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Say we have a family {$A_\alpha$} of disjoint Baire spaces. Also suppose that each $A_\alpha$ is disjoint from the closure of the union of the other sets. Show that $\bigcup_{\alpha} A_\alpha$ is a Baire space.

I think we can prove this by transfinite induction. Suppose the property holds for all $\beta$ < $\alpha$, show it holds for $\alpha$.

If $\alpha$ is a limit ordinal: we want to show that for every sequence of open dense subsets of $\bigcup_{\beta<\alpha} A_\alpha$ their intersection is open dense. This sequence of open dense sets in the union is defined as the union of all open dense sets in each of the sets since they all are Baire spaces and by induction hypothesis the union up to $\beta$ is a Baire space. Suppose this is not the case. Then there exists an open set $U$ such that the intersection of these open dense sets misses $U$. There is some ordinal $\beta$ such that $U \in A_\beta$, but this is a contradiction since we supposed that the union up to $\beta$ was a Baire space.

Now how do I handle the successor case? it seems trickier.

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I'm pretty sure it's simpler and doesn't require transfinite induction or anything nearly as difficult. In this case the $A_\alpha$ are open sets (as the complement of the closure of the union of the others). So suppose $U_1, \dots,$ are open dense sets in this union. Then each intersection $U_i \cap A_\alpha$ is open and dense in $A_\alpha$. The countable collection has intersection which is dense in each $A_\alpha$ by the Baire property.

This means that the countable intersection, call it $F$, is dense in the union $\bigcup A_\alpha$. (Any point in the union belongs one of the $A_\alpha$, and there is a point of $F \cap A_\alpha$ nearby, hence a point of $F$ nearby.)

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