# Cardinality of $H(\kappa)$

Again I have trouble with some exercises in Kunen's set theory.

In the following, let $\kappa > \omega$ a cardinal. Then I want to show that

1) $|H(\kappa)| = 2^{<\kappa}$

2) $H(\kappa)=R(\kappa)$ if and only if $\kappa = \beth_{\kappa}$

In 1) the inequality $\geq$ is easy, but I cannot prove $\leq$. In 2) I have no idea. It is always true that $H(\kappa) \subseteq R(\kappa)$.

Can you give me a hint?

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1) To show that $|H(\kappa)| \leq 2^{<\kappa}$, we need to code elements of $H(\kappa)$ using subsets of ordinals $< \kappa$.

Let $x \in H(\kappa)$ and let $a$ be the transitive closure of $\{x\}$. Fix a bijection $f:|a|\to a$ with $f(0) = x$ and let $E \subseteq |a|^2$ be defined by $\xi \mathrel{E} \zeta$ iff $f(\xi) \in f(\zeta)$. Then $x$ is recovered as the value of $0$ in the transitive collapse of $(|a|,E)$. The pair $(|a|,E)$ can be encoded as a subset of $|a|^2 < \kappa$. This encoding process shows that $|H(\kappa)| \leq 2^{<\kappa}$.

2) To see that $H(\kappa) \subseteq R(\kappa)$, note that the rank of an element $x \in H(\kappa)$ is smaller than $|a|^+$, where $a$ is once again the transitive closure of $\{x\}$. Since $|R(\kappa)| = \beth_\kappa$, we see that $H(\kappa) = R(\kappa)$ can only hold when $2^{<\kappa} = \beth_\kappa$. Note that $2^{\lambda} < \beth_\kappa$ for every $\lambda < \beth_\kappa$. Therefore, $H(\kappa) = R(\kappa)$ entails that $\kappa = \beth_\kappa$.

To see that $\kappa = \beth_\kappa$ entails that $H(\kappa) = R(\kappa)$ it suffices to check that every element of $R(\kappa)$ has transitive closure of size less than $\beth_\kappa$. In fact, this holds for every limit ordinal $\kappa$: if $x \in R(\kappa)$ then $x \in R(\alpha)$ for some $\alpha < \kappa$ and hence the transitive closure of $x$ has size at most $|R(\alpha)| \leq \beth_\alpha < \beth_\kappa$.

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Thanks. This answers 1). –  Martin Brandenburg Sep 27 '10 at 12:37
@Martin: I just expanded the hint for part 2. –  François G. Dorais Sep 27 '10 at 13:20
Ah ok, thanks. Little correction: $|R(\omega+\alpha)|= \beth_{\alpha}$ (this does not really change the arguments). –  Martin Brandenburg Sep 27 '10 at 15:37

I do not know if I am right, but I think it is possible to proove that if $k$ is regular then :

giving $R_{\alpha}=\cup_{\beta < \alpha} P_{<k}(R_{\beta})$ (with $R_0=\emptyset$) then $H(k)=\cup_{\alpha \in ORD} R_{\alpha}=\cup_{\alpha < k} R_{\alpha}$

where $P_{<k}$ is the set of all subset of cardinality $< k$

At least I proved it someday, but nobody checked if I was right... (and I am often wrong ;) )

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