Rank of a set as the supremum of ranks its elements.

Question

In general, for a set $X$, why is it true that $\mathrm{rank}(X)=\sup\{\mathrm{rank}(y)+1\mid y\in X\}$?

The definition of rank I have is that $\text{rank}(x)=\text{ the least }\alpha\text{ such that }x\in V_{\alpha+1}$. I suppose these must be the same thing?

As @AsafKaragila said, this is generally the definition. If, by chance, it's not the definition you've been given, but stated somewhere as a consequence, you should provide the definition you have. Otherwise, the question would be better as something like "why does this definition make sense?". — Quinn Culver, Dec 12 '11 at 21:36

Asaf Karagila · Accepted Answer · 2013-03-06T08:29:12.773

10

This is the definition of the rank of the set $X$. $\newcommand{rank}{\operatorname{rank}}$

We use the fact that $\in$ is well-founded, and that the ordinals are well-ordered. These properties allow us to define recursively this function. $$\rank:V\to\operatorname{Ord}$$

And allows us to talk about the sets $V_\alpha=\{x\in V\mid \rank (x)<\alpha\}$, for an ordinal $\alpha$. This is known as the von Neumann hierarchy.

Given the definition of $\rank$ as the least $\alpha$ such that $x\subseteq V_\alpha$, we can show the equivalence by induction.

Suppose that for an ordinal $\alpha$ we have that if $\rank(y)<\alpha$ then the equivalence holds.

If $\sup\{\rank(y)+1\mid y\in x\}=\beta<\alpha$ then we have that $y\in x$ implies that $y\in V_\beta$ and therefore $x\in V_{\beta+1}\subseteq V_\alpha$ and thus $\rank(x)<\alpha$, and the equivalence holds.

Otherwise, $\sup\{\rank(y)+1\mid y\in x\}=\alpha$, and given $\beta<\alpha$ we can find $y\in x$ such that $y\notin V_\beta$. In particular this means that $x\nsubseteq V_\beta$ therefore $x\notin V_{\beta+1}$. And indeed $\alpha$ is the minimal ordinal such that $x\in V_{\alpha+1}$, as we wanted.

edited Mar 06 '13 at 08:29

answered Dec 12 '11 at 21:28

Asaf Karagila

393,674

Thanks for adding that part. – rankled Dec 13 '11 at 00:12
No problem, it is my pleasure! :-) – Asaf Karagila Dec 13 '11 at 00:16
In the first half of your answer: don't we need that the ordinals are a strict well-order? – Rudy the Reindeer Mar 06 '13 at 07:58
If the set that constitutes the range of the function is not a strict linear order the $\sup$-definition goes broken because the resulting function is not well-defined. – Rudy the Reindeer Mar 06 '13 at 07:59
@Matt, well obviously. :-) But if the order wasn't strict there was some $\alpha\lea\alpha$, which is $\alpha\in\alpha$ which means ${\alpha}$ has no $\in$-minimal element. – Asaf Karagila Mar 06 '13 at 08:04
But you wrote that we use the well-foundedness of the ordinals to define the function. – Rudy the Reindeer Mar 06 '13 at 08:06
@Matt: Yes. My point is that the order defined by $\in$, and therefore is irreflexive. So it has to be strict. – Asaf Karagila Mar 06 '13 at 08:07
It has got to be both: strict and linear. – Rudy the Reindeer Mar 06 '13 at 08:15
@Matt: No, it has to be strict and have a well-defined $\sup$ function. It happens that well-orders have this property. Are you suggesting that I should have prefaced my answer with a proof that ordinals are well-ordered? – Asaf Karagila Mar 06 '13 at 08:17
No you are writing that the set in the range has to be well-founded. I am saying it has to be well-ordered. – Rudy the Reindeer Mar 06 '13 at 08:23
@Matt: Maybe now we can end this futile back-and-forth, and return to less-futile discussions? – Asaf Karagila Mar 06 '13 at 08:29

Rank of a set as the supremum of ranks its elements.

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