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.

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

Hindman's Theorem If \mathbb{N} = A_ 0 \dot\cup A_ 1, then \exists j \exists (x_ i)_ {i\in \omega} such that

Imagine you'd like to prove this with an ultrafilter:

What do we need? We will build (x_ i) inductively!

Pick x_ 0 \in A -- we can't really choose better than that (except maybe by shrinking the set first).

If we're looking for our result, we need

• x_ 1 \in A such that x_ 1, x_ 0 and x_ 0+x_ 1 \in A.
• i.e. x_ 1 \in -x_ 0 + A.
• so we need -x_ 0 + A \in p!

In other words, we need x_ 0 \in \{ x: -x+A \in p \} to begin with, i.e., \{ x: -x+A \in p \} \in p -- for any A\in p!

Galvin in 1970: p \in \beta \mathbb{N} is almost left-translation invariant iff \forall A\in p: \{ x : -x +A\in p\} \in p.

Is this enough?

• Pick x_ 0 \in A \cap \{ x: -x+A \} \in p
• Then choose x_ 1 \in -x_ 0 + A \cap A.

But to continue the process, we need more!

We need x_ 2 such that:

• x_ 2 \in A -- A\in p, check
• x_ 0 + x_ 2 \in A -- x_ 2 \in -x_ 0 +A \in p, check
• x_ 1 + x_ 2 \in A -- x_ 2 \in -x_ 1 + A \in p -- possible if we picked x_ 1 \in \{x: -x+A\in p\} \in p, check.
• x_ 0 +x_ 1 + x_ 2 \in A -- x_ 2 \in -(x_ 0+x_ 1) +A \in p???

What does this mean? -(x_ 0 +x_ 1) + A = -x_ 1 + (-x_ 0 +A) by associativity.

Ah! But we have seen this before!

We needed x_ 1 \in \{ x: -x + (-x_ 0 +A) \in p\}, so we needed \{ x: -x + (-x_ 0 +A) \in p\}\in p!

But that's ok!! -x_ 0 + A \in p & \forall B\in p: \{x : -x+B \in p \} \in p!

How do we get to the end?

• Inductively, assume we have x_ 0,\ldots, x_ n with FS(x_ 0,\ldots, x_ n) \subseteq A and\bigcap_ {z \in FS(x_ 0,\ldots,x_ n)} -z + A \in p.
• Pick x_ {n+1} from( \bigcap_ {z \in FS(x_ 0,\ldots,x_ n)} -z + A ) \cap A \cap \{ x: -x+ (\bigcap_ {z \in FS(x_ 0,\ldots,x_ n)} -z + A) \in p\} -- this intersection is in p!
• Note that-x_ {n+1} + ( \bigcap_ {z \in FS(x_ 0,\ldots,x_ n)} -z + A \cap A) = \bigcap_ {z \in FS(x_ 0,\ldots,x_ n)} -x_ {n+1} (-z + A) \cap -x_ {n+1} A \in p.
• So\bigcap_ {z \in FS(x_ 0,\ldots,x_ {n+1})} -z + A = \bigcap_ {z \in FS(x_ 0,\ldots,x_ n)} (-z + A) \cap \bigcap_ {z \in FS(x_ 0,\ldots,x_ n)} (-(z+x_ {n+1}) + A) \cap (-x_ {n+1} +A which is in p -- 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?

• (\mathbb{N}, +) is a semigroup

• \mathbb{N} is discrete, so \beta \mathbb{N}, the Stone-Čech compactification of \mathbb{N} exists, in fact \beta \mathbb{N} \cong the set of ultrafilters on \mathbb{N} with a topological basis \hat A = \{ p \in \beta \mathbb{N} : A \in p \} for A\subseteq \mathbb{N}.

• \beta \mathbb{N} is compact (exactly by the Boolean Prime Ideal Theorem)

• \beta \mathbb{N} is Hausdorff (p\neq q \Rightarrow \exists A\in p, B\in q: A\cap B = \emptyset).

• \beta \mathbb{N}has a semigroup structure extending (\mathbb{N}, +)

  • From \beta (\mathbb{N} \times \mathbb{N}:
  • p,q \in \beta \mathbb{N}\mapsto p \otimes q \in \beta(\mathbb{N}^2)
    • try (A\times B)_ {A\in p, B\in q} -- not an ultrafilter
    • if you try to prove ultraness:
    • \bigcup_ {a\in A} \{a\} \times B_ a for some A\in p, all B_ a \in q
    • generates an ultrafilter!

• Then p + q = + (p\otimes q)

  • i.e., generated by \{ \bigcup_ {a \in A} a + B_ a : A\in p, B_ a \in q\} [check: n+k = +(n \times k)]

• Properties

  • \forall q\in \beta \mathbb{N}: \rho_ q: \beta \mathbb{N} \rightarrow \mathbb{N}, p \mapsto p+q is continuous.
    • Why? X\in p+q iff \exists A\in p \exists (B_ a)_ {a \in A} , B_ a \in q: \bigcup_ {a\in A} a+ B_ a \subseteq X iff \{a: -a + X \in q\} =: X^{-q} \in p.
    • But X^{-q} only depends on q!

• associativity: check it -- use

The first set is in p+(q+r), the second in (p+q)+r.

Now remember: what did Galvin need?

I.e., A\in p \Rightarrow A^{-p} \in p \Rightarrow A \in p+p, so p \subseteq p+p
I.e. p+p = p (since ufs)

Ellis 1958 (X,\cdot) compact, Hausdorff, right-topological semigroup \Rightarrow \exists x\in X: x\cdot x =x.

Proof.

• Think: x\cdot x = x \Rightarrow \{x\} is a closed semigroup - a minimal one!
• \{ Y \subseteq X: Y \text{ compact, non-empty, semigroup} \}
• By Zorn's Lemma, \exists minimal, non-empty, compact semigroup Y.
• Think: that should be \left\vert Y \right\vert = 1!
• We'll show \forall y \in Y: y\cdot y = y (therefore Y = \{y\} by minimality)
• How? We only have continuity and associativity
• Y \cdot y = \rho_ y [Y] compact, non-empty
• (Y\cdot y) \cdot (Y\cdot y) \subseteq Y\cdot y, i.e., a semigroup.
• By minimality of Y, Y\cdot y = Y
• Great! We'd expect that if y\cdot y = y
• Y\cdot y = Y \Rightarrow \exists z \in Y: z\cdot y = y.
• Then \{ z \in Y : zy=y\} = \rho^{-1}_ y (y) \subseteq Y
• (z_ 0 z_ 1) y = z_ 0 (z_ 1 y) = z_ 0 y= y, i.e., semigroup.
• compact? Yes \rho^{-1}_ y[ \{y\}] closed.
• Y minimal, so \{z \in Y: zy=y \} = Y \Rightarrow y\cdot y = y.