2^{\aleph_0}$ is the cardinality of the set of functions from $\omega$ to $2 = {0,1}$. This cardinality equals some $\aleph_\alpha$ (if we assume the axiom of choice, which I do). Exactly which $\alpha$ is undetermined by the axioms. – Henno Brandsma Sep 28 '14 at 14:18