# $\varepsilon$-numbers

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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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 $\omega_1$.
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.