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updated proof on diagonal intersection
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Josia Pietsch 2024-01-16 00:10:27 +01:00
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inputs/lecture_14.tex
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@ -162,15 +162,17 @@ We have shown (assuming \AxC to choose contained clubs):
We aim to show that $\gamma \in \diagi_{\beta < \kappa} D_{\beta}$. We aim to show that $\gamma \in \diagi_{\beta < \kappa} D_{\beta}$.
Let $\beta_0 < \gamma$. Let $\beta_0 < \gamma$.
We need to see that $\gamma \in D_{\beta_0}$. We need to see that $\gamma \in D_{\beta_0}$.
For each $\beta_0 \le \beta' < \gamma$ For each $\beta_0 \le \beta' < \gamma$
there is some $\beta'' \in \diagi_{\beta < \kappa} D_\beta$ there is some $\beta'' \in \diagi_{\beta < \kappa} D_\beta$
such that $\beta' \le \beta'' < \gamma$, such that $\beta' \le \beta'' < \gamma$,
since $\gamma = \sup((\diagi_{\beta < \kappa} D_\beta) \cap \gamma)$. since $\gamma = \sup((\diagi_{\beta < \kappa} D_\beta) \cap \gamma)$.
In particular $\beta'' \in D_{\beta_0}$. In particular $\beta'' \in D_{\beta_0}$.
We showed that $D_{\beta_0} \cap \gamma$ So $D_{\beta_0} \cap \gamma$
is unbounded in $\gamma$, is unbounded in $\gamma$.
so $\gamma \in D_{\beta_0}$, since $D_{\beta_0}$ is closed. Since $D_{\beta_0}$ is closed
it follows that $\gamma \in D_{\beta_0}$.
%As $\beta_0 < \gamma$ was arbitrary, %As $\beta_0 < \gamma$ was arbitrary,
%this shows that $\gamma \in \diagi_{\beta < n} D_\beta$. %this shows that $\gamma \in \diagi_{\beta < n} D_\beta$.