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There is a theorem on $\Delta$-system in Jech's set theory above; let us observe the last sentence in the proof. For every $\alpha$, how does it always hold? In other words, how is the $Z$ constructed? Any help will be appreciated. Thanks ahead:)

Am I right: Since $|\{X \in W: a \in X\}|< \omega_1$, for any $a$, hence for any $\xi<\alpha$, given $X_\xi$, we have $|St(\bigcup{X_\xi}, W)|<\omega_1$. Therefore we have some $X \in W\setminus St(\bigcup{X_\xi,W})$ such that $X_\alpha=X$, which disjoint from all $X_\xi$.

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1 Answer 1

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If each $a$ belongs to at most countably many $X\in W$, then if we've chosen countably many $X_\xi$ for $\xi\lt\alpha$, then altogether these have used only countably many $a$'s, namely, those in $\bigcup_{\xi\lt\alpha}X_\xi$. Each of these $a$'s appears in only countably many further $X\in W$, so this rules out altogether only countably many $X\in W$, and so there are uncountably many $X\in W$ remaining that are disjoint from our previous choices, and so we may find $X_\alpha$ disjoint from all $X_\xi$ for $\xi\lt\alpha$, as desired.

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We only can make some $a$ not in $X_\alpha$ by your answer, however, can we find $X_\alpha$ disjoint from all $X_\xi$ for $\xi<\alpha$? –  Paul Jan 17 '12 at 13:15
    
The point is that each $a$ appears in only countably many $X\in W$. Since we have only countably many $a$ so far, namely those in $\bigcup_{\xi\lt\alpha}X_\xi$, these $a$'s appear altogether in only countably many $X\in W$. Those are the bad $X$'s, the ones containing an $a$ we already have. All the rest of the $X\in W$, which is uncountably many and in fact co-countably many, are good, since they contain no $a$'s that we've already got. In other words, the good $X$'s are disjoint from our current $X_\xi$'s. –  JDH Jan 17 '12 at 13:26
    
Could you express your idea by formulas, but not just by language? It's more unclear for me. –  Paul Jan 17 '12 at 13:43
    
To say it differently: each $a$ individually rules out countably many $X\in W$. So a countable set of $a$'s collectively rule out a countable union of countably many $X\in W$, so still countably many $X$ ruled out collectively. The remaining $X$ have no $a$'s from what we've got so far. –  JDH Jan 17 '12 at 13:44
    
Am I right: Since $|\{X \in W: a \in X\}|< \omega_1$, for any $a$, hence for any $\xi<\alpha$, given $X_\xi$, we have $|St(\bigcup{X_\xi}, W)|<\omega_1$. Therefore we have some $X \in W\setminus St(\bigcup{X_\xi,W})$ such that $X_\alpha=X$, which disjoint from all $X_\xi$. –  Paul Jan 17 '12 at 14:10

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