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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

  • Pick $x_ {n+1}$ from – this intersection is in $p$!

  • Note that

  • So 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!
  • $\{ Y \subseteq X: Y \mbox{ compact, non-empty, semigroup} \}$
  • 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 $\{z \in Y: zy=y \} = Y \Rightarrow y\cdot y = y$.
Part 1 Idempotent Ultrafilters
Part 1 Idempotent Ultrafilters, an introduction (Michigan Logic Seminar Nov 09, 2011)
Part 2 Idempotent Ultrafilters
Part 2 Idempotent Ultrafilters, an introduction (Michigan Logic Seminar Nov 09, 2011)
Part 3 Idempotent Ultrafilters
Part 3 Idempotent Ultrafilters, an introduction (Michigan Logic Seminar Nov 09, 2011)
Part 4 Idempotent Ultrafilters
Part 4 Idempotent Ultrafilters, an introduction (Michigan Logic Seminar Nov 09, 2011)