Idempotent Ultrafilters, an introduction (Michigan Logic Seminar Nov 09, 2011)
15 Nov 2011
Because of a power outage at the department my talk announced for October 29th was postponed by a week.
Idempotent Ultrafilters: An Introduction (University of Michigan Logic Seminar 20111108) from Peter Krautzberger on Vimeo.
Here are transcripts of my notes (as well as the originals at the end).
Hindman’s Theorem
Hindman’s Theorem If , then such that
Imagine you’d like to prove this with an ultrafilter:
What do we need? We will build inductively!
Pick – we can’t really choose better than that (except maybe by shrinking the set first).
If we’re looking for our result, we need
 such that and .
 i.e. .
 so we need !
In other words, we need to begin with, i.e., – for any !
Galvin in 1970: is almost lefttranslation invariant iff .
Is this enough?
 Pick
 Then choose .
But to continue the process, we need more!
We need such that:
 – , check
 – , check
 – – possible if we picked , check.
 – ???
What does this mean? by associativity.
Ah! But we have seen this before!
We needed , so we needed !
But that’s ok!! & !
How do we get to the end?

Inductively, assume we have with and

Pick from – this intersection is in !

Note that

So which is in – as desired.
Question: Do “almost lefttranslation invariant” ultrafilters exist?
Glazer, ~1975: Yes of course! These are the idempotent ultrafilters! We know these exist since Ellis 1958.
What does this mean?
 is a semigroup

is discrete, so , the StoneČech compactification of exists, in fact the set of ultrafilters on with a topological basis for .

is compact (exactly by the Boolean Prime Ideal Theorem)

is Hausdorff ().
 has a semigroup structure extending
 From :

 try – not an ultrafilter
 if you try to prove ultraness:
 for some , all
 generates an ultrafilter!
 Then
 i.e., generated by [check: ]
 Properties
 is continuous.
 Why? iff iff .
 But only depends on !
 associativity: check it – use The first set is in , the second in .
Now remember: what did Galvin need?
I.e., , so
I.e. (since ufs)
Ellis 1958 compact, Hausdorff, righttopological semigroup .
Proof.
 Think: is a closed semigroup  a minimal one!
 By Zorn’s Lemma, minimal, nonempty, compact semigroup .
 Think: that should be !
 We’ll show (therefore by minimality)
 How? We only have continuity and associativity
 compact, nonempty
 , i.e., a semigroup.
 By minimality of ,
 Great! We’d expect that if
 .
 Then
 , i.e., semigroup.
 compact? Yes closed.
 minimal, so .