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There is a standard notation $\mathrm{ZF}[n]$ for Zermelo Fraenkel set theory with the power set axiom restricted to saying the set of natural numbers has $n$ successive power sets $\beth_0\dots\beth_n$.

Is there a similarly standard notation for the extension of $\mathrm{ZF}[n]$ by an axiom saying every set has an hereditary embedding in $\beth_n$?

flag	asked Feb 21 at 16:54 Colin McLarty 1,959●2●14

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I had not known that ZF[$n$] is standard notation for this theory. In fact, if someone had used this notation without any explanation, my first guess would be that it means ZF with replacement limited to $\Sigma_n$ formulas. – Andreas Blass Feb 23 at 14:07

I suspect the notation $\mathrm{ZF}[n]$ comes from Harvey Friedman, in the context of saying that theory has the strength of order n+2 arithmetic. That is where I got it. I have seen it, likely on FOM, but I can't search it by Google since Google refuses to believe I want the square brackets! It gives me pages where virtually every variant possible of ZF by dropping some axiom has been named with a 0 somehow. – Colin McLarty Feb 23 at 14:35

Ali Enayat · Accepted Answer · 2013-02-23 02:05:44Z UTC

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In modern set theory $H(\kappa)$ denotes the collection of sets $X$ such that $X$ is hereditarily of cardinality strictly less than $\kappa$.

So, the statement you are asking for can be conveniently expressed as $V=H(\kappa)$, where $\kappa=\beth_{n}^+$.

answered Feb 23 at 2:05

Ali Enayat
4,459●1●9●23

That could work as a notation. But it seems odd to invoke cardinal successor in a context where we do not have choice, and especially to invoke the cardinal successor of a set, $\beth_n$, which the theory proves cannot have one. – Colin McLarty Feb 23 at 12:57

Notation for upperbound power sets.

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