Gap ordinal: Difference between revisions

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(Created page with "A gap ordinal is an ordinal \(\alpha\) such that \((L_{\alpha+1}\setminus L_\alpha)\cap\mathcal P(\omega)=\varnothing\).<ref name="MarekSrebrny73">W. Marek, M. Srebrny, "[https://www.sciencedirect.com/science/article/pii/0003484374900059 Gaps in the Constructible Universe]" (1973). Annals of Mathematical Logic vol. 6, pp.359--394.</ref><sup>p.364</sup> An ordinal \(\alpha\) is said to start a gap if \(\alpha\) is a gap ordinal but for all \(\beta<\alpha\), \((L_\alpha\se...")
 
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A gap ordinal is an ordinal \(\alpha\) such that \((L_{\alpha+1}\setminus L_\alpha)\cap\mathcal P(\omega)=\varnothing\).<ref name="MarekSrebrny73">W. Marek, M. Srebrny, "[https://www.sciencedirect.com/science/article/pii/0003484374900059 Gaps in the Constructible Universe]" (1973). Annals of Mathematical Logic vol. 6, pp.359--394.</ref><sup>p.364</sup> An ordinal \(\alpha\) is said to start a gap if \(\alpha\) is a gap ordinal but for all \(\beta<\alpha\), \((L_\alpha\setminus L_\beta)\cap\mathcal P(\omega)\neq\varnothing\).<ref name="MarekSrebrny73" /><sup>p.368</sup>
A gap ordinal is an ordinal \(\alpha\) such that \((L_{\alpha+1}\setminus L_\alpha)\cap\mathcal P(\omega)=\varnothing\).<ref name="MarekSrebrny73">W. Marek, M. Srebrny, "[https://www.sciencedirect.com/science/article/pii/0003484374900059 Gaps in the Constructible Universe]" (1973). Annals of Mathematical Logic vol. 6, pp.359--394.</ref><sup>p.364</sup> An ordinal \(\alpha\) is said to start a gap if \(\alpha\) is a gap ordinal but for all \(\beta<\alpha\), \((L_\alpha\setminus L_\beta)\cap\mathcal P(\omega)\neq\varnothing\).<ref name="MarekSrebrny73" /><sup>p.368</sup>

Gap ordinals are very large. This is because, if \( \alpha \) is a gap ordinal, then \( L_\alpha \cap \mathcal{P}(\omega) \) satisfies second-order arithmetic, despite not containing ''all'' subsets of \( \omega \). Therefore, if \( \alpha \) is a gap ordinal, it is admissible, recursively inaccessible, recursively Mahlo, nonprojectible, and more. However, there can still be countable gap ordinals.

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==
==Longer gaps==
Given any ordinal \(\gamma<\omega_1^L\), it is possible to find an \(\alpha<\omega_1^L\) such that \((L_{\alpha+\gamma}\setminus L_\alpha)\cap\mathcal P(\omega)=\varnothing\).<ref name="MarekSrebrny73" /><sup>p.365</sup>
Given any ordinal \(\gamma<\omega_1^L\), it is possible to find an \(\alpha<\omega_1^L\) such that \((L_{\alpha+\gamma}\setminus L_\alpha) \cap \mathcal{P}(\omega) = \emptyset\).<ref name="MarekSrebrny73" /><sup>p.365</sup> If such an \( \alpha \) starts a gap, then it is said to start a gap of length \( \gamma \). It is possible for \( \alpha \) to start a gap of length \( > \alpha \): for example, the least \( \alpha \) so that \( \alpha \) starts a gap of length \( \alpha^+ \) is equal to the least admissible which is not locally countable.

There can also be second-order gap, and more. An ordinal \( \alpha \) is said to start an \( \eta \)th-order gap of length \( \gamma \) if \( (L_{\alpha+\gamma} \setminus L_\alpha) \cap \mathcal{P}^\eta(\omega) = \emptyset \) and, for all \( \beta < \alpha \), \((L_\alpha \setminus L_\beta) \cap \mathcal {P}^\eta(\omega) \neq \emptyset\). The least ordinal which starts a second-order gap is greater than the least \( \alpha \) which starts a first-order gap of length \( \alpha \), and more.

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 \( \omega_1 \).

Revision as of 14:00, 30 August 2023

A gap ordinal is an ordinal \(\alpha\) such that \((L_{\alpha+1}\setminus L_\alpha)\cap\mathcal P(\omega)=\varnothing\).[1]p.364 An ordinal \(\alpha\) is said to start a gap if \(\alpha\) is a gap ordinal but for all \(\beta<\alpha\), \((L_\alpha\setminus L_\beta)\cap\mathcal P(\omega)\neq\varnothing\).[1]p.368

Gap ordinals are very large. This is because, if \( \alpha \) is a gap ordinal, then \( L_\alpha \cap \mathcal{P}(\omega) \) satisfies second-order arithmetic, despite not containing all subsets of \( \omega \). Therefore, if \( \alpha \) is a gap ordinal, it is admissible, recursively inaccessible, recursively Mahlo, nonprojectible, and more. However, there can still be countable gap ordinals.

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.[2]

Longer gaps

Given any ordinal \(\gamma<\omega_1^L\), it is possible to find an \(\alpha<\omega_1^L\) such that \((L_{\alpha+\gamma}\setminus L_\alpha) \cap \mathcal{P}(\omega) = \emptyset\).[1]p.365 If such an \( \alpha \) starts a gap, then it is said to start a gap of length \( \gamma \). It is possible for \( \alpha \) to start a gap of length \( > \alpha \): for example, the least \( \alpha \) so that \( \alpha \) starts a gap of length \( \alpha^+ \) is equal to the least admissible which is not locally countable.

There can also be second-order gap, and more. An ordinal \( \alpha \) is said to start an \( \eta \)th-order gap of length \( \gamma \) if \( (L_{\alpha+\gamma} \setminus L_\alpha) \cap \mathcal{P}^\eta(\omega) = \emptyset \) and, for all \( \beta < \alpha \), \((L_\alpha \setminus L_\beta) \cap \mathcal {P}^\eta(\omega) \neq \emptyset\). The least ordinal which starts a second-order gap is greater than the least \( \alpha \) which starts a first-order gap of length \( \alpha \), and more.

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 \( \omega_1 \).

  1. 1.0 1.1 1.2 W. Marek, M. Srebrny, "Gaps in the Constructible Universe" (1973). Annals of Mathematical Logic vol. 6, pp.359--394.
  2. R.Björn Jensen, The fine structure of the constructible hierarchy, Annals of Mathematical Logic, Volume 4, Issue 3, 1972, Pages 229-308