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An ordinal $\xi$ is an $\varepsilon$-number (where does this name come from?) if $\omega^\xi = \xi$. Is the set of all countable $\varepsilon$-numbers closed in $\omega_1$?

In other words, does an increasing sequence of $\varepsilon$-numbers converge to an $\varepsilon$-number?

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up vote 6 down vote accepted

Yes. This is because ordinal exponentiation is continuous on the exponent: If $\beta$ is limit, then $\omega^\beta=\sup_{\alpha<\beta}\omega^\alpha$. This in fact shows that the $\varepsilon$-numbers form a closed (and, of course, unbounded) class of the ordinals, not just of $\omega_1$.

As for the reason for the name, there is not much of a story here. According to Dauben's book, Cantor first called them Giganten (in correspondence with Goldscheider in 1886). When they saw print (in the second part of his "Beiträge zur Begründung der transfiniten Mengenlehre", in 1897), they were part of his theory of indecomposable ordinals.

Cantor talks about "irreducible" ordinals, and shows that every nonzero ordinal is a product of finitely many irreducibles, in a unique way. Considering these irreducibles, he is led to studying indecomposable ordinals, and had already called $\delta$-numbers those that were multiplicatively indecomposable. The $\varepsilon$-numbers came next, as the ordinals that are exponentially indecomposable.

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Sure, thank you! – J. Lund Jan 13 '13 at 21:06

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