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 $$FS(x_i) \subseteq A_j.$$

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?

$$(\forall A \in p) \{ x: -x+A \in p\}\in p$$

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 $|Y|=1$!
* 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 $\{z \in Y: zy=y \} = Y \Rightarrow y\cdot y = y$.

2 thoughts on “Idempotent Ultrafilters, an introduction (Michigan Logic Seminar Nov 09, 2011)

  1. Pingback: One day in Colorado or Strongly summable ultrafilters are rapid | Peter Krautzberger

  2. Ari Brodsky

    Ellis’s Lemma is always stated for compact Hausdorff spaces. But it doesn’t use the full strength of Hausdorff; only the fact that compact subsets are closed. I’ve seen spaces with this property called “KC-spaces”. Why isn’t Ellis’s Lemma stated with this weaker hypothesis, so that it can be automatically applied to this larger class of spaces? (KC does not imply Hausdorff, even for compact spaces. In fact, KC compact topologies are precisely the maximal compact topologies on any set.)

    Reply

Leave a Reply

Your email address will not be published. Required fields are marked *


six × = 6

You may use these HTML tags and attributes: <a href="" title=""> <abbr title=""> <acronym title=""> <b> <blockquote cite=""> <cite> <code> <del datetime=""> <em> <i> <q cite=""> <strike> <strong>