Author Archives: Peter Krautzberger

Red workbook, p15

Source

red workbook, p15-1

red workbook, p15-2

Transcript

Left page

  • Forcing moeglich?
  • Fuer strongly summable? kaputtmachen???????
  • Bem (* [[circled]]) [[boxed]]

Right page

  • [4.] Eine “elementare” Charakterisierung von “zentral”
  • 4.1 Def. $A\subseteq S$ Setze
    • (a) $T_A = \{ (a_o, \ldots, a_{n-1} ) \in S^{< \omega} : FP(a_i)_{i=0}^{n-1} \subseteq A \}$
    • $T_A$ Teilbaum von $S^{< \omega}$.
    • Notiz:
      • $A$ IP <=> $T_A$ hat unendlichen Zweig
      • tatsaechlich bei konstruktion von IP-Menge zeigt man,
      • dass es sehr viele unendliche Zweige gibt
      • SK: sogar perfekter Baum
    • (b) Fuer $R \subseteq S^{<\omega}$ Teilbaum, $r \in R$ setze
      • $N_r := N_r^R := \{ a \in S : r^a \in R \} (\subseteq S)$ [[a diagram: the tree $R$ and the set of successors $N_r$]]
  • 4.2 Satz [14.25 in HS]
    • Fuer $A\subseteq S$ aequivalent: (a) $A$ zentral
      • (b) $\exists R \subseteq T(A)$ Teilbaum mit (2) $\{ N_r^R: r \in R\}$ cwpws
        • (1) $r\in R, a \in N_r \Rightarrow a\cdot N_{r \hat{} a} \subseteq N_r$
      • Solches $R$ heisst $\star$-tree [in HS].

partial Translation

Left page

  • Forcing possible?
  • For strongly summable? destroying???????

Right page

  • [4.] An “elementary” Characterization of “central”
  • 4.1 Definition. For $A\subseteq S$ define:
    • (a) $T_A = \{ (a_o, \ldots, a_{n-1} ) \in S^{< \omega} : FP(a_i)_{i=0}^{n-1} \subseteq A \}$
    • $T_A$ subtree of $S^{< \omega}$.
    • Note:
      • $A$ IP <=> $T_A$ includes an infinite branch
      • in fact, in the construction of IP-set one shows that there are many infinite branches
      • Sabine Koppelberg: in fact, a perfect subtree.
    • (b) For $R \subseteq S^{<\omega}$ subtree, $r \in R$ define
      • $N_r := N_r^R := \{ a \in S : r^a \in R \} (\subseteq S)$ [[the successor set]]
  • 4.2 Theorem [14.25 in HS]
    • For $A\subseteq S$ TFAE:
      • (a) $A$ central
      • (b) $\exists R \subseteq T(A)$ subtree with
        • (2) $\{ N_r^R: r \in R\}$ cwpws (collectionwise piecewise syndetic)
        • (1) $r\in R, a \in N_r \Rightarrow a\cdot N_{r \hat{} a} \subseteq N_r$
      • Such $R$ is called $\star$-tree [in HS].

Notes

We’re getting to some serious results here. The “tree characterization” of centrality is, I think, not known (or not appreciated) widely enough. It might be a lot to wrap your mind around as a student but this might be one of the better ways of providing some insights into the notion of cwpws sets.

This page is very amusing. The random note on destroying strongly summable ultrafilters is what occupied a large part of my postdoctoral research. Apparently it took me a while to realize this is an interesting question. Come to think of it, Francois and I also spent quite a bit of time on the tree characterization; makes me want to skip ahead to a postdoc notebook…

Red workbook, p14

Source

red workbook, p14

Transcript

  • Beweis: (a) => (b): $\mathfrak{B} \subseteq q \Rightarrow \mathfrak{D} = \{ D_e : e \in [ \mathfrak{B} ]^{< \omega} \} \subseteq q$
    • Nimm $g_e \in [S^{< \omega}]$ mit $\beta S \cdot q \subseteq \bigcup_{x \in g_e} \widehat{ x^{-1} D_e} = \widehat{C_e}$
    • [ Verfeinerung von pws: $A$ pws $\in q \in K(\beta S) \Rightarrow \exists e \in [S^{<\omega}]: \beta S \cdot q \subseteq \bigcup x^{-1} A$]
    • $\Rightarrow \beta S \cdot q \subseteq \bigcap_{C \in \mathfrak{C}} \widehat{C_e} = Y_e \stackrel{3.3}{\Rightarrow} $ Beh.
  • (b) => (a): seien $g_e, C_e, \mathfrak{C}$ wie in (b).
    • Nimm $p \in \beta S$ mit $L := \beta S \cdot p \subseteq Y_{\mathfrak{C}}$ oBdA $p\in K(\beta S)$.
    • $\Rightarrow p\in L$.
    • Fuer $e \in [\mathfrak{B}]^{<\omega}: p \in \widehat{C_e} = \bigcup_{x\in g_e} \widehat{x^{-1} D_e}$
    • $\Rightarrow \exists x_e \in g_e: p \in \widehat{ x_e^{-1}D_e}$.
  • ((?) wieso eDe?) Nimm $w\in \beta S$ mit $S_e := \{ x_f : f \supseteq e, f \in [ \mathfrak{B}]^{<\omega} \} \underset{?}{\in} \omega (\forall e \in \beta S)$
  • Setze $q = w \cdot p \Rightarrow q \in \beta S \cdot p \subseteq K(\beta S)$ und
  • es ist $\mathfrak{B} \subseteq q$ [$B\in \mathfrak{B}$ zeige: $B\in q = w\cdot p$. Aber $e:{B} \in [\mathfrak{B}]^{\omega}$
    • $D_e = B, S_e \in \omega$; Fuer alle $f\supseteq e: x_f \cdot p \in \widehat{D_f}$
    • $\widehat{D_f} \subseteq \widehat{D_e} = \widehat{B} \Rightarrow S_e \cdot p \subseteq \widehat{B}$
    • $\Rightarrow w \cdot p \in \widehat{B}$]
  • Also folgt die Behauptung.□

partial Translation

  • Proof:
  • (a) => (b):
    • $\mathfrak{B} \subseteq q \Rightarrow \mathfrak{D} = \{ D_e : e \in [ \mathfrak{B} ]^{< \omega} \} \subseteq q$
    • Take $g_e \in [S^{< \omega}]$ with $\beta S \cdot q \subseteq \bigcup_{x \in g_e} \widehat{ x^{-1} D_e} = \widehat{C_e}$
      • [$A$ pws, $A \in q \in K(\beta S) \Rightarrow \exists e \in [S^{<\omega}]: \beta S \cdot q \subseteq \bigcup_{x\in e} x^{-1} A$]
    • $\Rightarrow \beta S \cdot q \subseteq \bigcap_{C \in \mathfrak{C}} \widehat{C_e} = Y_e \stackrel{3.3}{\Rightarrow} $ the claim.
  • (b) => (a): let $g_e, C_e, \mathfrak{C}$ as in (b).
    • Then take $p \in \beta S$ mit $L := \beta S \cdot p \subseteq Y_{\mathfrak{C}}$; without loss $p\in K(\beta S)$.
    • $\Rightarrow p\in L$.
    • For $e \in [\mathfrak{B}]^{<\omega}: p \in \widehat{C_e} = \bigcup_{x\in g_e} \widehat{x^{-1} D_e}$
    • $\Rightarrow \exists x_e \in g_e: p \in \widehat{ x_e^{-1}D_e}$.
    • Take $w\in \beta S$ with $S_e := \{ x_f : f \supseteq e, f \in [ \mathfrak{B}]^{<\omega} \} \in \omega (\forall e \in \beta S)$
    • Now define $q = w \cdot p \Rightarrow q \in \beta S \cdot p \subseteq K(\beta S)$ and
    • since $\mathfrak{B} \subseteq q$
      • [$B\in \mathfrak{B}$ show: $B\in q = w\cdot p$. But $e:{B} \in [\mathfrak{B}]^{\omega}$
      • $D_e = B, S_e \in \omega$; For all $f\supseteq e: x_f \cdot p \in \widehat{D_f}$
      • $\widehat{D_f} \subseteq \widehat{D_e} = \widehat{B} \Rightarrow S_e \cdot p \subseteq \widehat{B}$
      • $\Rightarrow w \cdot p \in \widehat{B}$]
    • The claim follows.

Notes

This page contains the proof of Theorem 3.4 of the previous part (I guess I should’ve included that yesterday). I can’t really make much of it. It’s the dull of writing up a new notion. But if you look closer, you might stumble over a few details (as I did when I took these notes). Writing this up just now I find the choice of $w$ quite striking.

Red workbook, p13

Source

red workbook, p13-1

red workbook, p13-2

Transcript

First page

  • 14. Sept. 2006
  • Fortsetzung Vortrag SK
  • 3. collectionwise thick (cwt, cwdick), collectionwise pws (cwpws)
  • 3.2 Notation. $\mathfrak{B} \subseteq \mathcal{P}(S), \mathfrak{V} := \{ D_e : e \in [\mathcal{B}]^{<\omega} \}$
    • $\mathfrak{D}$ (-Suetterlin?), $D_e = \bigcap_{B\in e} B$
    • Also: $\mathcal{B} \subseteq \mathfrak{V}$
  • 3.1 Definition
    • Fuer $\mathfrak{A} \subseteq \mathcal{P}(S): Y_{\mathfrak{A}} = \bigcap_{A\in\mathfrak{A}} \widehat{A} \subseteq \beta S$ abgeschlossen
      • [in 3.2, dann $Y_\mathfrak{B} = Y_\mathfrak{D}$]
  • 3.3 Satz & Def. Aequivalent:
    • (a) $\exists q \in \beta S: \beta S \cdot q \subseteq Y_\mathfrak{B}$ (natuerlich oBdA $q\in K(\beta S)$)
    • (b) $\forall D \in \mathfrak{D}$ $D$ dick.
    • Dann: heisst $\mathfrak{B}$ cwdick (cwd)
  • Beweis
    • =>: $\beta S \cdot q \subseteq Y_{\mathfrak{B}} = Y_\mathfrak{D} \subseteq \widehat{D}$ fuer all $D\in \mathfrak{D}$
    • <=: Fuer $e\in [\mathfrak{B}]^{<\omega}$ nimm $q_e \in \beta S, \beta S \cdot q_e \subseteq \widehat{D_e}$
      • OBdA, $q\in K(\beta S)$, ($q_e \in \beta S \cdot q_e \in \widehat{D_e}$ gilt ✓)
      • Setze $ X_e := \{ q_f : e \subseteq f \in [ \mathfrak{B} ]^{< \omega} \} \subseteq \beta S$.

Second page

  • Damit $X_e \subseteq \widehat{D_e}$ [ $f\supseteq e \Rightarrow q_f \in \widehat{D_f} \subseteq \widehat{D_e}$]
  • $\{ X_e : e \in [\mathfrak{B}]^{< \omega} \}$ hat eDE (? check)
  • nimm $q\in \bigcap_{e \in [ \mathfrak{B} ]^{<\omega}} cl_{\beta S}(X_e)$
  • Beh. $\forall D \in \mathfrak{D}: \beta S \cdot q \subseteq \widehat{D}$
    • [ $D = D_e, e\in [ \mathfrak{B}]^{< \omega} \Rightarrow X_e \subseteq \widehat{D_e}, q\in cl(X_e) \Rightarrow q \in D_e$
    • $\forall S \in S, e \subseteq f, s\cdot q_f \underset{\beta S \cdot q_f \subseteq\widehat{D_f}}{\in} \widehat{D_f} \subseteq \widehat{D_e} \Rightarrow s\cdot X_e \subseteq \widehat{D_e}$
    • $\Rightarrow s\cdot q \in \widehat{D_e} \Rightarrow \beta S \cdot q \subseteq \widehat{D_e}$]
    • Damit folgt die Behauptung □
  • ["Aufgabe": konstruiere $q$ durch $p$-limiten?]
  • 3.4 Satz & Def. Aequivalent
    • (a) $\exists q\in K(\beta S)$ mit $\mathfrak{B} \subseteq q$
    • (b) $\forall e \in [\mathfrak{B} ]^{<\omega} \exists g_e \in [S]^{< \omega}$ mit $\mathfrak{C} = { C_e: e \in [\mathfrak{B}]^{<\omega} }$ cwdick; hierbei $C_e = \bigcup_{x \in g_e x^{-1}} D_e$.
    • Nenne $\mathfrak{B}$ dann cwpws.

partial Translation

First page

  • September 14, 2006
  • Continuation: Talk by Sabine Koppelberg
  • 3. collectionwise thick (cwt), collectionwise piecewise syndetic (cwpws)
  • 3.2 Notation. $\mathfrak{B} \subseteq \mathcal{P}(S), \mathfrak{V} := \{ D_e : e \in [\mathcal{B}]^{<\omega} \}$
    • $D_e = \bigcap_{B\in e} B$
    • In particular, $\mathcal{B} \subseteq \mathfrak{V}$
  • 3.1 Definition
    • For $\mathfrak{A} \subseteq \mathcal{P}(S)$ let $Y_{\mathfrak{A}} := \bigcap_{A\in\mathfrak{A}} \widehat{A} \subseteq \beta S$ (closed)
      • [in the setup of 3.2, then $Y_\mathfrak{B} = Y_\mathfrak{D}$]
  • 3.3 Theorem & Definition. TFAE:
    • (a) $\exists q \in \beta S: \beta S \cdot q \subseteq Y_\mathfrak{B}$ (without loss of generality, $q\in K(\beta S)$)
    • (b) $\forall D \in \mathfrak{D}$ $D$ thick.
    • We then call $\mathfrak{B}$ collectionwise thick (cwthick, cwt)
  • Proof:
    • =>: $\beta S \cdot q \subseteq Y_{\mathfrak{B}} = Y_\mathfrak{D} \subseteq \widehat{D}$ for any $D\in \mathfrak{D}$
    • <=: For $e\in [\mathfrak{B}]^{<\omega}$ take $q_e \in \beta S, \beta S \cdot q_e \subseteq \widehat{D_e}$
      • Without loss $q\in K(\beta S)$,
        • (since $q_e \in \beta S \cdot q_e \in \widehat{D_e}$ holds ✓)
      • Let $ X_e := \{ q_f : e \subseteq f \in [ \mathfrak{B} ]^{< \omega} \} \subseteq \beta S$.

Second page

  • Then $X_e \subseteq \widehat{D_e}$
    • [since $f\supseteq e \Rightarrow q_f \in \widehat{D_f} \subseteq \widehat{D_e}$]
  • $\{ X_e : e \in [\mathfrak{B}]^{< \omega} \}$ has the finite intersection property.
  • So take $q\in \bigcap_{e \in [ \mathfrak{B} ]^{<\omega}} cl_{\beta S}(X_e)$
  • Claim: $\forall D \in \mathfrak{D}: \beta S \cdot q \subseteq \widehat{D}$
    • [proof]
    • $D = D_e, e\in [ \mathfrak{B}]^{< \omega} \Rightarrow X_e \subseteq \widehat{D_e}, q\in cl(X_e) \Rightarrow q \in D_e$
    • $\forall S \in S, e \subseteq f, s\cdot q_f \underset{\beta S \cdot q_f \subseteq\widehat{D_f}}{\in} \widehat{D_f} \subseteq \widehat{D_e} \Rightarrow s\cdot X_e \subseteq \widehat{D_e}$
    • $\Rightarrow s\cdot q \in \widehat{D_e} \Rightarrow \beta S \cdot q \subseteq \widehat{D_e}$
    • The claim follows. □
  • ["Exercise": construct $q$ as $p$-limit]
  • 3.4 Theorem & Definition. TFAE:
    • (a) $\exists q\in K(\beta S)$ with $\mathfrak{B} \subseteq q$
    • (b) $\forall e \in [\mathfrak{B} ]^{<\omega} \exists g_e \in [S]^{< \omega}$ mit $\mathfrak{C} = { C_e: e \in [\mathfrak{B}]^{<\omega} }$ cwthick; where $C_e = \bigcup_{x \in g_e x^{-1}} D_e$.
    • We then call $\mathfrak{B}$ collectionwise piecewise syndetic (cwpws).

Notes

We’re back to Sabine Koppelberg’s talks about basic $\beta S$ results (with four more pages to come). This time, tackling the not-so-basic notions of collectionwise thick/pws sets. These notions are cricital for analysing sets the minimal ideal — and equally elusive.

I’m not very happy with notation here; it seems to sacrifice accessibility over corrrectness. A sloppier notation might be helpful. In addition, “collectionwise” is a cumbersome prefix. I’d go for “uniformly” or “coherently” as they are often used in the context of filters (and this is what “collectionwise” is all about). But it probably wouldn’t help to add yet another terminology.

Funny thing. I actually spent my last few weeks in Michigan thinking about these notions.


Red workbook, p12

Source

red workbook, p12

Transcript

  • WA $\stackrel{p = q + r}{\Longrightarrow}$ (i) $\forall k \in \mathbb{N}, W_v \in r: | W_v \cap W_v+k| < \omega$
  • [struck through] (ii) $\forall k \in \mathbb{N}: k \in V \rightarrow 2k \notin V$ ($k$ und $2k$ haben versch. $W_v$)
  • Beweis: $ p = q+r$ beide in $\mathbb{N}^*$, $A\in p$.
    • => $ \displaystyle A = \bigcup_{\underset{\in p}{v\in V}} v + \underset{\in r}{W_v}$ => $\exists v, k: v, v+k \in V$
      • [ $q \in \mathbb{N}^*$]
    • => $W_v \cap W_{v+k} =: W \in r$, also unendlich
    • => $\underbrace{v+W} ,v+k+W \in A$
      • => $v+k+W \in A+k$ => Beh.
  • Notiz: es gibt kofinal viele solche $k$’s! □
  • (2) Ein solches $k$ liefert mit Aufzaehlung von $A \cap A+k$
    • eine kofinale Folge, so dass $|a_{n_k} – a_{n_j -1}| \leq k$.
    • Also kann keine Aufzaehlung jede Schranke auf Endstuecken uebertrefeen.
  • Gilt <= ? d.h. $p\in \mathbb{N}^* \forall A\in p: \underset{\text{unendl}}{ \{ k: A \cap A+k \text{ unendl.} \}}$
    • => $p \in \mathbb{N}^* + \mathbb{N}^*$

partial Translation

  • Assume to the contrary. $\stackrel{p = q + r}{\Longrightarrow}$ (i) $\forall k \in \mathbb{N}, W_v \in r: | W_v \cap W_v+k| < \omega$
  • [struck through] (uu) $\forall k \in \mathbb{N}: k \in V \rightarrow 2k \notin V$ ($k$ and $2k$ have different. $W_v$)
  • Proof: $ p = q+r$ both in $\mathbb{N}^*$, $A\in p$.
    • => $ \displaystyle A = \bigcup_{\underset{\in p}{v\in V}} v + \underset{\in r}{W_v}$ => $\exists v, k: v, v+k \in V$ [[since]] $q \in \mathbb{N}^*$ [[ $V$ is infinite, thus contains two elements]]
    • => $W_v \cap W_{v+k} =: W \in r$, hence infinite
    • => $\underbrace{v+W} ,v+k+W \subseteq A$
      • and $v+k+W \in A+k$ => Claim.
  • Remark: there are cofinal many such $k$! □
  • (2) With such $k$ we can find an enumeration $A \cap A+k$ and some cofinal enumeration $|a_{n_k} – a_{n_j -1}| \leq k$.
    • Hence no enumeration can exceed an arbitrary bound on end pieces
  • Does <= hold? d.h. $p\in \mathbb{N}^* \forall A\in p: \underbrace{ \{ k: A \cap A+k \text{ infinite}\}}_{\text{infinite}}$
    • => $p \in \mathbb{N}^* + \mathbb{N}^*$

Notes

This finishes the attempts to solve 4.1.7 from Hindman&Strauss (successfully). Given the nice write up of the solution, I’m guessing I worked the proof out someplace else (blackboard, separate piece of paper etc). This reminds me that in the office I was working in at the time I found this wonderful stack of thick letter size paper (letter size! in Germany!). I loved writing on the paper for rough drafts, preparing talks/lecture notes etc. But it clashed with my desire to keep notebooks.

This page is extremely fascinating for me because of the final question. It’s always easy to ask yourself if the reverse of a proposition holds; that’s just standard. In this case, the answer should be a pretty straight forward “no”; however, I don’t think I ever worked out a counterexample.

But that’s not what makes this so fascinating for me. What is fascinating is that I spent a lot of time during my postdoc to solve a very similar problem (and failed) which I consider one of the most interesting questions about idempotent filters. Unfortunately, I was unable to solve the question. I don’t want to go into detail here and it will take months until we get to that (a teaser never hurts, right?). It’s fascinating to see that I was very nearly thinking about the very same problem this early in my PhD (and, not surprisingly, missed the actually interesting question at this point).

Open questions

  • If $p\in \mathbb{N}^*$ and $\forall A\in p: \underbrace{ \{ k: A \cap A+k \text{ infinite}\}}_{\text{infinite}}$, does it follow that $p \in \mathbb{N}^* + \mathbb{N}^*$?
    • Probably no — there needs to be more additive structure, in a coherent/filter fashion; just infinite seems too weak.

Red workbook, p11

Source

red workbook, p11-1

red workbook, p11-2

Transcript

Left page

  • Bestanden bei: 60% Uebungen
  • Korrektur d. Klausur mit SK
  • Hindman & Strauss:
  • 4.1.7: $A \in p \in \mathbb{N}^*+\mathbb{N}^*$
    • => $\exists k: |A \cap (A+k)| = \omega$
    • ?=> $A$ kann nicht aufgezaehlt werden, so dass $a_{n+1} – a_n \to \infty (n \to \infty)$
  • vgl. van Douwen Artikel zu $\beta \mathbb{N}$.

Right page

  • Aufgabe 4.1.7
  • $p \in \mathbb{N}^*+\mathbb{N}^* (\subseteq \mathbb{N}^*), A\in p \Rightarrow \exists k \in \mathbb{N}: |A \cap (A+k)| = \omega$
    • WA: $\forall k \in \mathbb{N}: |A \cap A+k| < \omega$
    • [struck through: $\underset{p \in \mathbb{N}^*}{\Rightarrow} \mathbb{N} \setminus A' \in p$] $\Rightarrow \mathbb{N} – (A+k) \in p$
    • $\Rightarrow \exists V^k, (W_v^k)_{v \in V^k}$ unendlich $\bigcup_{v\in V_k} v+ W_v^k \subseteq N – (A-k)$
      • ?? Nutzen? $\bigcap V_k$? $\bigcap W_v^k$?
  • $\mathbb{N} \rightarrow \mathbb{Z}_n \overset{\text{Homo}}{\rightarrow} \rightarrow \beta \mathbb{N} \rightarrow \mathbb{Z}_n$ Homo
    • $\Rightarrow p = q+r \Rightarrow p \bmod n = q \bmod n + r \bmod n$
    • [struck out: [illegible] $= -p +\bmod n \in \mathbb{Z}_p = {0, \ldots, n-1}$
  • $A \in p \in \mathbb{N}^*+\mathbb{N}^* \Rightarrow p = \overset{\overset{\mathbb{N}^*}{\ni \ \ \in}}{q+r} \Rightarrow A \supseteq \bigcup_{v\in V \in q} v+ \overset{\in r}{W_v}$
  • [struck out: WA $\forall k\ in \mathbb{N}: |A \cap A+k| < \omega \Rightarrow \forall k \in \mathbb{N}: k+p \notin \widehat{A}$]
  • WA $\forall k\ in \mathbb{N}: |A \cap A+k| < \omega \underset{p = q+r}{\Longrightarrow} \forall k \in \mathbb{N}: |k+W_v \cap W_v | < \omega$
  • $\Rightarrow \forall k \in \mathbb{N} k+r \notin \widehat{W_v}$ [struckout: $\Rightarrow: \forall k\in \mathbb{N}: k+r \notin \bigcup_{v\in V} \widehat{W_v}$]
  • $\Rightarrow q + r \notin \widehat{W_v} \Rightarrow q+r \notin \bigcup \widehat{W_v}$

partial Translation

Left page

  • [some notes on grading a course]
  • Hindman & Strauss:
  • 4.1.7: $A \in p \in \mathbb{N}^*+\mathbb{N}^*$
    • => $\exists k: |A \cap (A+k)| = \omega$
    • ?=> $A$ can not be enumerated such that $a_{n+1} – a_n \to \infty (n \to \infty)$
  • cf. van Douwen article about $\beta \mathbb{N}$.

Right page

  • Exercise 4.1.7
  • [first try] $p \in \mathbb{N}^*+\mathbb{N}^* (\subseteq \mathbb{N}^*), A\in p \Rightarrow \exists k \in \mathbb{N}: |A \cap (A+k)| = \omega$
    • Assume to the contrary: $\forall k \in \mathbb{N}: |A \cap A+k| < \omega$
    • [struck through: $\underset{p \in \mathbb{N}^*}{\Rightarrow} \mathbb{N} \setminus A' \in p$] $\Rightarrow \mathbb{N} – (A+k) \in p$
    • $\Rightarrow \exists V^k, (W_v^k)_{v \in V^k}$ infinite $\bigcup_{v\in V_k} v+ W_v^k \subseteq N – (A-k)$
      • ?? Useful? $\bigcap V_k$? $\bigcap W_v^k$?
  • [second try] $\mathbb{N} \rightarrow \mathbb{Z}_n \overset{\text{Homomorphism}}{\Longrightarrow} \beta \mathbb{N} \rightarrow \mathbb{Z}_n$ Homomorphism
    • $\Rightarrow p = q+r \Rightarrow p \bmod n = q \bmod n + r \bmod n$
    • [struck out: [something illegible] $= -p +\bmod n \in \mathbb{Z}_p = {0, \ldots, n-1}$
  • [third try] $A \in p \in \mathbb{N}^*+\mathbb{N}^* \Rightarrow p = q+r (\text{both in } \mathbb{N}^*) \Rightarrow A \supseteq \bigcup_{v\in V \in q} v+ \overset{\in r}{W_v}$
  • [struck out: Assume to the contrary $\forall k\ in \mathbb{N}: |A \cap A+k| < \omega \Rightarrow \forall k \in \mathbb{N}: k+p \notin \widehat{A}$]
  • [fourth try] Assume to the contrary $\forall k\ in \mathbb{N}: |A \cap A+k| < \omega \underset{p = q+r}{\Longrightarrow} \forall k \in \mathbb{N}: |k+W_v \cap W_v | < \omega$
  • $\Rightarrow \forall k \in \mathbb{N} k+r \notin \widehat{W_v}$ [struckout: $\Rightarrow: \forall k\in \mathbb{N}: k+r \notin \bigcup_{v\in V} \widehat{W_v}$]
  • $\Rightarrow q + r \notin \widehat{W_v} \Rightarrow q+r \notin \bigcup \widehat{W_V}$

Notes

I find this double page (and the one following it) quite interesting. Mathematically speaking, there’s very little going. If I recall correctly, it was Stefan Geschke (or else Sabine Koppelberg) who had mentioned the fact to me that sets in ultrafilters that are sums have too many small gaps, i.e., the size of gaps in their enumeration does not have an (improper) limit. So I found the exercise in Hindman&Strauss and tried to solve it.

What’s interesting is how I went about solving it. I would call this the “formalist approach”, i.e., by manipulation of symbols following simple logic since I have no intuition of the subject. Of course, I fail, repeatedly; the solution will be found on the next page.

By the way, the first two lines are about the grading of a set theory course (the previous page contains more but I did not reproduce it). I will skip a rant about how PhD students are often forced into TA duties without being paid; in a logical twist, they often “cannot” be paid because they are on grant money and most grants directly prohibit teaching duties.