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

  • 4
    $\begingroup$ $2^{\aleph_0} = \aleph_1$ is the Continuum Hypothesis, which is independent of ZFC. $\endgroup$
    – smackcrane
    Commented Jun 7, 2012 at 6:27
  • 7
    $\begingroup$ Cardinality of the ordinal number $\omega^\omega$ is $\aleph_0$, see e.g. here or here. $\endgroup$ Commented Jun 7, 2012 at 6:30
  • $\begingroup$ Yes, you and Scott are correct, thank you. $\endgroup$
    – Popopo
    Commented Jun 7, 2012 at 8:21
  • 4
    $\begingroup$ You need to be careful with the different kinds of exponentiation. In some set theory books you can see (on different pages) $\omega=\aleph_0$, $2^\omega=\omega$, $2^{\aleph_0}\gt \aleph_0$. $\endgroup$ Commented Sep 17, 2012 at 17:45

2 Answers 2


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).

  • 3
    $\begingroup$ 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. $\endgroup$ Commented Jun 7, 2012 at 6:56
  • $\begingroup$ Brian Scott says $\omega^{\omega}$ is countable while you seem to imply it's isomorphic to the reals. How do these two reconcile? $\endgroup$ Commented Aug 16, 2018 at 8:40
  • $\begingroup$ @Robert: Ordinal arithmetic vs. cardinal arithmetic. $\endgroup$
    – Asaf Karagila
    Commented Aug 16, 2018 at 8:41
  • $\begingroup$ 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. $\endgroup$ Commented Aug 16, 2018 at 8:41
  • 1
    $\begingroup$ @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. $\endgroup$
    – Asaf Karagila
    Commented Aug 16, 2018 at 8:44

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .