# 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.