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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?

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  • $\begingroup$ @Asaf: Why did you change the (large-cardinals) tag to (cardinals)? Since $\kappa$ is a strongly inaccessible cardinal, which ZFC can't prove exists, doesn't that meet the criteria for (large-cardinals)? $\endgroup$ – Travis Bemrose Aug 25 '12 at 22:25
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    $\begingroup$ $H(\kappa)$ is not reserved for inaccessible cardinals. In fact, when talking about inaccessible cardinals this is the same as talking about $V_\kappa$. On the other hand, $H(\omega_1)$ is used throughout set theory, as well $H(\kappa)$ for other non-large cardinals. Since you only asked about the definition I don't see why it should be tagged as [large-cardinals]. $\endgroup$ – Asaf Karagila Aug 25 '12 at 22:28
  • $\begingroup$ Ah. I've been stuck in large-cardinal land for the past several days. Good catch. $\endgroup$ – Travis Bemrose Aug 25 '12 at 23:43
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    $\begingroup$ Large cardinals are fun. $\endgroup$ – Asaf Karagila Aug 26 '12 at 0:10
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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 $\kappa=\omega$ or $\kappa$ is strongly inaccessible.

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  • $\begingroup$ @tomasz: Probably a good idea, yes; done. $\endgroup$ – Brian M. Scott Aug 25 '12 at 19:00
  • $\begingroup$ @BrianM.Scott, I'm new to TeX. Is there a way to add more space between the two rows of your piecewise definition for ${\bigcup}^n(x)$? $\endgroup$ – Travis Bemrose Aug 25 '12 at 21:21
  • $\begingroup$ @Travis: Is that better now? $\endgroup$ – Brian M. Scott Aug 25 '12 at 21:27
  • $\begingroup$ @BrianM.Scott: Much easier to read. :-) I was struggling to see what change you made to the TeX, but then I figured out that somehow with SE's dynamic updating, the page had refreshed but view-source was still giving me the old source until I did a manual refresh. $\endgroup$ – Travis Bemrose Aug 25 '12 at 21:40
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    $\begingroup$ To conclude that the transitive closure of a set $x$ that is hereditarily of cardinality $< \kappa$ is of cardinality $< \kappa$, you need to assume that the cardinal $\kappa$ is regular, I think. So your definition of $H(\kappa)$ is only "correct" for regular $\kappa$. $\endgroup$ – Daniel Gerigk Jul 26 '16 at 12:57
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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
  • $\bigcup^{(n+1)} x = \bigcup \left( \bigcup^{(n)} x \right)$.
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    $\begingroup$ To conclude that the transitive closure of a set $x$ that is hereditarily of cardinality $< \kappa$ is of cardinality $< \kappa$, you need to assume that the cardinal $\kappa$ is regular, I think. $\endgroup$ – Daniel Gerigk Jul 26 '16 at 12:43
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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 $\operatorname{cl}(A)\subseteq B$
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