4
$\begingroup$

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?

2 Answers 2

3

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.
  • 0
    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:)2012-08-07
  • 0
    (*bangs forehead against desk*)2012-08-07
  • 0
    @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 $\alpha,\beta$.2012-08-07
  • 0
    @Arthur: If you keep banging your head against the desk you'll end up forgetting *more* simple ways! :-)2012-08-07
  • 0
    @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.2012-08-07
  • 0
    @Asaf Yes, it will be more easier. Thanks again for your answer.2012-08-07
  • 0
    @Arthur Don't be frustrate. Your answer, i think, feed me also very much! I've upvoted your answer.2012-08-07
  • 0
    @Asaf: On the other hand, if I bang my head enough my permanent state will be blissful ignorance... ;)2012-08-07
  • 0
    @Paul: I'm really not that frustrated (about this). There are many much more important things to be frustrated about.2012-08-07
  • 0
    @Asaf I still have an question: How to use your lemma to solve my second question, i.e., How to show there are $\kappa$ epsilon numbers below $\kappa$?2012-08-08
  • 0
    @Paul: First show that above any ordinal $\alpha$ there is some $\varepsilon$ number of the same cardinality. Now you have that this is a cofinal subset of $\kappa$. If $\kappa$ is regular then there are at least $\kappa$ many of those; if $\kappa$ is singular then it is the limit of an increasing sequence of regular cardinals and you can use the sum of the cofinal subsets to show that the cardinality of all $\epsilon$ numbers is as that of $\kappa$.2012-08-08
  • 0
    @Asaf Thanks. Maybe I will take some time to take it in. It is a little difficult for me.2012-08-08
4

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.)

  • 0
    I tried to think of an easier way...2012-08-07