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I would like to prove the equality (last property of Ordinals $\varepsilon$ numbers)

$$ \varepsilon_{\omega_\alpha} = \omega_\alpha, $$

where as it is usual, $\varepsilon$ are the fixed-points of the exponential map

$$ f(\alpha)=(\omega_0)^\alpha . $$

Here there is a related question about the cardinality of the class $\{\varepsilon_\alpha:\alpha<\omega_1\}$. For me the answer proves $\varepsilon_{\omega_1}\geq \omega_1$, but not the equality (just for me). Any way, I would like to know the general proof.

You are allow to use AC if it was necessary.

I would like post my efforts, but I have no idea. If you explain me the case $\alpha=1$ maybe I could try to generalize the proof. for arbitary $\alpha$.

Thanks in advance

EDIT

Following Asaf Karagila's answer and this tips I have made some progress:

Proving $\mbox{card }\alpha\beta =\mbox{card }\alpha+\mbox{card }\beta$ is very easy, as he says, because $\aleph_\alpha+\aleph_\beta=\max\{\aleph_\alpha,\aleph_\beta\}$.

Now I need to prove $\mbox{card }\alpha^\beta = \mbox{card }\alpha\beta$. I set $\alpha=\omega$ and I prove it for all $\beta$ by induction (later I'll use Cantor normal form to extend the result to all $\alpha$): $\omega^\omega$ is countable because is a countable union of countables.

$$\mbox{card }\omega^{\omega+1} = \mbox{card }\omega^\omega\omega = \mbox{card }\omega^\omega+\mbox{card }\omega $$

and the results follows again.

I suppose the result up to $\alpha$. If it is a succesor ordinal the proof follows the same computation than in the last case. So let $\alpha$ a limit ordinal. Then

$$ \mbox{card }\omega^\alpha = \mbox{card } \bigcup\{\omega^\xi:\xi<\alpha\} = \sum_{\xi<\alpha}\mbox{card }\omega^\xi = \sum_{\xi<\alpha}(\mbox{card }\omega + \mbox{card }\xi) $$ and I don't know how to continue. In fact I'm not sure the second equality is correct.

Once I get this result I think I can extend it to all ordinals $\alpha^\beta$, since

$$ \omega^{\alpha'}\leq\alpha=\sum_{i=1}^n k_i\omega^{\alpha_i}<\omega^{\alpha'+1}, $$ where $\alpha'$ is de maximum of the $\alpha_i$'s. Because $\mbox{card }\omega^\alpha = \mbox{card }\omega^{\alpha+1}=\mbox{card }\alpha $ $$ \mbox{card }\alpha^\beta =\mbox{card } \omega^{(\alpha'\beta)}= \mbox{card }\alpha'\beta=\mbox{card }\alpha+\mbox{card }\beta . $$

I don't know (yet) how to prove from this $\mbox{card }\varepsilon_\alpha=\mbox{card }\varepsilon_{\alpha+1}$ and why that implies $\mbox{card }\varepsilon_\alpha = \mbox{card } \alpha$.

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  • $\begingroup$ I agree. It's totally hot to show that $\varepsilon_{\omega_\alpha}$ is $\omega_\alpha$. $\endgroup$
    – Asaf Karagila
    Commented Apr 17, 2018 at 17:50
  • $\begingroup$ I didn't understand your comment until I have read again the title. Now, do you think I should correct it? $\endgroup$
    – Dog_69
    Commented Apr 17, 2018 at 20:50
  • $\begingroup$ Sure... why not? $\endgroup$
    – Asaf Karagila
    Commented Apr 17, 2018 at 20:55
  • $\begingroup$ Because seems funny. $\endgroup$
    – Dog_69
    Commented Apr 17, 2018 at 20:57
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    $\begingroup$ The correct way, by the way, is not to \mbox{card } everywhere, but either \operatorname{card} or once at the start do \DeclareMathOperator{\card}{card} and then just use \card everywhere. $\endgroup$
    – Asaf Karagila
    Commented Apr 18, 2018 at 10:33

1 Answer 1

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Prove by induction:

For all $\alpha$, $|\varepsilon_\alpha|=|\varepsilon_{\alpha+1}|$.

First note that for two infinite ordinals, $|\alpha\cdot\beta|=|\alpha|\cdot|\beta|$. Now by induction, we prove that $|\alpha^\beta|=|\alpha|\cdot|\beta|$. Suppose that for a fixed, infinite $\alpha$, $|\alpha^\beta|=|\alpha|\cdot|\beta|$ for all $\beta<\delta$.

  • If $\delta$ is a limit, then $\alpha^\delta=\sup\{\alpha^\beta\mid\beta<\delta\}$, so $|\alpha^\delta|=\sup\{|\alpha^\beta|\mid\beta<\delta\}$, which is exactly $|\delta|\cdot\sup\{|\alpha^\beta|\mid\beta<\delta\}$, and since for $\beta<\delta$, we know that $|\alpha^\beta|=|\alpha|\cdot|\beta|$ we get the wanted result.

  • If $\delta=\beta+1$, then $\alpha^\delta=\alpha^\beta\cdot\alpha$, and the result follows.

This, in turn, implies that $|\omega^\alpha|=|\alpha|$ for all $\alpha\geq\omega$. Now, $\varepsilon_{\alpha+1}=\sup\{\varepsilon_\alpha+1,\omega^{\varepsilon_\alpha+1},\ldots\}$, this is a countable supremum, and by the above we know that $\omega^{\varepsilon_\alpha+1}$ has the same cardinality as $\varepsilon_\alpha$, so by induction the whole sequence leading to $\varepsilon_{\alpha+1}$ also have the same cardinality, and therefore so does $\varepsilon_{\alpha+1}$.

Finally, this means that $|\varepsilon_\alpha|=|\alpha|+\aleph_0$. Again, we prove this by induction on $\alpha$. For $\alpha=0$, yes, this is well-known. For $\alpha=\beta+1$, this follows from the previous claim. Therefore for a limit ordinal, the only case we need to check is when $\alpha$ is itself a cardinal. But then we get that $\varepsilon_\alpha=\sup\{\varepsilon_\beta\mid\beta<\alpha\}$. This is a union of $\alpha$ many ordinals, and each has cardinality at most $\alpha$ (or strictly less, if $\alpha>\omega$). And again the conclusion follows.

This means that the $\omega_\alpha$th $\varepsilon$-number is the supremum of an increasing sequence of length $\omega_\alpha$, of ordinals which are all smaller than $\omega_\alpha$ itself. So it has to be $\omega_\alpha$.

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  • $\begingroup$ OK. I'll try. If I have problems I'll write a comment. Thanks. $\endgroup$
    – Dog_69
    Commented Apr 17, 2018 at 18:35
  • $\begingroup$ By accepting this, I guess that you managed to get the proof to work? $\endgroup$
    – Asaf Karagila
    Commented Apr 17, 2018 at 21:00
  • $\begingroup$ No... :( In fact I was going to demand you help to prove the equality $|\alpha^\beta|=|\alpha|+|\beta|$. $\endgroup$
    – Dog_69
    Commented Apr 17, 2018 at 21:10
  • $\begingroup$ Demand is a big word. This follows by induction from $|\alpha\cdot\beta|=|\alpha|+|\beta|$ (for infinite ordinals). $\endgroup$
    – Asaf Karagila
    Commented Apr 17, 2018 at 21:11
  • $\begingroup$ Sorry, ask. Mmmm... As I expected. Ok. Thanks again. $\endgroup$
    – Dog_69
    Commented Apr 17, 2018 at 21:13

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