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 20111108)](http://vimeo.com/32109926) from [Peter Krautzberger](http://vimeo.com/pkrautzberger) on [Vimeo](http://vimeo.com).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 lefttranslation 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 lefttranslation 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 StoneČech 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 \(q\)!
 \(\forall q\in \beta \mathbb{N}: \rho_ q: \beta \mathbb{N} \rightarrow \mathbb{N}, p \mapsto p+q\) is continuous.

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, righttopological 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!
 \(\\{ Y \subseteq X: Y \mbox{ compact, nonempty, semigroup} \\}\)
 By Zorn's Lemma, $\exists $ minimal, nonempty, 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, nonempty
 \((Y\cdot y) \cdot (Y\cdot y) \subseteq Y\cdot y\), 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 \(\\{z \in Y: zy=y \\} = Y \Rightarrow y\cdot y = y\).