# Idempotent Ultrafilters, an introduction (Michigan Logic Seminar Nov 09, 2011)

Because of a power outage at the department my talk announced for October 29th was postponed by a week.

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 left-translation 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 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!
• 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, right-topological semigroup .

Proof.

• 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
• .
• Then
• , i.e., semigroup.
• compact? Yes closed.
• minimal, so .