Constructible hierarchy: Difference between revisions
Jump to navigation
Jump to search
no edit summary
RhubarbJayde (talk | contribs) No edit summary |
|||
(2 intermediate revisions by one other user not shown) | |||
Line 2:
== Definition ==
Say a subset \(X\) of \(Y\) is definable if there are some \(z_0, z_1, \cdots, z_n \in Y\) and some formula \(\varphi\) in the language of set theory so that the elements of \(X\) are precisely the \(x\) so that \(Y\) satisfies \(\varphi(x, z_0, z_1, \cdots, z_n)\). For example, under the von Neumann
Like with the von Neumann hierarchy, the constructible hierarchy is built up in stages, denoted \(L_\alpha\).<ref>K. J. Devlin, "[https://core.ac.uk/download/pdf/30905237.pdf An introduction to the fine structure of the constructible hierarchy]" (1974)</ref>
Line 10:
* For limit \(\alpha\), \(L_\alpha\) is the union of \(L_\beta\) for \(\beta < \alpha\).
Note that this is a cumulative hierarchy, and thus the [[reflection principle]] applies.
This is always contained in the respective rank of the von Neumann hierarchy: \(L_\alpha \subseteq V_\alpha\). This can be shown by a transfinite induction argument. It initially completely actually agrees with \(V\): all subsets of a finite set are definable, therefore \(L_\alpha = V_\alpha\) for \(\alpha \leq \omega\). However, while \(V_{\omega+1}\) is uncountable, there are (as we mentioned) only countably many subsets of a countable subset, and thus \(L_{\omega+1}\) is countable and a proper subset of \(V_{\omega+1}\). In general, \(|L_\alpha| = |\alpha|\) for \(\alpha \geq \omega\).<ref>Most set theory texts</ref>
|