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

As we know, $2^{\aleph_0}$ is a cardinal number, so it is a limit ordinal number. However, it must not be $2^\omega$, since $2^\omega=\sup\{2^\alpha|\alpha<\omega\}=\omega=\aleph_0<2^{\aleph_0}$, and even not be $\sum_{i = n<\omega}^{0}\omega^i\cdot a_i$ where $\forall i \le n[a_i \in \omega]$. Since $\|\sum_{i = n<\omega}^{0}\omega^i\cdot a_i\| \le \aleph_0$ for all of them.

Besides, $\sup\{\sum_{i = n<\omega}^{0}\omega^i\cdot a_i|\forall i \le n(a_i \in \omega)\}=\omega^\omega$, and $\|\omega^\omega\|=2^{\aleph_0}$ since every element in there can be wrote as $\sum_{i = n<\omega}^{0}\omega^i\cdot a_i$ where $\forall i \le n[a_i \in \omega]$ and actually $\aleph_{0}^{\aleph_0}=2^{\aleph_0}$ many.

Therefore $\omega^\omega$ is the least ordinal number such that has cardinality $2^{\aleph_0}$, and all ordinal numbers below it has at most cardinality $\aleph_0$. Hence $\omega^\omega=2^{\aleph_0}=\aleph_1$?

• $2^{\aleph_0} = \aleph_1$ is the Continuum Hypothesis, which is independent of ZFC. Jun 7, 2012 at 6:27
• Cardinality of the ordinal number $\omega^\omega$ is $\aleph_0$, see e.g. here or here. Jun 7, 2012 at 6:30
• Yes, you and Scott are correct, thank you. Jun 7, 2012 at 8:21
• You need to be careful with the different kinds of exponentiation. In some set theory books you can see (on different pages) $\omega=\aleph_0$, $2^\omega=\omega$, $2^{\aleph_0}\gt \aleph_0$. Sep 17, 2012 at 17:45

Your notation confuses cardinal and ordinal exponentiation, which are two very different things. If you’re doing cardinal exponentiation, $2^\omega$ is exactly the same thing as $2^{\aleph_0}$, just expressed in a different notation, because $\omega=\aleph_0$. If you’re doing ordinal exponentiation, then as you say, $2^\omega=\omega$.

But if you’re doing ordinal exponentiation, then $$\omega^\omega=\sup_{n\in\omega}\omega^n=\bigcup_{n\in\omega}\omega^n\;,$$ which is a countable union of countable sets and is therefore still countable; it doesn’t begin to reach $\omega_1$. Similarly, still with ordinal exponentiation, $\omega^{\omega^\omega}$ is countable, $\omega^{\omega^{\omega^\omega}}$ is countable, and so on. The limit of these ordinals, known as $\epsilon_0$, is again countable, being the limit of a countable sequence of countable ordinals, and so is smaller than $\omega_1$. (It’s the smallest ordinal $\epsilon$ such that $\omega^\epsilon=\epsilon$.)

Now back to cardinal exponentiation: for that operation you have $2^\omega\le\omega^\omega\le(2^\omega)^\omega=2^{\omega\cdot\omega}=2^\omega$, where $\omega\cdot\omega$ in the exponent is cardinal multiplication, and therefore $2^\omega=\omega^\omega$ by the Cantor-Schröder-Bernstein theorem. The statement that this ordinal is equal to $\omega_1$ is known as the continuum hypothesis; it is both consistent with and independent of the other axioms of set theory.

There are very different definitions for cardinal and ordinal exponentiation. Ordinal exponentiation is defined in a way which allows us to generate well-orderings of a particular set; where as cardinal exponentiation strips out the ordering and deals with cardinality of all functions from one set to another.

This is why some authors differentiate the two by using $^\omega\omega$ for cardinal exponentiation and $\omega^\omega$ for ordinal exponentiation (at least where context is ambiguous). Personally I am not a big fan of this approach, despite the fact it may clear some possible confusion.

Lastly, as commented, $2^{\aleph_0}$ need not be equal to $\aleph_1$. This is known as The Continuum Hypothesis which was proved unprovable from ZFC.

To add on the confusion, let me give a short list of some common uses for $\omega^\omega$:

• The first limit ordinal which is a limit of limit ordinals each a limit of limit ordinals which are not limits of limit ordinals.

• The set of all sequences of natural numbers (which also form the underlying set for the following uses).

• The Baire space.

• The real numbers (in some contexts).

• I’m not a fan of it either, since for me $^\omega\omega$ is the set of functions from $\omega$ to $\omega$, not the cardinality of that set. Jun 7, 2012 at 6:56
• Brian Scott says $\omega^{\omega}$ is countable while you seem to imply it's isomorphic to the reals. How do these two reconcile? Aug 16, 2018 at 8:40
• @Robert: Ordinal arithmetic vs. cardinal arithmetic. Aug 16, 2018 at 8:41
• So people using the form you describe here are doing cardinal arithmetic with what appear to be ordinal numbers? Looks like something I'll need to watch out for. Aug 16, 2018 at 8:41
• @Robert: People, myself included, use $A^B$ to denote the set of all functions from $B$ to $A$. Brian prefers ${}^BA$ for that set. The reason being that the set of all functions from $\omega$ to $\omega$ is denoted $\omega^\omega$, but so does the ordinal exponentiation of $\omega^\omega$. This ends up being confusing, so people have two different notations. I just prefer clarifying the context. Aug 16, 2018 at 8:44