disjoint union of Baire rooms which is a Baire room

Claim we have a family members $A_\alpha$ of disjoint Baire rooms. Additionally intend that each $A_\alpha$ is disjoint from the closure of the union of the various other collections. Show that $\bigcup_{\alpha} A_\alpha$ is a Baire room.

I assume we can confirm this by transfinite induction. Intend the building holds for all $\beta$ < $\alpha$, show it holds for $\alpha$.

If $\alpha$ is a restriction ordinal: we intend to show that for every single series of open thick parts of $\bigcup_{\beta<\alpha} A_\alpha$ their junction is open thick. This series of open thick embed in the union is specified as the union of all open thick embed in each of the collections given that they all are Baire rooms and also by induction theory the union approximately $\beta$ is a Baire room. Intend this is not the instance. After that there exists an open set $U$ such that the junction of these open thick collections misses out on $U$. There is some ordinal $\beta$ such that $U \in A_\beta$, yet this is an opposition given that we intended that the union approximately $\beta$ was a Baire room.

Currently just how do I take care of the follower instance? it appears more difficult.

2019-05-07 00:39:18
Source Share
Answers: 1

I'm rather certain it's less complex and also does not call for transfinite induction or anything virtually as hard. In this instance the $A_\alpha$ are open collections (as the enhance of the closure of the union of the others). So intend $U_1, \dots,$ are open thick embed in this union. After that each junction $U_i \cap A_\alpha$ is open and also thick in $A_\alpha$. The countable collection has junction which is thick in each $A_\alpha$ by the Baire building.

This suggests that the countable junction, call it $F$, is thick in the union $\bigcup A_\alpha$. (Any factor in the union belongs among the $A_\alpha$, and also there is a factor of $F \cap A_\alpha$ close by, therefore a factor of $F$ close by.)

2019-05-08 21:19:17