# 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 $\mathbb{N} = A_ 0 \dot\cup A_ 1$, then $\exists j \exists (x_ i)_ {i\in \omega}$ such that $FS(x_ i) \subseteq A_ j.$

Imagine you’d like to prove this with an ultrafilter: $p \in \beta \mathbb{N} \Rightarrow \exists j A_ j=:A \in p.$

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 $\bigcup_ {a\in A} a+ (\bigcup_ {b\in B_ a} b + C_ b)= \bigcup_ {c\in \bigcup_ {a\in A} a+ (\bigcup_ {b\in B_ a} a+ b)} c + C_ c.$ 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!
• $\{ Y \subseteq X: Y \mbox{ compact, non-empty, semigroup} \}$
• 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 $\{z \in Y: zy=y \} = Y \Rightarrow y\cdot y = y$.