Constructible hierarchy: Difference between revisions

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 interpretationdefinition of ordinal, the set of even numbers, the set of odd numbers, the set of prime numbers, the set of perfect squares greater than 17, and so on, are all definable. Using elementary cardinal arithmetic, youthere can note that thereare \(\max(\aleph_0, |Y|) = |Y|\) definable subsets of an infinite set \(Y\), and thus "almost all" subsets of an infinite set are not definable. The parameters \(\vec{z}\) aren'tare of importance inwhen the\(Y\) contextis ofuncountable, definableto subsetsensure ofthat thethere naturalare numbers,more sincethan all\(\aleph_0\) elementsdefinable subsets of the\(Y\), naturalbut numbersthey aredo definable,not buthave theyany willeffect be ifwhen \(Y\supseteq\mathbb N\) is uncountablecountable, becausesince noall uncountableelements beof pointwisethe definable,natural andnumbers ensure that there aren't just always \(\aleph_0\)are definable subsets of a set.

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>