# About the notions of Grothendieck Universe and Tarski Universe

I assume ZFC.

Let $U$ a set with the following (1), (2), (3):

1) $\omega\in U$

2) $x\in U\ \Rightarrow x\subset U$

3) $x\in U\ \Rightarrow \mathcal{P}(x)\in U$ (where $\mathcal{P}(x):=${$y| y\subset x$})

Then by these premises, I wish prove that $(4)\Leftrightarrow (4')$ where:

4) if $x\subset U$ and $|x|<|U|$ then $x\in U$ (where $|a|$ is the cardinality of the set $a$)

4') If $f: a\to U$ is function and $a\in U$ then $\bigcup_{s\in a}f(s)\in U$ .

The Book of Monk "Introduction to set theory' claim this equivalence as a exercise.

and the implication $(4)\Rightarrow (4')$ is immediate from the book above..

I tried to prove by induction that $|U|\subset U$, or tried to generalize the Mostowsky theorem for get a inijection $|U|\to U$ which preserves the relation '$\in$' (unsuccessfully). I have see also some posts here about this problem, but I hope exist a (relatively) simple answere..

THen I ask: How to prove $(4')\Rightarrow (4)$ ?

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Why are you trying to prove $|U| \subset U$? –  David Roberts Jul 4 '12 at 23:45
Property 4 appears not to follow. Assume GCH; then $U:=V_{\aleph_\omega}$ satisfies (1)-(3), but the set $\{\aleph_n: n\in\omega\}$ is a set with cardinality $<\vert U\vert$ but not in $U$. Is there more to the exercise? –  Noah S Jul 5 '12 at 2:24
This also provides a counterexample to 4'. –  Noah S Jul 5 '12 at 2:25
A simpler counterexample to (4) and (4') is $V_{\omega+\omega}$. But the question is not to infer (4) and/or (4') from (1,2,3) but rather to infer that (4) is equivalent to (4'). –  Andreas Blass Jul 5 '12 at 4:27
David ROberts: From $|U|\subset U$ if $|x|<|U|$ i.e. $|x|\in |U|$ follow $|x|\in U$ and from (4') follow $x\in U$. –  Buschi Sergio Jul 5 '12 at 5:29

I don't see a really slick proof, so here's a rough (and fairly standard) one. Assume that $U$ satisfies (1,2,3,4'). Notice that, by (2) and (3), $U$ is closed under subsets, so the cardinalities of elements of $U$ form an initial segment of the cardinal numbers; let $\kappa$ be the smallest cardinal not in this initial segment. It follows from (1) that $\kappa$ is uncountable, from (3) that whenever $\mu<\kappa$ then $2^\mu<\kappa$, and from (4') that $\kappa$ is a regular cardinal. So $\kappa$ is strongly inaccessible. If $U$ contained any sets of rank $\geq\kappa$, then the lowest-rank such sets would, by (2), have rank exactly $\kappa$; but this is impossible, because (by regularity of $\kappa$) any set of rank exactly $\kappa$ must have cardinality at least $\kappa$. The conclusion, therefore, is that all sets in $U$ have rank $<\kappa$. With the usual notation for the levels of the cumulative hierarchy, we have $U\subseteq V_\kappa$.
Next, notice that for all $\alpha<\kappa$, $V_\alpha$ has cardinality $<\kappa$. (This is proved by induction on $\alpha$, using at successor stages that $\mu<\kappa$ implies $2^\mu<\kappa$, and using at limit stages that $\kappa$ is regular and uncountable.) Therefore, $V_\kappa$, being the union of $\kappa$ sets $V_\alpha$ each smaller than $\kappa$, has cardinality only $\kappa$. In view of the result in the preceding paragraph, we have $|U|\leq \kappa$.
This means that, if $x$ is as in the statement (4) that we want to prove, then $|x|<\kappa$, and therefore there exists some $a\in U$ with $|a|=|x|$. Fix such an $a$ and fix a bijection $f:a\to x$. Applying (4'), we find that the union $y$ of all the members of $x$ is an element of $U$. But each member of $x$ is a subset of $y$ and therefore an element of the power set of $y$. So we have $x\subseteq\mathcal P(y)$ and, by (3), $\mathcal P(y)\in U$. We already noted above that $U$ is closed under subsets, so it follows that $x\in U$, as desired.