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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 $(x)$....

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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.

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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

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