Gap ordinal: Difference between revisions

There is a nice analogy between gap ordinals and cardinals. Note that \( \alpha \) is a cardinal if, for all \( \gamma < \alpha \), there is no surjection \( \pi: \gamma \longrightarrow \alpha \). If \( \alpha \) is infinite, we have \( \pi \subseteq \gamma \times \alpha \subseteq \alpha \times \alpha \subseteq V_\alpha^2 \subseteq V_\alpha \) and thus \( \pi \in V_{\alpha + 1} \). Thus, "for all \( \gamma < \alpha \), there is no surjection \( \pi: \gamma \longrightarrow \alpha \) with \( \pi \in V_{\alpha + 1} \)" is equivalent to being a cardinal. Meanwhile, the least ordinal satisfying the similar but weaker condition "for all \( \gamma < \alpha \), there is no surjection \( \pi: \gamma \longrightarrow \alpha \) with \( \pi \in L_{\alpha + 1} \)" is equal to the least gap ordinal, since it's equivalent to \( L_\alpha \) satisfying separation.<ref>R.Björn Jensen, The fine structure of the constructible hierarchy, Annals of Mathematical Logic, Volume 4, Issue 3, 1972, Pages 229-308</ref>

==Longer gaps==

If \( 0^\sharp \) exists, then, for any countable \( \eta\) and any \( \gamma \) at all, there is a countable \( \delta \) which starts an \( \eta \)th-order gap of length \( \gamma \). Meanwhile, if \( V = L \), then there is no countable ordinal starting a first-order gap of length \( \omega_1 \).