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An epsilon number is an ordinal $\epsilon$ such that $\epsilon=\omega^\epsilon.$ What is the cardinality of the set of all epsilon numbers less than $\omega_1$?

I'm asking this because of a proof I've just read that seems to presuppose that there are countably many such ordinals, and it seems to me intuitively that there should be uncountably many (although I don't know how to prove it).

Added. OK, I've just understood that the proof I mentioned is OK even if there are uncountably many such ordinals, but I still don't see how I can find their number.

Bartek
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    $\omega_1$.${}$ – Andrés E. Caicedo May 27 '13 at 18:37
  • @Andres Thank you, that's what I thought. But how can I prove it? – Bartek May 27 '13 at 18:41
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    OK, finally found it. The reason why I do not like the standard answer (as the ones below) is that it uses choice (in the form: The countable union of countable sets is countable), but the fact that there are $\omega_1$-many $\varepsilon$-numbers below $\omega_1$ does not need choice. This was asked on this site before, here is the link: http://math.stackexchange.com/q/181424/462 – Andrés E. Caicedo May 27 '13 at 18:58
  • Now, this is mostly a pet peeve of mine, but I think it is useful to be aware of these things. Let me go further: There is a standard trick to show that some results using choice do not require it. It is metamathematical, we move between the universe and an inner model of choice, and check that whatever relevant facts verified in the inner model hold in the universe. This is the shape of Joel's answer in the link above, which is why I prefer tomasz's, since it is purely combinatorial. For another example, see here: http://mathoverflow.net/questions/40507/distinct-well-orderings-of-the-same-set – Andrés E. Caicedo May 27 '13 at 19:03
  • Andres, the problem in this case with moving up and down between $L$ and $V$ is that the only time we need this is when $\omega_1$ is singular, in which case $\omega_1^L\neq\omega_1$, so the proof requires some augmentation. I think, whoever, that one can probably prove an interesting lemma that if the ordinals are an iterative sequence like the one in my answer, then we can uniformly enumerate them, in which case no choice is needed in proving their limit is countable. – Asaf Karagila May 27 '13 at 19:12
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    @Andres: Ah, actually no trick is needed. Ordinal arithmetics are absolute, so even if $\omega_1^L\neq\omega_1$, in $L$ there is a bijection between the $\varepsilon$ numbers which are smaller than $\omega_1^V$ and $\omega_1^V$, which would witness there are $\aleph_1$ countable $\varepsilon$ numbers in $V$. – Asaf Karagila May 27 '13 at 21:22

3 Answers3

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Note that $\varepsilon_0$ is countable. So if there are only countably many countable $\varepsilon$ numbers, they would have a countable supremum, $\alpha$. Consider now the same construction as $\varepsilon_0$, starting $\alpha+1$. That is: $$\sup\{\omega^{\alpha+1},\omega^{\omega^{\alpha+1}},\ldots\}$$

The result is itself an $\varepsilon$ number, and it is countable (as the countable limit of countable ordinals). But all the countable $\varepsilon$ numbers were assumed to be below $\alpha$, which is a contradiction.

Asaf Karagila
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Your intuition is correct: $\omega_1=\epsilon_{\omega_1}$. See the discussion in Wikipedia: the constructions of $\epsilon_{\alpha+1}$ from $\epsilon_\alpha$ and of $\epsilon_\alpha$ for a countable limit ordinal $\alpha$ preserve countability, so there must be $\omega_1$ countable $\epsilon$-numbers.

kimchi lover
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Brian M. Scott
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Another answer that uses choice:

Since $\varepsilon_\alpha$ is countable if $\alpha$ is countable by this other SE question or by the arguments in the other answers (and obviously $\varepsilon_\alpha\ge\alpha$), the map $f:\beta\to\omega_1$ defined by $f(\alpha)=\varepsilon_\alpha$ (where $\beta$ is the number of countable $\varepsilon$ numbers) is a cofinal map. But by the axiom of choice, $\omega_1$ is regular. Thus there are $\beta=\omega_1$ countable $\varepsilon$ numbers.

Community
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Mario Carneiro
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