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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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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 $\xi<\alpha$?2012-01-17
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    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.2012-01-17
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    Could you express your idea by formulas, but not just by language? It's more unclear for me.2012-01-17
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    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.2012-01-17
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    **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 $X_\xi$.2012-01-17
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    No, you aren't using the notation correctly, since the star of the countable set refers to the *union* of the sets, rather than the family of sets themselves. The point is that for every countable set $A$, the set $\{X\in W\mid \exists a\in A\cap X\}$ is countable. So most of the sets in $W$ have empty intersection with $A$.2012-01-17