# Is there a “hereditary” construction for $L$?

Recall that $L$, Godel's constructible universe is constructed by defining the following hierarchy:

$L_0=\varnothing$, for a limit ordinal $\delta$, $L_\delta=\bigcup_{\alpha<\delta}L_\alpha$, and $L_{\alpha+1}=\operatorname{Def}^1(\langle L_\alpha,\in\rangle)$ (where the $^1$ denotes first-order definability).

On the other end of spectrum, we define $\sf HOD$ the class of hereditary ordinal-definable sets as the class of sets which are ordinal definable, i.e. there is some formula $\varphi(x,\alpha_1,\ldots,\alpha_n)$ such that only a single $x$ satisfies with a given sequence of ordinal parameters, and that their transitive closure is made entire of ordinal definable sets.

But Scott and Myhill showed that if you replace $\operatorname{Def}^1$ by $\operatorname{Def}^2$ (second order definability) in the constructible hierarchy, you get $\sf HOD$ as well. So we can construct $\sf HOD$ in two ways:

1. Take a "definability closure" for second-order logic.
2. Define a property, then consider those which have this property in a hereditary fashion.

Question. Is there some natural definable class which is similar to $\sf OD$, that when taking the "hereditary" subclass we get $L$?

• I’m shooting in the dark, but how do hereditary ordinal $\Sigma_1$-definable sets look like? – Emil Jeřábek Aug 21 '14 at 19:05
• Emil, what precisely do you mean by that? – Asaf Karagila Aug 21 '14 at 19:07
• @EmilJeřábek The class of hereditarily ordinal $\Sigma_1$-definable sets can be more than $L$. For example, start in $L$ and let $T$ be a Suslin tree with the unique branch property (see jdh.hamkins.org/degreesofrigidity), which means that if we force to add an $L$-generic branch $b$, then $L[b]$ has only one branch though $T$. In $L[b]$, the branch $b$ is hereditarily ordinal $\Sigma_1$-definable, since $T$ is ordinal $\Sigma_1$-definable (being in $L$) and $b$ is the unique branch through it. So in $L[b]$, we get $\text{HOD}_{\Sigma_1}=L[b]$. – Joel David Hamkins Aug 21 '14 at 20:27
• Well, $\Sigma_n$-definable elements are $\Delta_n$-definable by the usual argument, aren’t they? – Emil Jeřábek Aug 21 '14 at 21:18
• @Joel: Yes, I meant definability in $V$. The argument, which works for arbitrary first-order structures, is that if $a$ is definable by a formula $\phi(x)$, it is also definable by the formula $\forall y\,(\phi(y)\to y=x)$. – Emil Jeřábek Aug 22 '14 at 10:45