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For what spaces does every countable Borel cover have a finite subcover?
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A topological space $X$ is countably compact if every countable open cover of $X$ has a finite subcover. But I'm interested in a stronger condition. Suppose that every countable cover of $X$ by Borel sets has a finite subcover. Then what properties must X have?
And what are examples of spaces that do and don't have this property?
general-topology compactness examples-counterexamples borel-sets
asked 3 hours ago
Keshav Srinivasan
2,34611340
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up vote
3
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favorite
A topological space $X$ is countably compact if every countable open cover of $X$ has a finite subcover. But I'm interested in a stronger condition. Suppose that every countable cover of $X$ by Borel sets has a finite subcover. Then what properties must X have?
And what are examples of spaces that do and don't have this property?
general-topology compactness examples-counterexamples borel-sets
asked 3 hours ago
Keshav Srinivasan
2,34611340
add a comment |
up vote
3
down vote
favorite
up vote
3
down vote
favorite
A topological space $X$ is countably compact if every countable open cover of $X$ has a finite subcover. But I'm interested in a stronger condition. Suppose that every countable cover of $X$ by Borel sets has a finite subcover. Then what properties must X have?
And what are examples of spaces that do and don't have this property?
general-topology compactness examples-counterexamples borel-sets
asked 3 hours ago
Keshav Srinivasan
2,34611340
A topological space $X$ is countably compact if every countable open cover of $X$ has a finite subcover. But I'm interested in a stronger condition. Suppose that every countable cover of $X$ by Borel sets has a finite subcover. Then what properties must X have?
And what are examples of spaces that do and don't have this property?
general-topology compactness examples-counterexamples borel-sets
general-topology compactness examples-counterexamples borel-sets
asked 3 hours ago
Keshav Srinivasan
2,34611340
asked 3 hours ago
Keshav Srinivasan
2,34611340
asked 3 hours ago
Keshav Srinivasan
2,34611340
asked 3 hours ago
Keshav Srinivasan
2,34611340
asked 3 hours ago
Keshav Srinivasan
2,34611340
2,34611340
add a comment |
add a comment |
1 Answer
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This is an extraordinarily strong property: it is equivalent to $X$ having only finitely many open sets. Indeed, it is easy to see that if $X$ has finitely many open sets, then it has only finitely many Borel sets, and so your condition holds trivially.
Conversely, suppose $X$ has infinitely many open sets. Then in particular $X$ has infinitely many Borel sets. Let us call a set $Asubseteq X$ large if $A$ is Borel and has infinitely many Borel subsets. Note that if $A$ is large and $A$ is the disjoint union of two Borel sets $B$ and $C$, then at least one of $B$ and $C$ is large, since every Borel subset of $A$ is the union of a Borel subset of $B$ and a Borel subset of $C$. It follows that any large set has a large proper subset (pick some nonempty proper Borel subset, and either it or its complement must be large).
We can now use this to construct a countable Borel cover of $X$ with no finite subcover. Starting with $A_0=X$, we can pick a large proper subset $A_1subset A_0$, and then a large proper subset $A_2subset A_1$, and so on. The sets $A_0setminus A_1, A_1setminus A_2,dots$ and $bigcap_n A_n$ are then all Borel and cover $X$, and have no finite subcover (they are disjoint and all except possibly $bigcap_n A_n$ are nonempty).
(This construction more generally shows that any infinite Boolean algebra has an infinite sequence of nonzero pairwise disjoint elements.)
answered 3 hours ago
Eric Wofsey
173k12201324
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
accepted
This is an extraordinarily strong property: it is equivalent to $X$ having only finitely many open sets. Indeed, it is easy to see that if $X$ has finitely many open sets, then it has only finitely many Borel sets, and so your condition holds trivially.
Conversely, suppose $X$ has infinitely many open sets. Then in particular $X$ has infinitely many Borel sets. Let us call a set $Asubseteq X$ large if $A$ is Borel and has infinitely many Borel subsets. Note that if $A$ is large and $A$ is the disjoint union of two Borel sets $B$ and $C$, then at least one of $B$ and $C$ is large, since every Borel subset of $A$ is the union of a Borel subset of $B$ and a Borel subset of $C$. It follows that any large set has a large proper subset (pick some nonempty proper Borel subset, and either it or its complement must be large).
We can now use this to construct a countable Borel cover of $X$ with no finite subcover. Starting with $A_0=X$, we can pick a large proper subset $A_1subset A_0$, and then a large proper subset $A_2subset A_1$, and so on. The sets $A_0setminus A_1, A_1setminus A_2,dots$ and $bigcap_n A_n$ are then all Borel and cover $X$, and have no finite subcover (they are disjoint and all except possibly $bigcap_n A_n$ are nonempty).
(This construction more generally shows that any infinite Boolean algebra has an infinite sequence of nonzero pairwise disjoint elements.)
answered 3 hours ago
Eric Wofsey
173k12201324
add a comment |
up vote
5
down vote
accepted
This is an extraordinarily strong property: it is equivalent to $X$ having only finitely many open sets. Indeed, it is easy to see that if $X$ has finitely many open sets, then it has only finitely many Borel sets, and so your condition holds trivially.
Conversely, suppose $X$ has infinitely many open sets. Then in particular $X$ has infinitely many Borel sets. Let us call a set $Asubseteq X$ large if $A$ is Borel and has infinitely many Borel subsets. Note that if $A$ is large and $A$ is the disjoint union of two Borel sets $B$ and $C$, then at least one of $B$ and $C$ is large, since every Borel subset of $A$ is the union of a Borel subset of $B$ and a Borel subset of $C$. It follows that any large set has a large proper subset (pick some nonempty proper Borel subset, and either it or its complement must be large).
We can now use this to construct a countable Borel cover of $X$ with no finite subcover. Starting with $A_0=X$, we can pick a large proper subset $A_1subset A_0$, and then a large proper subset $A_2subset A_1$, and so on. The sets $A_0setminus A_1, A_1setminus A_2,dots$ and $bigcap_n A_n$ are then all Borel and cover $X$, and have no finite subcover (they are disjoint and all except possibly $bigcap_n A_n$ are nonempty).
(This construction more generally shows that any infinite Boolean algebra has an infinite sequence of nonzero pairwise disjoint elements.)
answered 3 hours ago
Eric Wofsey
173k12201324
add a comment |
up vote
5
down vote
accepted
up vote
5
down vote
accepted
This is an extraordinarily strong property: it is equivalent to $X$ having only finitely many open sets. Indeed, it is easy to see that if $X$ has finitely many open sets, then it has only finitely many Borel sets, and so your condition holds trivially.
Conversely, suppose $X$ has infinitely many open sets. Then in particular $X$ has infinitely many Borel sets. Let us call a set $Asubseteq X$ large if $A$ is Borel and has infinitely many Borel subsets. Note that if $A$ is large and $A$ is the disjoint union of two Borel sets $B$ and $C$, then at least one of $B$ and $C$ is large, since every Borel subset of $A$ is the union of a Borel subset of $B$ and a Borel subset of $C$. It follows that any large set has a large proper subset (pick some nonempty proper Borel subset, and either it or its complement must be large).
We can now use this to construct a countable Borel cover of $X$ with no finite subcover. Starting with $A_0=X$, we can pick a large proper subset $A_1subset A_0$, and then a large proper subset $A_2subset A_1$, and so on. The sets $A_0setminus A_1, A_1setminus A_2,dots$ and $bigcap_n A_n$ are then all Borel and cover $X$, and have no finite subcover (they are disjoint and all except possibly $bigcap_n A_n$ are nonempty).
(This construction more generally shows that any infinite Boolean algebra has an infinite sequence of nonzero pairwise disjoint elements.)
answered 3 hours ago
Eric Wofsey
173k12201324
This is an extraordinarily strong property: it is equivalent to $X$ having only finitely many open sets. Indeed, it is easy to see that if $X$ has finitely many open sets, then it has only finitely many Borel sets, and so your condition holds trivially.
Conversely, suppose $X$ has infinitely many open sets. Then in particular $X$ has infinitely many Borel sets. Let us call a set $Asubseteq X$ large if $A$ is Borel and has infinitely many Borel subsets. Note that if $A$ is large and $A$ is the disjoint union of two Borel sets $B$ and $C$, then at least one of $B$ and $C$ is large, since every Borel subset of $A$ is the union of a Borel subset of $B$ and a Borel subset of $C$. It follows that any large set has a large proper subset (pick some nonempty proper Borel subset, and either it or its complement must be large).
We can now use this to construct a countable Borel cover of $X$ with no finite subcover. Starting with $A_0=X$, we can pick a large proper subset $A_1subset A_0$, and then a large proper subset $A_2subset A_1$, and so on. The sets $A_0setminus A_1, A_1setminus A_2,dots$ and $bigcap_n A_n$ are then all Borel and cover $X$, and have no finite subcover (they are disjoint and all except possibly $bigcap_n A_n$ are nonempty).
(This construction more generally shows that any infinite Boolean algebra has an infinite sequence of nonzero pairwise disjoint elements.)
answered 3 hours ago
Eric Wofsey
173k12201324
answered 3 hours ago
Eric Wofsey
173k12201324
answered 3 hours ago
Eric Wofsey
173k12201324
answered 3 hours ago
Eric Wofsey
173k12201324
173k12201324
add a comment |
add a comment |
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