# 905: Foundations of Large Cardinals/3

Harvey Friedman hmflogic at gmail.com
Wed Oct 13 15:17:10 EDT 2021

```1. Conceptual Remarks concerning 903 - #2
2. Second Thoughts about Elementary Extensions versus Elementary
Equivalence - #2
3. Two Universe Elementary Extension - #2
4. Two Extended Universe Restricted Elementary Extension - here
5. Double Completion Elementary Extension - foundations of ZF - #4
6. Multiple Universes Elementary Extension - next posting - #5
7. Two Universe Elementary Extension/complex - #6

4. Two Extended Universe Restricted Elementary Extension

Here as in section 3, V_0,V_1 are universes in the sense of ranks on
strongly inaccessible cardinals kappa < lambda. We got a weak form of
strongly Mahlo cardinals in section 3, from elementary extension, and
failed to achieve a strongly Mahlo cardinal. Here we stay with two
universes, rather than go to three universes, and take the
intermediate step of using classes of V_0 and classes of V_1. I.e.,
using V(kappa + 1) and V(lambda + 1).

Now obviously V(lambda + 1) cannot be an elementary extension of
(V(kappa + 1), because of the parameter kappa. But we can insist that
the parameters be from V(kappa) - just like they are when we used
V(kappa),(lambda) in section 3.

We will get second order indescribable cardinals from this, much
bigger than weakly compact which is much bigger than strongly Mahlo -
but only in the constructible universe L. Thus we might as well just
assume lambda < kappa are ordinals and derive that they are strongly
inaccessible cardinals in L.

THEOREM 4.1. Let V(lambda + 1) be a proper elementary extension of
V(kappa + 1) with parameters from V(kappa), where kappa < lambda are
ordinals.
i. kappa and lambda are strong limit cardinals, and strongly
inaccessible cardinals in L.
ii. kappa and lambda are definably second order indescribable
cardinals in V(kappa), V(lambda) respectively.
iii. kappa is a second order indescribable cardinal in L(lambda).
iv. In L(kappa), L(lambda), there are arbitrarily large second order
indescribable cardinals.
In ii, definably refers to allowing *elements* of V(kappa) and
L(lambda) as parameters in the definition of the subset of V(kappa)
and V(lambda) being reflected downwards, and where the definitions are
over V(kappa +1) and V(lambda + 1), respectively..

Proof: kappa <= omega is immediate from even elementary equivalence
between V(kappa) and V(lambda). kappa, lambda are limit ordinals by
proper elementary extension. Suppose f:V(alpha) onto kappa, where
alpha < kaap. We get a well ordering of a subset of V(alpha) of type
kappa. The well ordering can be used as a parameter in elementary
extension, obtaining also that this well ordering is of order type
lambda, which is impossible. So kappa is a strong limit cardinal.
Hence lambda is a strong limit cardinal. However proving regularity is
an issue. But not in L. Suppose kappa is not regular in L. Let f:alpha
into kappa be strictly increasing and unbounded, in L. Assume f is
constructibly least. Then f is definable with parameter alpha over
V(kappa + 1), and so it retains this property over V(lambda + 1),
which is impossible.

For ii, let A contained in V(kappa) be definable over V(kappa) with
parameters from V(kappa) where (V(kappa),A) satisfies phi, defined
over V(kappa + 1). Use restricted elementary extension to extend to A'
contained in V(lambda) with the same definition, also (V(lambda +
1),A') satisfying phi. Then reflecting down to V(kappa), we see that
according to V(lambda + 1), A' reflects down to A for phi. This is a
statement about V(lambda + 1) and hence by restricted elementary
extension, A also reflects down somewhere for phi. This establishes
that kappa is definably second order indescribable in V(kappa). By
restricted elementary extension, lambda is also definably second order
indescribable in V(lambda).

For iii, let n be given. Suppose kappa is not Pi-1-n indescribable in
the sense of L(lambda). Let A contained in V(kappa) be constructibly
least such that A does not reflect downwards. Then A is definable over
L(kappa + 1) with no parameters. This contradicts i.

For iv, let alpha be such that in L(kappa), there are no second order
indescribable cardinals > alpha. By elementary extension, in
L(lambda), there are no second order indescribable cardinals > alpha.

COROLLARY 4.2. ZFC + there exists arbitrarily large second order
indescribable cardinals is interpretable in KP + there exists ordinals
alpha < beta such that V(beta + 1) is an elementary extension of
V(alpha + 1) with parameters from V(alpha).

We cannot go much farther because third order indescribability dominates.

THEOREM 4.3. (Known) Let mu be the least third order indescribable
cardinal. There exists strongly inaccessible cardinals kappa < lambda
< mu such that V(lambda + 1) is an elementary extension of V(kappa +
1) with parameters from V(kappa). We can take kappa to be arbitrarily
large below mu.

Proof: Let A contained in V(mu) be the set of all (phi,x1,...,xn) such
that V(mu) satisfies phi(x1,...,xn), where phi is a second order
formula with n free variables, and x1,...,xn in V(mu). Let psi be the
third order statement about A that A is as described with mu replaced
by the union of A. Then psi holds of A intersect V(alpha) for
arbitrarily large alpha < mu, and these alpha must be strongly
inaccessible. QED

It is clear that we can go further using V(alpha + n), V(beta + n),
for interpreting (many)  (n+1)-th order indescribable cardinals, n >=
1.

To make the next big leap with the V(alpha)'s, we need to consider
three or more V(alpha)'s at once with the proper way of handling
elementary extensions for pairs of V(alpha)'s. This is taken up in
section 6.

Meanwhile in section 5 we work with two relations instead of two set
theoretic universes. We regard this as a kind of foundations for ZF.

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This is the 905th in a series of self contained numbered
postings to FOM covering a wide range of topics in f.o.m. The list of
previous numbered postings #1-899 can be found at