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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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@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)? – Travis Bemrose Aug 25 '12 at 22:25
$H(\kappa)$ 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]. – Asaf Karagila Aug 25 '12 at 22:28
Ah. I've been stuck in large-cardinal land for the past several days. Good catch. – Travis Bemrose Aug 25 '12 at 23:43
Large cardinals are fun. – Asaf Karagila Aug 26 '12 at 0:10
up vote 7 down vote accepted

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 $x$:

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


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

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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@tomasz: Probably a good idea, yes; done. – Brian M. Scott Aug 25 '12 at 19:00
@BrianM.Scott, I'm new to TeX. Is there a way to add more space between the two rows of your piecewise definition for ${\bigcup}^n(x)$? – Travis Bemrose Aug 25 '12 at 21:21
@Travis: Is that better now? – Brian M. Scott Aug 25 '12 at 21:27
@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. – Travis Bemrose Aug 25 '12 at 21:40
@Travis: You can also right-click on a formula and select Show Math As and then TeX Commands from the context menu. (Or click on the time stamp where it says ‘edited 15 mins ago’ to see the revision history, though in general I find that less useful.) – Brian M. Scott Aug 25 '12 at 21:45

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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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. – Daniel Gerigk yesterday

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 $$\begin{align}\operatorname{cl}_0(A)&:=A\\ \operatorname{cl}_{n+1}(A)&:=\bigcup\operatorname{cl}_n(A)\\ \operatorname{cl}(A)&:=\bigcup_{n<\omega}\operatorname{cl}_n(A).\end{align}$$


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