What does $H(\kappa)$ mean?

Question

As the typical references (Wikipedia, Mathworld, etc.) don't seem to address this satisfactorily, I figured this would be a good place to put a nice formal definition. Hence:

I've heard that if $\kappa$ is a strongly inaccessible cardinal, then $H(\kappa)$ (or sometimes $H_\kappa$) equals $V_\kappa$. What does $H(\kappa)$ mean in this instance and how is it defined?

Answer 1

The elements of $H(\kappa)$ are the sets that are hereditarily of cardinality less than $\kappa$. If $x\in H(\kappa)$, then $|x|<\kappa$, $|y|<\kappa$ for every $y\in x$, $|z|<\kappa$ whenever there are $x$ and $y$, such that $z\in y\in x$, and so on.

This gives the intuitive idea, but it’s not really a definition. For that it’s easiest to start by defining the transitive closure of a set $x$:

$$\operatorname{tr cl}(x)=\bigcup_{n\in\omega}{\bigcup}^n(x)\;,$$

where

$${\bigcup}^n(x)=\begin{cases} x,&\text{if }n=0\\\\ {\bigcup}\left({\bigcup}^{n-1}(x)\right),&\text{if }n>0\;. \end{cases}$$

Then $$H(\kappa)=\{x:|\operatorname{tr cl}(x)|<\kappa\}\;.$$

(Although it doesn’t have the form required by the axiom schema of comprehension, this definition can be justified by showing that $H(\kappa)\subseteq V_\kappa$ for every infinite cardinal $\kappa$.)

Assuming AC, $H(\kappa)=V_\kappa$ iff either $\kappa=\omega$ or $\kappa$ is a beth fixed point; this is proved in the answer by François G. Dorais to this question. (Beth fixed points exist in ZFC.)

Answer 2

Given an infinite cardinal $\kappa$, by $H(\kappa)$ we denote the family of all sets hereditarily of cardinality less than $\kappa$. Of course, this might beg the question: What do we mean by hereditarily of cardinality less than $\kappa$?

Well, in order for $x$ to be hereditarily of cardinality less than $\kappa$ we demand that

• $x$ itself has cardinality less than $\kappa$; and
• every element of $x$ has cardinality less than $\kappa$; and
• every element of every element of $x$ has cardinality less than $\kappa$; and
• $\ldots$

I think you get the picture. More succinctly, $x \in H(\kappa)$ iff its transitive closure, $\mathop{TC}(x)$, has cardinality $< \kappa$. This set is defined to be either the smallest transitive set including $x$, or, equivalently, to be $\bigcup_{n \in \omega} \bigcup^{(n)} x$ where for each $n \in \omega$ we inductively define $\bigcup^{(n)}x$ by

• $\bigcup^{(0)} x = x$; and
• $\bigcup^{(n+1)} x = \bigcup \left( \bigcup^{(n)} x \right)$.

Answer 3

I just thought I'd leave an alternate version of transitive closure in case readers might find it more intuitive/useful.

The transitive closure of a set $A$ is defined as $$\begin{align}\operatorname{cl}_0(A)&:=A\\ \operatorname{cl}_{n+1}(A)&:=\bigcup\operatorname{cl}_n(A)\\ \operatorname{cl}(A)&:=\bigcup_{n<\omega}\operatorname{cl}_n(A).\end{align}$$

Note:

• $A=\operatorname{cl}_0(A)\subseteq\operatorname{cl}(A)$
• If $A\subseteq B$ and $B$ is transitive then $\operatorname{cl}(A)\subseteq B$