1. Infinite exponents
Last lecture we showed that for any
Later, we will see that for any
and any
there is
such that
On the other hand (recall we are assuming choice), there are no nontrivial positive partition relations
with
infinite.
(To avoid vacuously true statements, we are always assuming implicitly that requires
for at least one
)
Proof: It is enough to show that In effect, if
holds and
we may assume that
is regular, so any countable subset
of
is bounded, and the ordinal
is still strictly below
For let
denote the subset of
consisting of its first
many members.
Now, given let
be the function
If
is homogeneous for
and of size
then
is homogeneous for
In effect, let
so
If
then
by homogeneity of
This shows that for
infinite implies that there is some
such that
so it is enough to show that the latter never holds. In fact, we cannot even have
(Recall our convention that
denotes the order type and
the size.)
For this, let be an arbitrary infinite cardinal, and let
be a well-ordering of
Define
by
iff
is the
-least member of
and
otherwise.
If is homogeneous for
and
is the
-least member of
then
so
is
-homogeneous. Now consider any infinite sequence
of infinite subsets of
Since
for all
we must have that
contradicting that
is a well-order.
In the presence of the axiom of choice, Theorem 1 completely settles the question of what infinite exponent partition relations hold. Without choice, the situation is different.
For example, Mathias showed that is consistent and, in fact, holds in Solovay’s model. It follows that in the presence of large cardinals, or under determinacy,
holds in
Amusingly, it is still open whether
suffices to obtain
(not just in
) This seems related to the question of whether
implies Woodin’s
Kleinberg and Seiferas studied the question of whether well-ordered choice suffices to prove Theorem 1. I believe the question is still open, but they showed that there is essentially only one instance that requires investigation, namely, whether Mathias’s partition relation is compatible with well-ordered choice. For the remainder of this lecture, we work without choice, and use the arrow notation in the sense of order types; unless otherwise stated, all variables denote ordinals. By a cardinal I mean an initial ordinal.
For an ordinal, let
be the largest limit ordinal (or zero) not larger than
and let
be the unique natural number such that
Definition 2 Let
be an infinite cardinal.
![]()
-well-ordered choice, is the statement that every
-sequence of nonempty sets admits a choice function.
well-ordered choice, is
![]()
For
a binary relation and
an ordinal, an
-chain of length
is a function
with domain
such that
whenever
![]()
![]()
-dependent choice, is the statement that for every binary relation
if every
-chain of length less than
can be extended to a longer
-chain, then there is an
-chain of length
![]()
It is easy to see that, for any
implies
which implies
However,
does not imply
is strictly weaker than
but
implies
Also,
implies
but no
suffices.
Homework 13 (Jensen) Show that implies
Recall the following monotonicity properties of the arrow notation; the only one that may need arguing is the last one, that can be proved as in the opening paragraphs of the proof of Theorem 1.
Lemma 3 Let
and
be cardinals, and and let
and
be ordinals such that
Assume that
![]()
- If
then
![]()
- If
then
![]()
- If
then
![]()
- If
and
then
![]()
![]()
Theorem 4 (Kleinberg-Seiferas) Assume
and suppose that
![]()
- If
or
or
then for all
![]()
![]()
- If
![]()
and
then
iff there is some
such that
![]()
By the proof of Theorem 1, it follows that plus the existence of a well-ordering of
implies that no infinite exponent partition relations hold.
The main use of in the proof below is as follows: Assuming
we can assign to each
a set
with
We also define
this way if
In what follows, we will always assume such an assignment has taken place.
Given any and
write
for the
-th element of
in increasing order. Suppose
Define
as the least
such that
The following obvious remark will prove useful:
Lemma 5 Assume
Given any
and any
there is some
with
![]()
![]()
We need another preliminary result; its proof will be given in another lecture; the notation is defined in the obvious way for
an arbitrary set:
Theorem 6 (
-Rado) For any set
any ordinal
and any nonzero
there is some
such that
![]()
![]()
The -Rado theorem was originally proved using choice. Proofs of the choiceless version above, that we need, have been rediscovered a few times. Apparently the first such proof is due to Keisler and Morley, from 1967. Kleinberg and Seiferas also provide a proof. Another proof (for
finite) is due to Thomas Forster, from 2007.
We are now ready to begin the proof of Theorem 4.
Proof: The proof divides naturally in several cases.
Suppose first that By monotonicity, if
then
which implies that
and therefore
Given denote by
the two sets of order type
with
such that
Define
by
iff
It is enough to show that for any there is some
such that
This shows that
does not admit homogeneous sets of order type
and therefore
to begin with.
To see the claim, suppose so
Let
be such that
and take
Then
The argument is similar if
Suppose now that By the previous case, we may assume that
so
By monotonicity, if
then
so
and therefore
Define by
iff
for all
where
It is enough to show that
does not admit homogeneous sets of order type
For this, it suffices to show that given any
there are
with
Let Define
so that
is the least member of
larger than
for all
and also larger than
By construction,
As before, let be the subsets of
of order type
such that
and
Let
be such that
and define
by
Then
as witnessed by
Suppose that As before, if
then
so
and therefore
Define by
Since for any
there is some
with
it follows that
does not admit infinite homogeneous sets, and we are done.
Suppose that and
Now we prove that
iff
Suppose first that By monotonicity,
so
and therefore
Towards a contradiction, suppose that and let
be a witnessing coloring. Given
let
and set
Note that is infinite, so
is defined. Now define a coloring
by
It is enough to check that admits no infinite homogeneous sets. But this is clear, since given any
any infinite subset
of
induces an infinite subset
of
with
Since
admits no infinite homogeneous sets, there is
such that
But then
is not constant on
Conversely, suppose that Let
be such that
and
We need to check that if
then
First, notice that by an obvious induction. Second, note that Ramsey’s theorem
for any
holds; for example, the amount of choice used in the first proof given last lecture is
which follows from
by Homework 13.
We will show that which clearly suffices. For this, consider a coloring
and let
enumerate
Use
to produce a sequence
where
-
are infinite sets, and
-
for all
To do this, assume is given, and set
for
Since
we can find an infinite
and an
such that
is
-homogeneous for
Now define by
By Ramsey’s theorem, there is an infinite set
and an
such that
is
-homogeneous for
Let
consist of the first
elements of
Let be largest such that
We claim that
is
-homogeneous. To see this, note that any
-sized subset of
is
for some
If
then
because
since the
are decreasing. But
since
This shows that
is constantly equal to
on
Finally, suppose that
and
We need to show that
iff there is some
such that
First, suppose that By monotonicity,
so
and therefore
As shown above, this implies that
Conversely, suppose that Let
and
We need to show that there is some
such that
For this, we use the choiceless
-Rado Theorem 6.
We claim that we can take such that
To see this, consider a coloring
For
define
by
There is a canonical bijection between and
By
it follows that we can associate to each
a pair
such that
and
is
-homogeneous for
By choice of
we can find
an
and an
such that
is
-homogeneous for the map
Let Then
is homogeneous for
since any
has the form
for some
and
so
2. Partition cardinals
By further restricting the amount of choice in the universe, it becomes possible that several of the infinite exponent relations shown inconsistent in the arguments above now hold. In fact, the study of infinite exponent relations under the axiom of determinacy has proved to be very fruitful, and a detailed theory has emerged. I illustrate this with a result of Kleinberg that shows that infinite exponents give rise to measurable cardinals. In what follows, whenever the elements of a linearly ordered set are displayed, as in it is understood that
Lemma 7 Suppose that
Then
is a regular cardinal.
Proof: It is easy to check that must be a limit ordinal. Let
be such that there is some
with
Define by
iff
Let
be a homogeneous subset of
of order type
Then
since
is unbounded.
Now define by
Then
is strictly increasing. Hence
Proof: It is easy to see that so
is regular.
Let be given. Given
let
and
denote the subsets of
consisting, respectively, of its first
elements and of its remaining
Now define
by
iff
Let be homogeneous for
of type
Then in fact
is
-homogeneous. To see this, slice
off into successive sets
of order type
Clearly,
cannot be injective on
since it only takes
many values.
It follows that is also homogeneous for
Because if
we can find
with
and since
we must have
Lemma 9 If
and
then
for any
![]()
Proof: We argue as above: Let be given, where
is the set of functions from
into
and define
by
iff
where
are as before. Let
be homogeneous for
of order type
and define
as before. If
is 0-homogeneous for
then it is homogeneous for
as before, and we are done.
Suppose towards a contradiction that is 1-homogeneous for
so the map
given by
is 1-1.
By lemma 8 and an easy argument, Let
be given by
least
such that
Then we can find
and
that is
-homogeneous for
Then
is a
-sized subset of
contradiction.
Theorem 10 (Kleinberg) Suppose that
for some limit ordinal
Then
is measurable, and there is a normal ultrafilter over
concentrating on ordinals of the same cofinality as
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Proof: is measurable.
As before, is regular. First, we generalize the notion of
-club and identify the ultrafilter that witnesses the measurability of
For a limit ordinal and
let
denote the set of all suprema of increasing
-sequences of elements of
Say that
is
-closed iff
If
is
-closed and unbounded in
say that
is a
-club. Clearly, for any
the set
is
-club.
Let denote the collection of subsets of
that contain a
-club. Clearly,
is nonempty, is closed under supersets, and does not contain the empty set.
We want to show that if then in fact
is a
-complete filter over
To see this claim, suppose that
and that
is a sequence of members of
Since we are not assuming
we cannot automatically pick
-clubs
to conclude that
which is again a
-club. Instead, we use that
For and
a limit ordinal, let
denote the initial segment of
of order type
Let
and
be unbounded subsets of
By an obvious back-and-forth construction, we can find
and
unbounded subsets of
and
respectively, such that for all limit ordinals
Define as follows: Given
and
let
iff
Let
be homogeneous for
of size
Then in fact
for all
and
To see this, suppose that is such that
for all
Let
be
-club, and consider
and
as above. Then
so
contradicting that
Now let so
is
-club. It is enough to show that
To see this, let
so there is
with
Then
for all
so
for all
and the claim follows.
A similar argument shows that is in fact an ultrafilter. To see this, let
be given. We use that
and consider the partition
given by
iff
If
is homogeneous for
of size
then
is
-club, and either
or
depending on whether
is
-homogeneous or
-homogeneous.
is normal.
First we show that Clearly,
To see the other containment, it is enough to see that if
is
-club, then
is
-closed. Let
and let
be its increasing enumeration.
Let so
Fix
strictly increasing, continuous, and cofinal, and let
such that
For let
be the initial segment of
such that
Also, let
have order type
Then
has order type
and the same supremum as
Thus,
Note that implies
This is all we require below, so it follows that we may assume that
is a regular cardinal.
To show that is normal, consider a function
regressive on a
-measure 1 set, and we need to argue that
is constant on a
-measure 1 set.
For this, fix a -club
where
is regressive, and let
be given by
iff
Let
be homogeneous for
of size
Given
let
Then
and
so
Hence
is
-homogeneous.
Given any and any
with
let
It follows that
By
-additivity of
it follows that there is some
such that
Definition 11 A strong partition cardinal is a cardinal
such that
![]()
A weak partition cardinal is a cardinal
such that
for all
![]()
In the presence of the existence of uncountable strong partition cardinals
implies the existence of large cardinals
Also, uncountable strong partition cardinals satisfy stronger partition relations than the one required by the definition. For example, one has:
Definition 12
means that for any partition
there is a set
such that
is constant on
for all
![]()
Note that fails; for example, consider
given by
Also, as noted by
and Rado,
fails; for example, consider
given by
iff
Definition 13 A Ramsey cardinal is a cardinal
such that
![]()
Theorem 14 (Henle)
- If
is an uncountable cardinal,
is a cardinal, and
then
In particular, an uncountable strong partition cardinal is both Ramsey and a weak partition cardinal.
- If
then
for all
![]()
- If
then
implies that the least
such that
admits a
-complete nonprincipal ultrafilter. If
holds, then
(and therefore
is measurable.) If
holds, then the ultrafilter can be taken to be normal.
- If
is uncountable and
then
![]()
![]()
On the other hand, it is not the case that if then
for arbitrary sets
For example,
holds under
and therefore also
However,
and it follows that
To state the main result about partition cardinals, we need two additional large cardinal notions:
Definition 15 An algebra on
is a structure of the form
where for each
there is an
such that
![]()
A cardinal
is Jónsson iff every algebra on
admits a proper subalgebra of size
![]()
For example, that is not Jónsson can be witnessed by letting
if
and setting
and
arbitrary for
We will return to Jónsson cardinals next lecture.
We also need the notion of Rowbottom cardinal, that requires a significant weakening of the ordinary partition relation for ordinals. We briefly encountered this weakening when discussing Chang’s conjecture in lecture II.7.
Definition 16 Given ordinals
say that
iff for any
there is an
such that
Similarly, by replacing
with
![]()
Say that
iff for any
there is an
such that
Similarly by replacing
with
![]()
If
say that
is
-Rowbottom iff
for any
and say that
is Rowbottom iff it is
-Rowbottom.
Theorem 17 (Kleinberg) Assume
Let
be an uncountable strong partition cardinal. Then there are cardinals
such that:
for all
In particular,
and
are measurable.
- All the
![]()
are singular Jónsson cardinals of cofinality
and
is a Rowbottom cardinal.
![]()
Theorem 18 (Kleinberg) In the situation of Theorem 17, assume moreover that there is a normal ultrafilter
over
such that
has order type
Then:
for
![]()
is a weak partition cardinal.
- Every normal ultrafilter over
or
is an
as defined in the proof of Theorem 10.
![]()
It is in the context of that Kleinberg results have proved useful. In part, this may be accidental, in that we do not know any ways of obtaining partition cardinals other than via determinacy. On the other hand, Kechris and Woodin have shown that, in
determinacy is equivalent to the existence of unboundedly many strong partition cardinals below
so determinacy seems to be the natural setting for studying these cardinals.
As mentioned above,
holds under This is a theorem of Martin.
In one can make sense of ultrapowers of ordinals by
-complete ultrafilters; this is not necessarily the case in general without any amount of choice, since one cannot prove
‘s theorem for the full ultrapower of
so some care is required.
Theorem 19 (Martin) Assume
If
and
is uncountable, then for any
-complete ultrafilter
over
the image
of
under the induced ultrapower map, is a cardinal.
![]()
From Martin’s theorem 19 it follows that where
is the
-club ultrafilter over
Hence from Theorems 17 and 18 it follows that
and that each
is a Jónsson cardinal of cofinality
and
is Rowbottom.
Under a very detailed analysis of measures has been carried out by Steve Jackson and others, and so many cardinals with strong partition properties have been identified. Steel proved, for example, that in
under
every regular cardinal below
is in fact measurable. Also, it follows from Jackson’s analysis that
are all strong partition cardinals and their successors are weak partition cardinals; here, exponentiation is ordinal exponentiation.
The precise result shown by Kleinberg is that if is a strong partition cardinal and
is a normal ultrafilter over
then we can define
as in Theorem 17 by setting
A careful analysis of measures under
then allows us to study many cardinals above a given
For example, the only regular cardinals under
between
and
are
for
and
the
-club ultrafilter over
These cardinals are precisely
and
Theorem 20 (Jackson-Khafizov) Assume
Suppose that
Let
where
be the Cantor normal form of
Then the following hold:
- If
then
![]()
- If
is a successor ordinal, then
![]()
- If
is a limit ordinal, then
![]()
![]()
Löwe exploited Kleinberg’s result and the Jackson-Khafizov analysis thanks to the following “shifting lemma:”
Theorem 21 (Löwe) Assume
Let
and let
be a
-complete ultrafilter over
Let
Suppose that for all cardinals
either
is a successor of cofinality strictly larger than
or else
is a limit of cofinality strictly smaller than
Then
![]()
![]()
For example, under if
then there are only three normal ultrafilters over
namely
for
It follows that the Kleinberg sequence associated to
is:
- If
then
for
- If
then
for
- If
then
for
Bibliography
Here are some references consulted while preparing this note:
- James Henle, Some consequences of an infinite-exponent partition relation, The Journal of Symbolic Logic, 42 (4), (Dec., 1977), 523–526.
- Steve Jackson, Structural consequences of
, in Handbook of set theory, Matthew Foreman, Akihiro Kanamori, eds., to appear.
- Akihiro Kanamori, The higher infinite, Springer (1994).
- Eugene Kleinberg, Strong partition properties for infinite cardinals, The Journal of Symbolic Logic, 35 (3), (Sep., 1970), 410–428.
- Eugene Kleinberg, Infinitary combinatorics and the axiom of determinateness, Springer (1977).
- Eugene Kleinberg and J. Seiferas, Infinite exponent partition relations and well-ordered choice, The Journal of Symbolic Logic, 38 (2), (Jun., 1973), 299–308.
- Benedikt Löwe, Kleinberg sequences and partition cardinals below
, Fundamenta Mathematicae, 171 (1), (2002), 69–76.
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[…] again, assume choice throughout. Last lecture, we showed that for any The results below strengthen this fact in several ways. Definition 1 […]
[…] proof of Theorems 1 and 3 above require the axiom of choice. However, the proof of Theorem 4 in lecture III.2 requires a choiceless version. I now proceed to establish this version. Theorem 5 () For any set […]
Woodin has extended significantly the results about Jónsson cardinals mentioned in the context of
namely, he has shown using the directed systems analysis of
that (assuming determinacy) in
every uncountable cardinal below
is Jónsson. It is not clear whether the argument can be extended to strengthen the conclusion to say that every such cardinal is Rowbottom.
Steve Jackson has shown (from determinacy) that this is the case for all uncountable cardinals below
See https://andrescaicedo.files.wordpress.com/2008/04/jonsson1.pdf for a sketch of this result.
[…] from Definition 15 on lecture III.2 that a cardinal is Jónsson iff every algebra on admits a proper subalgebra of size Here, an […]
[…] of partition properties and the computation of ultrapowers of associated measures. See also here for extensions of this result, references, and some additional […]
[…] proofs of Theorems 1 and 3 above require the axiom of choice. However, the proof of Theorem 4 in lecture III.2 requires a choiceless version. I now proceed to establish this version. Theorem 5 () For any set […]
[…] from Definition 15 on lecture III.2 that a cardinal is Jónsson iff every algebra on admits a proper subalgebra of size Here, an […]
[…] again, assume choice throughout. Last lecture, we showed that for any The results below strengthen this fact in several […]