-2
$\begingroup$

Sometimes when one tries to capture the abstract aspect of some notion that is intuitively considered as being truly of abstract nature, in set theoretic terms, then this can extend the theory in a huge manner!

An obvious example of that is Mathias principle about ordered pairs, p2, see also: kanamori, p54, which simply states that the particular set theoretic implementation of "ordered pair" must not matter as far as using them to represent relations in set theoretic terms as sets of those ordered pairs, so any function that satisfy the characteristic property of ordered pairs can be used in representing relations, and so stipulate the existence of a Cartesian product between any two sets, defined in terms of any such function.

This captures the notion of relations being abstract entities.

When this principle is added to Zermelo, it blows it up to ZF.

Here I'll present a similar general line but of extending ZFC, by a principle the capture the abstract aspect of "hierarchy" by being indifferent to the indexing objects and indexing well ordering relation involved in the build up of the hierarchy.

So to ZFC add the following principle:

Principle of Indifferent Hierarchical Construction: If $\mathcal Q$ is a definable unary predicate and if $<^\mathcal Q$ [abbreviated hereafter as $<$] is some well ordering relation on sets satisfying $\mathcal Q$, then there is a hierarchical injective function $\mathcal H^{\mathcal Q, <}$ (abbreviated hereafter as $\mathcal H$), from sets satisfying $\mathcal Q$, defined recursively as:

$\mathcal H(i) = \emptyset \leftrightarrow i=min^< x: \mathcal Q(x)$

$\mathcal H (i+1) = \mathcal P (\mathcal H (i))$

$\mathcal H (i) = \bigcup (\{\mathcal H(j)|j<i\}) \leftrightarrow limit^<(i)$.

Where $i+1$ is the smallest (with respect to $<$) set $x$ fulfilling $\mathcal Q$ such that $i < x$.

The third condition stipulates that the set $\bigcup (\{\mathcal H(j)|j<i\})$ exists for every limit $i$ with respect to relation $<$.

This principle is simply stating that the cumulative hierarchy of sets can be built up using different indexing function as long as some well ordering is there. So it is indifferent to the particulars of sets indexing the hierarchy and the well ordering relation involved in doing so. In some sense this principle can be viewed as capturing the abstract concept of hierarchy!

However, this principle is not innocent as it may seem for the first glance, this principle actually PROVES the consistency of ZFC. Actually it is much stronger than ZFC.

To see that it implies the existence of an inaccessible, take $\mathcal Q$ to be the predicate Von Neumann ordinal and take $<$ to be the well ordering on Von Neumann ordinals defined as:

$\alpha < \beta \equiv_{df} \alpha> \emptyset \land (\beta > \alpha \lor \beta=\emptyset)$

Now according to the above principle $\mathcal H(\emptyset)$ must exist! And clearly this would be an inaccessible stage of the cumulative hierarchy.

Question 1: Is this principle inconsistent?

Question 2: If not, then what's its consistency strength?

$\endgroup$

1 Answer 1

2
$\begingroup$

This question can be more clearly phrased as:

Is the principle "We can iterate powerset along any (definable) class-well-ordering" (which is really a scheme, appropriately) consistent with ZFC?

Unless I'm missing something, the example you give in fact demonstrates inconsistency: $\mathcal{H}(0)$ would have to contain every set since we'll have $V_\alpha=\mathcal{H}(1+\alpha)\subseteq\mathcal{H}(0)$.

More generally, since there are definable class-well-orderings longer than $Ord$, there's no reason to believe we should be able to iterate a given operator along them. (In particular, any increasing operator which we can so iterate would have to eventually stabilize.)


Of course, the culprit here is that your "well-orderings" are allowed to be proper classes. If we restrict attention to genuine (set) well-orderings, then this is a consequence of the Replacement scheme.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.