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 2011-11-08) from Peter Krautzberger on Vimeo.
Here are transcripts of my notes (as well as the originals at the end).
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 left-translation invariant iff .
Is this enough?
- 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 !
So which is in – as desired.
Question: Do “almost left-translation 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 Stone-Čech 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!
- i.e., generated by [check: ]
- 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, right-topological semigroup .
- Think: is a closed semigroup - a minimal one!
- By Zorn’s Lemma, minimal, non-empty, compact semigroup .
- Think: that should be !
- We’ll show (therefore by minimality)
- How? We only have continuity and associativity
- compact, non-empty
- , i.e., a semigroup.
- By minimality of ,
- Great! We’d expect that if
- , i.e., semigroup.
- compact? Yes closed.
- minimal, so .