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I have some trouble with hereditarily countable sets.

Let $H_\kappa=\{x:|trcl(x)|< \kappa\}.$ Prove the following.

a) For every infinite cardinal $\kappa$, $H_\kappa$ is transitive.

b) For every infinite cardinal $\kappa$, $H_\kappa \cap ORD=\kappa$, where $ORD=\{x:x$ is an ordinal$\}$

c) For every infinite cardinal $\kappa$, $H_\kappa \subseteq V_\kappa$, where $V_\alpha$ is defined as follows :$\alpha$ is an ordinal $V_0=\emptyset, V_\beta = \mathcal P(V_\alpha)$, and if $\gamma$ is limit ordinal, $V_\gamma=\cup\{V_\alpha:\alpha<\gamma\}$.

d) For every regular uncountable cardinal $\kappa$, $H_\kappa = V_\kappa$ iff $\kappa$ is strongly inacessible.

This is my working so far.

For a). Let $z \in x$. Note that $|trcl(z)|< |trcl(x)| <\kappa$. Thus $z \in H_\kappa$ and hence $x \subseteq H_ \kappa$. Thus $H_\kappa$ is transitive. (Is this correct?)

For b). $ORD$ is transitive and since $x \subseteq ORD$ then $ trcl(x) \subseteq ORD$. Then I am stuck here.

I am stuck for c) and d) too. Help would be greatly appreciated.

taupi
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  • I have an initial problem with this question. It's not at all obvious to me that $H_{\kappa}$ is actually a set, and that definition of $V_{\kappa}$ definitely isn't a set. So something is missing here. – Robert Shore Sep 19 '19 at 01:34
  • Oh yes sorry about that. I got confused about the notation. I have edited it above. Thanks for pointing that out. – taupi Sep 19 '19 at 02:00
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    @RobertShore I guess (c) is the easiest way to see it is a set. – spaceisdarkgreen Sep 19 '19 at 17:23

1 Answers1

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(a) is correct except that it should be $|\operatorname{trcl}(z)|\le |\operatorname{trcl}(x)|.$

(b) If $\alpha$ is an ordinal then $\alpha$ is transitive, so $\operatorname{trcl}(\alpha)=\alpha.$

(c) Prove that for any $\alpha < \operatorname{rank}(x)$ there is a $y\in \operatorname{trcl}(x)$ with $\operatorname{rank}(y)=\alpha.$ It follows that $|\operatorname{rank}(x)|\le|\operatorname{trcl}(x)|.$

(d) If $\kappa$ is not a strong limit, it's easy to find an set in $V_\kappa\setminus H_\kappa.$ For the converse, show that if $\kappa$ is strongly inaccessible, $|V_\alpha|<\kappa$ for every $\alpha<\kappa.$

spaceisdarkgreen
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