Let me begin with a remark related to the question of whether . We showed that this is the case if
for some
, or if
is Dedekind-finite.
Theorem. The axiom of choice is equivalent to the statement that any Dedekind-infinite cardinal is a square.
Proof. Let be a set. Assuming that every D-infinite cardinal is a square, we need to show that
is well-orderable. We may assume that
. Otherwise, replace
with
. Let
. Assume that
is a square, say
. Then
. By Homework problem 2,
, so
for some
, and
.
Lemma. Suppose are D-infinite sets and
is an (infinite) initial ordinal. If
then either
or
.
Proof. Let be an injection. If there is some
such that
we are done, so we may assume that for all
there is some
such that
. Letting
be the least such
, the map
is an injection of
into
.
By the lemma, it must be that either or else
. The former is impossible since
, so
is well-orderable, and thus so is
, and since
, then
is well-orderable as well.
9.
-trichotomy.
This is a very recent result, see David Feldman, Mehmet Orhon, Generalizing Hartogs’s trichotomy theorem, preprint (2008), available at the ArXiv.
[Update, January 1, 2013: Asaf Karagila recently pointed out that the result is due to Tarski. It appears as principle on page 22 of “Equivalents of the Axiom of CHoice. II” by Rubin and Rubin, and was announced by Tarski (with a sketch of the argument) in the Notices of AMS, vol. 11, 1964.]
Definition. Trichotomy is the statement that any two sets are comparable.
Theorem. (Hartogs). Trichotomy is equivalent to choice.
Proof. If choice holds, any set is equipotent with an ordinal, and any two ordinals are comparable.
Conversely, if any two sets are comparable, in particular and
are. Since
, we must have
, and we are done.
Definition. Let .
-trichotomy is the statement that given any
sets, at least two are comparable.
Theorem. For any ,
-trichotomy is equivalent to choice.
Proof. (Blass). Assume -trichotomy. Let
be a set. We want to show that
is well-orderable. Let
.
Lemma. Let be an initial ordinal. If
, then
is well-orderable.
Here, is to be understood as disjoint union.
Proof. Let be 1-1. For each
, since
does not inject into
, it must be the case that the range of
meets
. Let
be the least member of this intersection. Then
is an injection of
into
.
Remark. If , then
, and it follows that
. However, it is not necessarily the case that
for
infinite. For example, this fails if
is D-finite.
Question. If for some initial ordinal
, does it follow that
is well-orderable?
Define a sequence of cardinals,
as follows:
.
.
(Then )
We claim that if is not well-orderable, the set
contradicts
-trichotomy. Or, positively,
-trichotomy applied to this set implies the well-orderability of
. Because if
and there is an injection
, then in particular
. If
, then
. Otherwise,
is defined, and
(since
is the aleph of the set that
injects into). Since the sequence of cardinals
is increasing, we must have
, so
and (since
) in fact
.
Let . Then
, using that
. Also,
, by assumption. But
, and we have that
. By the lemma,
is well-orderable and we are done, since
injects into
.
Open question. With the obvious definition, does -trichotomy imply choice?
Let -trichotomy be the statement that any infinite family of sets contains two that are comparable.
Open question. Does -trichotomy imply
-trichotomy? Does
-trichotomy imply choice?
10. The generalized continuum hypothesis.
The last topic of this part of the course is Specker’s theorem that implies
.
Definition. Let be a cardinal. The continuum hypothesis for
,
, is the statement that for any cardinal
, if
, then either
or else
.
Definition. The Generalized Continuum Hypothesis, , is the statement that
holds for all infinite cardinals
.
Theorem. (Specker). implies
, the axiom of choice.
Remark. In fact, it suffices that holds for all infinite initial ordinals. I am not sure who first proved this; it is a consequence of a result of Kanamori and Pincus in the paper mentioned last lecture.
Specker’s result is in fact a local argument. What one proves is the following:
Theorem. Assume and
. Then
is well-orderable and, in fact,
.
Next lecture will be devoted to the proof of this result.
[…] , so, if has size , then is equipotent with , as shown on lecture 4, section 9 (remark after the proof of the first […]
[…] , so, if has size , then is equipotent with , as shown on lecture 4, section 9 (remark after the proof of the first […]