Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Firstly, I give the definition of the epsilon number:

$\alpha$ is called an epsilon number iff $\omega^\alpha=\alpha$.

Show that if $\kappa$ is an uncountable cardinal, then $\kappa$ is an epsilon number and there are $\kappa$ epsilon numbers below $\kappa$; In particular, the first epsilon number, called $\in_0$, is countable.

I've tried, however I have not any idea for this. Could anybody help me?

share|cite|improve this question
up vote 3 down vote accepted

Here is a slightly easier way:

Lemma: For every $\alpha,\beta$ the ordinal exponentiation $\alpha^\beta$ has cardinality of at most $\max\{|\alpha|,|\beta|\}$.

Now use the definition of $\omega^\kappa=\sup\{\xi^\omega\mid\xi<\kappa\}$, since $|\xi|<\kappa$ we have that $\omega^\kappa\leq\kappa$, but since $\xi\leq\omega^\xi$ for all $\xi$, $\omega^\kappa=\kappa$.

Here is an alternative way (a variation on the above suggestion):

Lemma: If $\alpha$ is an infinite ordinal then there is some $\varepsilon_\gamma\geq\alpha$ such that $|\alpha|=|\varepsilon_\gamma|$

Hint for the proof: Use the fact that you need to close under countably many operations, and by the above Lemma none changes the cardinality.

Now show that the limit of $\varepsilon$ numbers is itself an $\varepsilon$ number, this is quite simple:

If $\beta=\sup\{\alpha_\gamma\mid\alpha_\gamma=\omega^{\alpha_\gamma}\text{ for }\gamma<\tau\}$ (for some $\tau$ that is) then by definition of ordinal exponentiation $$\omega^\beta=\sup\{\omega^{\alpha_\gamma}\mid\gamma<\tau\}=\sup\{\alpha_\gamma\mid\gamma<\tau\}=\beta$$

Now we have that below $\kappa$ there is a cofinal sequence of $\varepsilon$ numbers, therefore it is an $\varepsilon$ number itself; now by induction we show that there are $\kappa$ many of them:

  • If $\kappa$ is regular then every cofinal subset has cardinality $\kappa$ and we are done;
  • if $\kappa$ is singular there is an increasing sequence of regular cardinals $\kappa_i$, such that $\kappa = \sup\{\kappa_i\mid i\in I\}$. Below each one there are $\kappa_i$ many $\varepsilon$ numbers, therefore below $\kappa$ there is $\sup\{\kappa_i\mid i\in I\}=\kappa$ many $\varepsilon$ numbers.
share|cite|improve this answer
It do be a slightly easier. I'm very interesting the lamma which yor offer. Could you tell me where it from? (The proof is also very welcome:) – Paul Aug 7 '12 at 9:52
(bangs forehead against desk) – arjafi Aug 7 '12 at 9:55
@Paul: I don't remember. I can't find it on Jech... I think it might be from Introduction to Cardinal Arithmetic (Holz, Steffens, Weitz), but I'm not sure if it was fully proved there (they tend to just write things off as an exercise in transfinite induction). The proof is not hard, really. Go by the maximum of the pair $\alpha,\beta$. – Asaf Karagila Aug 7 '12 at 10:01
@Arthur: If you keep banging your head against the desk you'll end up forgetting more simple ways! :-) – Asaf Karagila Aug 7 '12 at 10:02
@Paul: In your case you might also just be interested in the case $\alpha=\omega$ and $\beta>\omega$. It would probably be a bit easier. – Asaf Karagila Aug 7 '12 at 10:05

The following is intended as a half-outline/half-solution.

We will prove by induction that every uncountable cardinal $\kappa$ is an $\epsilon$-number, and that the family $E_\kappa = \{ \alpha < \kappa : \omega^\alpha = \alpha \}$ has cardinality $\kappa$.

Suppose that $\kappa$ is an uncountable cardinal such that the two above facts are knows for every uncountable cardinal $\lambda < \kappa$.

  • If $\kappa$ is a limit cardinal, note that in particular $\kappa$ is a limit of uncountable cardinals. By normality of ordinal exponentiation it follows that $$\omega^\kappa = \lim_{\lambda < \kappa} \omega^\lambda = \lim_{\lambda < \kappa} \lambda = \kappa,$$ where the limit is taken only over the uncountable cardinals $\lambda < \kappa$.

    Also, it follows that $E_\kappa = \bigcup_{\lambda < \kappa} E_\lambda$, and so $| E_\kappa | = \lim_{\lambda < \kappa} | E_\lambda | = \kappa$.

  • If $\kappa$ is a successor cardinal, note that $\kappa$ is regular. Note, also, that every uncountable cardinal is an indecomposable ordinal. Therefore $\kappa = \omega^\delta$ for some (unique) ordinal $\delta$. As $\omega^\kappa \geq \kappa$, we know that $\delta \leq \kappa$. It suffices to show that $\omega^\beta < \kappa$ for all $\beta < \kappa$. We do this by induction: assume $\beta < \kappa$ is such that $\omega^\gamma < \kappa$ for all $\gamma < \beta$.

    • If $\beta = \gamma + 1$, note that $\omega^\beta = \omega^\gamma \cdot \omega = \lim_{n < \omega} \omega^\gamma \cdot n$. By indecomposability it follows that $\omega^\gamma \cdot n < \kappa$ for all $n < \omega$, and by regularity of $\kappa$ we have that $\{ \omega^\gamma \cdot n : n < \omega \}$ is bounded in $\kappa$.

    • If $\beta$ is a limit ordinal, then $\omega^\beta = \lim_{\gamma < \beta} \omega^\gamma$. Note by regularity of $\kappa$ that $\{ \omega^\gamma : \gamma < \beta \}$ must be bounded in $\kappa$.

    To show that $E_\kappa$ has cardinality $\kappa$, note that by starting with any ordinal $\alpha < \kappa$ and defining the sequence $\langle \alpha_n \rangle_{n < \omega}$ by $\alpha_0 = \alpha$ and $\alpha_{n+1} = \omega^{\alpha_n}$ we have that $\alpha_\omega = \lim_{n < \omega} \alpha_n < \kappa$ is an $\epsilon$-number. Use this fact to construct a strictly increasing $\kappa$-sequence of $\epsilon$-numbers less than $\kappa$.

(There must be an easier way, but I cannot think of it.)

share|cite|improve this answer
I tried to think of an easier way... – Asaf Karagila Aug 7 '12 at 9:42

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.