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I would appreciate help understanding and substantiating the quote in the title which comes from Stillwell's "Real Numbers," page 143.

A definition of an inaccessible set can be found here: An example of a set that is not inaccessible

And $V_{\alpha}$ is defined:

$V_0= \emptyset$

$V_{\alpha+1}= P(V_{\alpha})$ where $P$ is the power set

$V_{\lambda}=\bigcup_{\beta\lt \lambda}V_{\beta}$ for each limit ordinal $\lambda$

Following the text:

As a prelude, it is shown that if $V_{\alpha}$ is inaccessible, then it satisfies all the ZF axioms.

Then taking the $least$ $\alpha$, such that $V_{\alpha}$ is inaccessible, it follows that any $V_{\beta}$ in $V_{\alpha}$ is $not$ inaccessible.

So $V_{\alpha}$ satisfies the statement: "There is no inaccessible $V_{\beta}$." Existence of an inaccessible set is therefore not provable in ZF.

I understand the assertion that $V_{\alpha}$ satisfies the statement: There is no inaccessible $V_{\beta}$.

But how does this establish that the existence of an inaccessible set is therefore not provable in ZF?

Unfortunately there is no discussion of logic in the text, so I am a bit stymied.

Also: Is it necessary to explicitly define $V_{\alpha}$ as was done above, or is it adequate to only state that it is inaccessible?

Thanks

EDIT This is the exact statement from Stillwell:

"In Sect. 3.8 we claimed that there are “largeness” properties so extreme that sets with those properties cannot be proved to exist. We suggested that one such “largeness” property is inaccessibility, where an inaccessible set is one that has infinite members and is closed under the operations of power set and taking ranges of functions. It should now be apparent that if Vα is an inaccessible set, then Vα satisfies the ZF axioms

"Certainly, if Vα is large enough to have an infinite member, then it satisfies the empty set and infinity axioms. It satisfies power set and replacement by the hypothesis of closure under power set and taking ranges of functions. Closure under power set also guarantees that α is a limit ordinal, in which case Vα is also closed under pairing and union, so Vα satisfies the pairing and union axioms. Finally, any Vα satisfies foundation, so Vα satisfies all the ZF axioms. It follows that Vα also satisfies any logical consequence of the ZF axioms; that is, any proposition provable in ZF set theory. But now suppose we take the least α such that Vα is inaccessible. It follows that any Vβ in Vα is not inaccessible, so Vα satisfies the sentence “there is no inaccessible Vβ .” Existence of an inaccessible set is therefore not provable in ZF.This explains the surprising claim made at the end of Sect. 3.8 : if inaccessible sets exist, then their existence is not provable in ZF."

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    The usual terminology is that it's $\alpha$ rather than $V_\alpha$ that is inaccessible. (Though I can see this is not the case in the question you link as source). – hmakholm left over Monica Jul 14 '17 at 15:10
  • As $V_\alpha$ is a model of ZF, any proof in ZF that inaccessibles exist would imply that there actually exists an inaccessible $\in V_\alpha$. – Hagen von Eitzen Jul 14 '17 at 15:10
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    Would you agree that the conclusion follows if ZF proves: "If an inaccessible set exists, then there is an ordinal $\beta$ such that $V_\beta$ is inaccessible"? – hmakholm left over Monica Jul 14 '17 at 15:17
  • @HenningMakholm Well, I certainly agree with the quote itself. Yet I am particularly ignorant about distinctions between consistency and proof. I definitely see the point you are making. –  Jul 14 '17 at 15:49
  • I think it's true that ZF proves "If an inaccessible set exists, then there is an ordinal $\beta$ such that $V_\beta$ is inaccessible" -- but actually producing such a proof from ZF (given the definition of "inaccessible" in your earlier question) seems to be surprisingly tricky. – hmakholm left over Monica Jul 14 '17 at 15:54
  • For example, it is not the case that if $Z$ is "inaccessible", then $Z$ has some inaccessible $V_\beta$ as a subset -- since an "inaccessible" $Z$ doesn't have to contain $\varnothing$. – hmakholm left over Monica Jul 14 '17 at 15:56
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    Stillwell is being very ideosyncratic here. I've only ever seen ordinals described as inaccessible - namely, an ordinal $\alpha$ is (strongly) inaccessible if it is regular and $\vert2^\beta\vert<\vert\alpha\vert$ whenever $\beta<\alpha$. (It's easy to show that all inacessible ordinals are cardinals, so we speak of inaccessible cardinals instead.) The set $V_\alpha$ is inaccessible in Stillwell's sense iff $\alpha$ is (strongly) inaccessible in the usual sense. Stillwell's broadening to arbitrary sets leads to the subtle mistake you've observed. – Noah Schweber Jul 14 '17 at 17:21

1 Answers1

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You're right to be confused. The text you quote from Stilwell sweeps several subtleties under the carpet.

First:

It follows that any $V_\beta$ in $V_\alpha$ is not inaccessible, so $V_\alpha$ satisfies the sentence “there is no inaccessible $V_\beta$.”

Here the point is to use $V_\alpha$ as a model of ZF, but the argument tacitly assumes that $V_\alpha$ agrees with us about which sets are inaccessible -- that is, when the definition of "inaccessible" is interpreted within $V_\alpha$, such that the quantifiers in it range over $V_\alpha$ only, then it yields the same outcome as asking whether a set (such as $V_\beta$) is actually inaccessible.

If this were not the case, it might be that there is some $V_\beta\in V_\alpha$ that we can see is not inaccessible, but $V_\alpha$ thinks it is. This would invalidate the argument that "ZF cannot prove so-and-so because it is not true in $V_\alpha$".

Similarly, $V_\alpha$ needs to agree with us about which sets have the form $V_\beta$ in the first place.

As far as I can see it is actually true that $V_\alpha$ will agree with us about those points, but proving this to be the case is rather more subtle (in addition to somewhat tedious) than Stillwell makes it out to be.

Second, and much worse:

Even so, what this argument shows is only that ZF cannot prove the sentence "there is a $V_\beta$ that is inaccessible".

If we want to argue that ZF cannot prove the sentence "there is some set that is inaccessible", then we need more than the argument presented here.

One way forward that suggests itself would be to show that ZF proves "if there is some inaccessible set, then there is a $\beta$ such that $V_\beta$ is inaccessible". If that is the case, then if ZF were to prove "there is some inaccessible set", then it would also prove "there is an inaccessible $V_\beta$", which we now know it doesn't.

However ZF does not in fact prove "if there is some inaccessible set, then there is a $\beta$ such that $V_\beta$ is inaccessible". Namely, ZFC proves that there exists an "inaccessible" set: $H_{\beth_\omega}$, the set of all sets hereditarily of cardinality less than $\beth_\omega$, satisfies all of Stillwell's conditions. If this implied the existence of an inaccessible $V_\beta$, then ZFC would prove its own consistency, which since Gödel we know it doesn't. (Note that $H_{\beth_\omega}$ does not satisfy the axiom of union!)

Therefore, a better explanation seems to be that Stillwell was expressing himself sloppily and was not actually interested in defining "inaccessible" as a property of arbitrary sets but only of sets of the particular form $V_\beta$. In that case the "surprising claim" he's really aiming at is just that ZF cannot prove "there is an inaccessible $V_\beta$" even if that happens to be true.

(As Noah Schweber points out in a comment, it is not usual at all to use "inaccessible" about arbitrary sets. What one usually does is to define "inaccessible" to be a certain property of ordinals -- then $\alpha$ is called inaccessible iff $V_\alpha$ is "inaccessible" in Stillwell's sense. It is possible that Stillwell hadn't fully thought through his broadening of the concept to general sets).

hmakholm left over Monica
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