Peter Krautzberger on the web

Red workbook, p14

Source

red workbook, p14
Red Workbook, p.14

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 $\beta S \cdot q \subseteq \bigcup_{x \in g_e} \widehat{ x^{-1} D_e} = \widehat{C_e}$
    • $\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 $\beta S \cdot q \subseteq \bigcup_{x \in g_e} \widehat{ x^{-1} D_e} = \widehat{C_e}$
    • $\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.