Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

The class of all hereditarily countable sets can be proven to be a set from the axioms of Zermelo–Fraenkel set theory (ZF) without any form of the axiom of choice, and this set is designated . The hereditarily countable sets form a model of Kripke–Platek set theory with the axiom of infinity (KPI), if the axiom of countable choice is assumed in the metatheory. If $x \in H_{\aleph_1}$, then $L_{\omega_1}(x) \subset H_{\aleph_1}$.

What is $L_{\omega_1}(x)$? I heard of constructible universe, but that did not contain a thing like $(x)$....

share|cite|improve this question

When you see $L(x)$ or $L_\alpha(x)$, you’re dealing with relative constructibility. Instead of starting from the empty set ($L_0=\varnothing$), you start from $TC(x)$, the transitive closure of $x$ ($L_0(x)=TC(x)$). Then you proceed just as in the construction of the constructible hierarchy: $L_{\alpha+1}(x)$ is the set of first-order definable subsets of $L_\alpha(x)$, and $L_\alpha(x)=\bigcup_{\xi<\alpha}L_\xi(x)$ for limit $\alpha$. The last statement that you quote says that if $x$ is hereditarily countable, every set constructible from $x$ in at most countably many steps is still hereditarily countable.

share|cite|improve this answer
The last sentence should probably say "constructible from $x$..." – Trevor Wilson Sep 18 '12 at 5:03
@Trevor: I was using definable informally, but you’re right, I probably oughtn’t. – Brian M. Scott Sep 18 '12 at 5:11

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.