-1
$\begingroup$

I read that if some property $P(x)$ defined by a formula $\phi$ is absolute for some class $M$, then $\phi(P(x)) \leftrightarrow \phi^M(P(x))$, where $\phi(P(x))$ is interpreted in $V$, and $\phi^M(P(x))$ is relativized to $M$, which is a subclass of $V$.

In most texts in set theory (i.e. proof that $R$ (axiom of regularity) is consistent with $ZF-R$ or that $ZF$ is consistent with the axiom of constructibility), a lot of pages are devoted to showing that some property $P$ is absolute (i.e. that it is definable by a $\Delta_0$ formula for transitive models, or by a $\Delta_1$ formula for inner models - as defined in Jech (2006))...

But just to help me see the forest from the trees - I have two beginner-level questions, and I really appreciate the very helpful community here.

  1. The reason why we need to show that $P$ is absolute is because by default (or by assumption), our "domain of discourse" is $V$ right ? - so all formal proofs are done in $V$, and we need to take the extra step that we think is true in $V$ is also true in its subclass $M$.

  2. From what I understand, (1) above refers to "downward-absolute", i.e. what is true in the higher levels of the Levy heirarchy (particularly in $V$) also holds in the lower level $M$ ... however, for a property to be absolute in say, an inner model $M$, it has to be $\Delta_1$, which is both downward-absolute ($\Pi_1$) and upward-absolute $(\Sigma_1$). But if our universe of discourse is $V$, if we get to prove some property inside $V$ shouldn't we be concerned with downward-absolute only since we know it would hold in $M$ - why is there need for it to be upward-absolute ($\Sigma_1$) as well ?

$\endgroup$

1 Answer 1

3
+150
$\begingroup$

Both of your questions seem premised on this misconception: We don't need to prove a given property is absolute (upwards or downward) any more than we need to prove the Riemann hypothesis or that there are infinitely many primes. Some properties are absolute, and that can be useful, particularly when we're trying to analyzed relativized statements between transitive models. If it's true for some important property, we should prove it and make use of that fact. If it's not absolute, that's fine too... that can also be useful and interesting. The zoo of absoluteness statements proved in set theory books are for general purposes, not necessarily all necessary to use for the next theorem.

One of the most useful nontrivial properties that is absolute both ways is the property of being an ordinal. In other words, for any class $M$ such that ZF proves $M$ is transitive, and some modest finite fragment of ZF, relativized to $M$, ZF proves $$\forall x\in M((\mbox{$x$ is an ordinal})^M\leftrightarrow \mbox{$x$ is an ordinal})$$ (note however this does not hold for ZF - Regularity). We, use this, as well as more basic things like that the notion of subset, pair, union, etc. (but crucially not power set) are absolute for transitive models all the time, and it gets to the point of being unconscious.

An example of a notion that absolute downward but not absolute upward is being a cardinal. An example of one that is absolute upward but not downward absolute downward is being constructible.

We'll use the first example frequently when studying $L$ (or other inner models or transitive submodels)... it's useful to know that $\omega_1$ is definitely a cardinal in $L$ (though it's not necessarily the second infinite one... an interesting consequence of the lack of upward absoluteness that is worthy of exploration). And despite the second example I gave there, one example that really sticks out in highlighting downward absoluteness's usefulness as a concept for proving relativized statements comes from the constructible hierarchy. The statement that $L$ satisfies $V=L$ is just $$ \forall x(x\in L\to (\mbox{$x$ is constructible})^L)$$ and since $x\in L$ is just an abbreviation for "$x$ is constructible", this is exactly a statement of downward absoluteness of constructibility.

As mentioned, we can't prove this by showing constructibility is $\Pi_1,$ because it's not and there are instances of failure of downward absoluteness to classes that aren't $L$. The crucial feature of $L$ used here (in addition to the usual transitivity and satisfying some fragment of ZF) is that it contains all the ordinals. This works because the relation $x\in L_\alpha$ between a set $x$ and an ordinal $\alpha$ is absolute (both ways) and "$x$ is constructible" means $\exists \alpha\in Ord\; x\in L_\alpha$... so since Ord is absolute, constuctibility is absolute to inner models (i.e. transitive classes containing all ordinals). Where it can fail is in relativizing to transitive sets, where there might be a constructible set that the set doesn’t know is constructible because the witnessing level of the hierarchy is outside it.

As for the importance of upward absoluteness, we use the upward absoluteness of constructibility all the time. For instance, it's very useful to know that any inner model that is not $L$ does not satisfy $V=L,$ which is a consequence of upward absoluteness (coupled with the minimality of $L$).

(On a minor note, I think you have a misconception on what the Levy hierarchy is. It’s not a hierarchy of bigger and smaller models. The levels of the Levy Hierarchy are the definability classes $\Sigma_n$, $\Pi_n$, and $\Delta_n$.)

$\endgroup$
3
  • $\begingroup$ thank you ! I accepted your answer because it is comprehensive. I was re-reading Jech and $V=L$ holds in $L$ because $L$ has the ordinals as you pointed out. But just one more little clarification, suppose that we assume that $L$ has the ordinals. If I manage to show that the function $\alpha \rightarrow L_{\alpha}$ is downward-absolute (i.e. $\Pi_1$), is this enough to show that $V=L$ holds in $L$ ? $\endgroup$
    – Link L
    Commented May 22 at 6:36
  • 1
    $\begingroup$ @LinkL Yes it is. $\endgroup$ Commented May 22 at 15:11
  • $\begingroup$ thank you ! I had the impression that both upward and downward absoluteness are necessary to show that $V=L$ in $L$ as Jech used it in his book - which did not match my intuition. but your tips showed me the light $\endgroup$
    – Link L
    Commented May 23 at 0:21

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .