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sammycaps
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1. So the solutions claim a different and better proof of this, but I just wanted to see if mine made sense.
Question - Let X be a compact Hausdorff space; let {An} be a countable collection of closed sets of X. Show that if each set An has empty interior in X, then the union [itex]\bigcup[/itex]An has empty interior in X.
If X is compact Hausdorff with no isolated points, for any nonempty open set U of X and any point x of X, there exists a nonempty open set V contained in U such that x[itex]\notin[/itex][itex]\overline{V}[/itex].
Now that I've seen the solution it seems obvious, but here is what I did.
If we let Int([itex]\bigcup[/itex]An)=U and take a point xi of Ai, then there exists an open set Vi [itex]\subset[/itex] Int([itex]\bigcup[/itex]An) s.t. x[itex]\notin[/itex][itex]\overline{V}[/itex]i. We can form this construction because we can choose a point in Int([itex]\bigcup[/itex]An) different from x, because if not, then we would either have an empty interior (in the case where x[itex]\notin[/itex] Int([itex]\bigcup[/itex]An) ) or we would have a one-point open set (in the case where x[itex]\in[/itex] Int([itex]\bigcup[/itex]An) , which would mean that Ai for some i has non-empty interior. Then by the Hausdorff property, there is a W1 and W2 about x and y respectively, that are disjoint. Then W2[itex]\bigcap[/itex]Int([itex]\bigcup[/itex]An) is an open set contained in Int([itex]\bigcup[/itex]An), not containing x. It follow that the closure of this set, which I'll call V, does not contain x.
Then we can choose [itex]\overline{V}[/itex]1 [itex]\supset[/itex] [itex]\overline{V}[/itex]2 [itex]\supset[/itex] ...
And since X is compact, there exists an x in this infinite intersection. Then, since U is the union of {An}, x must be contained in Ai for some i. But then [itex]\overline{V}[/itex]i [itex]\cap[/itex] Ai is a closed set containing x but not xi. Then the complement of [itex]\overline{V}[/itex]i [itex]\cap[/itex] Ai in Ai is open and non empty in Ai, implying that Ai has non-empty interior, a contradiction.
Question - Let X be a compact Hausdorff space; let {An} be a countable collection of closed sets of X. Show that if each set An has empty interior in X, then the union [itex]\bigcup[/itex]An has empty interior in X.
Homework Equations
If X is compact Hausdorff with no isolated points, for any nonempty open set U of X and any point x of X, there exists a nonempty open set V contained in U such that x[itex]\notin[/itex][itex]\overline{V}[/itex].
The Attempt at a Solution
Now that I've seen the solution it seems obvious, but here is what I did.
If we let Int([itex]\bigcup[/itex]An)=U and take a point xi of Ai, then there exists an open set Vi [itex]\subset[/itex] Int([itex]\bigcup[/itex]An) s.t. x[itex]\notin[/itex][itex]\overline{V}[/itex]i. We can form this construction because we can choose a point in Int([itex]\bigcup[/itex]An) different from x, because if not, then we would either have an empty interior (in the case where x[itex]\notin[/itex] Int([itex]\bigcup[/itex]An) ) or we would have a one-point open set (in the case where x[itex]\in[/itex] Int([itex]\bigcup[/itex]An) , which would mean that Ai for some i has non-empty interior. Then by the Hausdorff property, there is a W1 and W2 about x and y respectively, that are disjoint. Then W2[itex]\bigcap[/itex]Int([itex]\bigcup[/itex]An) is an open set contained in Int([itex]\bigcup[/itex]An), not containing x. It follow that the closure of this set, which I'll call V, does not contain x.
Then we can choose [itex]\overline{V}[/itex]1 [itex]\supset[/itex] [itex]\overline{V}[/itex]2 [itex]\supset[/itex] ...
And since X is compact, there exists an x in this infinite intersection. Then, since U is the union of {An}, x must be contained in Ai for some i. But then [itex]\overline{V}[/itex]i [itex]\cap[/itex] Ai is a closed set containing x but not xi. Then the complement of [itex]\overline{V}[/itex]i [itex]\cap[/itex] Ai in Ai is open and non empty in Ai, implying that Ai has non-empty interior, a contradiction.
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