If a set is $F_sigma$ and is of first category then it is a countable union of closed nowhere dense sets
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We say a set is $F_sigma$ if it is a countable union of closed sets. We say a set is of first category if it is a countable union of nowhere dense sets.
The question is, if a set of $F_sigma$ and of first category, is it true that this set is a countable union of closed nowhere dense sets.
real-analysis functional-analysis analysis
asked Nov 30 '18 at 4:44
bbwbbw
47038
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add a comment |
$begingroup$
We say a set is $F_sigma$ if it is a countable union of closed sets. We say a set is of first category if it is a countable union of nowhere dense sets.
The question is, if a set of $F_sigma$ and of first category, is it true that this set is a countable union of closed nowhere dense sets.
real-analysis functional-analysis analysis
asked Nov 30 '18 at 4:44
bbwbbw
47038
$endgroup$
add a comment |
$begingroup$
We say a set is $F_sigma$ if it is a countable union of closed sets. We say a set is of first category if it is a countable union of nowhere dense sets.
The question is, if a set of $F_sigma$ and of first category, is it true that this set is a countable union of closed nowhere dense sets.
real-analysis functional-analysis analysis
asked Nov 30 '18 at 4:44
bbwbbw
47038
$endgroup$
We say a set is $F_sigma$ if it is a countable union of closed sets. We say a set is of first category if it is a countable union of nowhere dense sets.
The question is, if a set of $F_sigma$ and of first category, is it true that this set is a countable union of closed nowhere dense sets.
real-analysis functional-analysis analysis
real-analysis functional-analysis analysis
asked Nov 30 '18 at 4:44
bbwbbw
47038
asked Nov 30 '18 at 4:44
bbwbbw
47038
asked Nov 30 '18 at 4:44
bbwbbw
47038
asked Nov 30 '18 at 4:44
bbwbbw
47038
asked Nov 30 '18 at 4:44
bbwbbw
47038
47038
add a comment |
add a comment |
1 Answer
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Yes, an $F_sigma$ set of the first category is the union of countably many closed nowhere dense sets.
Since an $F_sigma$ set of the first category is the union of countably many closed sets of the first category, it will suffice to show that a closed set of the first category is the union of countably many closed nowhere dense sets. Well, a closed set of the first category is the union of countably many nowhere dense sets, and it is also the union of the closures of those nowhere dense sets, and the closure of a nowhere dense set is a closed nowhere dense set, Q.E.D.
Actually, a closed set of the first category (in a Baire space) is nowhere dense; so there is a shorter proof, if you're working with a Baire space.
answered Nov 30 '18 at 4:57
bofbof
51k457120
$endgroup$
$begingroup$
If a set $A$ is $F_sigma$ and is of first category, then it is a countable union of closed nowhere dense sets. In this case, is $A$ closed? The closure of an infinite union may not be equal to the infinite union of closures right?
$endgroup$
– bbw
Nov 30 '18 at 5:53
$begingroup$
Right, an $F_sigma$ set of first category is not necessarily closed. Any countable subset of $mathbb R$ is an $F_sigma$ set of first category, but a countable set of real numbers is not necessarily closed, for example, the set $mathbb Q$ of all rational numbers.
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– bof
Nov 30 '18 at 6:16
add a comment |
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1 Answer
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$begingroup$
Yes, an $F_sigma$ set of the first category is the union of countably many closed nowhere dense sets.
Since an $F_sigma$ set of the first category is the union of countably many closed sets of the first category, it will suffice to show that a closed set of the first category is the union of countably many closed nowhere dense sets. Well, a closed set of the first category is the union of countably many nowhere dense sets, and it is also the union of the closures of those nowhere dense sets, and the closure of a nowhere dense set is a closed nowhere dense set, Q.E.D.
Actually, a closed set of the first category (in a Baire space) is nowhere dense; so there is a shorter proof, if you're working with a Baire space.
answered Nov 30 '18 at 4:57
bofbof
51k457120
$endgroup$
$begingroup$
If a set $A$ is $F_sigma$ and is of first category, then it is a countable union of closed nowhere dense sets. In this case, is $A$ closed? The closure of an infinite union may not be equal to the infinite union of closures right?
$endgroup$
– bbw
Nov 30 '18 at 5:53
$begingroup$
Right, an $F_sigma$ set of first category is not necessarily closed. Any countable subset of $mathbb R$ is an $F_sigma$ set of first category, but a countable set of real numbers is not necessarily closed, for example, the set $mathbb Q$ of all rational numbers.
$endgroup$
– bof
Nov 30 '18 at 6:16
add a comment |
$begingroup$
Yes, an $F_sigma$ set of the first category is the union of countably many closed nowhere dense sets.
Since an $F_sigma$ set of the first category is the union of countably many closed sets of the first category, it will suffice to show that a closed set of the first category is the union of countably many closed nowhere dense sets. Well, a closed set of the first category is the union of countably many nowhere dense sets, and it is also the union of the closures of those nowhere dense sets, and the closure of a nowhere dense set is a closed nowhere dense set, Q.E.D.
Actually, a closed set of the first category (in a Baire space) is nowhere dense; so there is a shorter proof, if you're working with a Baire space.
answered Nov 30 '18 at 4:57
bofbof
51k457120
$endgroup$
$begingroup$
If a set $A$ is $F_sigma$ and is of first category, then it is a countable union of closed nowhere dense sets. In this case, is $A$ closed? The closure of an infinite union may not be equal to the infinite union of closures right?
$endgroup$
– bbw
Nov 30 '18 at 5:53
$begingroup$
Right, an $F_sigma$ set of first category is not necessarily closed. Any countable subset of $mathbb R$ is an $F_sigma$ set of first category, but a countable set of real numbers is not necessarily closed, for example, the set $mathbb Q$ of all rational numbers.
$endgroup$
– bof
Nov 30 '18 at 6:16
add a comment |
$begingroup$
Yes, an $F_sigma$ set of the first category is the union of countably many closed nowhere dense sets.
Since an $F_sigma$ set of the first category is the union of countably many closed sets of the first category, it will suffice to show that a closed set of the first category is the union of countably many closed nowhere dense sets. Well, a closed set of the first category is the union of countably many nowhere dense sets, and it is also the union of the closures of those nowhere dense sets, and the closure of a nowhere dense set is a closed nowhere dense set, Q.E.D.
Actually, a closed set of the first category (in a Baire space) is nowhere dense; so there is a shorter proof, if you're working with a Baire space.
answered Nov 30 '18 at 4:57
bofbof
51k457120
$endgroup$
Yes, an $F_sigma$ set of the first category is the union of countably many closed nowhere dense sets.
Since an $F_sigma$ set of the first category is the union of countably many closed sets of the first category, it will suffice to show that a closed set of the first category is the union of countably many closed nowhere dense sets. Well, a closed set of the first category is the union of countably many nowhere dense sets, and it is also the union of the closures of those nowhere dense sets, and the closure of a nowhere dense set is a closed nowhere dense set, Q.E.D.
Actually, a closed set of the first category (in a Baire space) is nowhere dense; so there is a shorter proof, if you're working with a Baire space.
answered Nov 30 '18 at 4:57
bofbof
51k457120
answered Nov 30 '18 at 4:57
bofbof
51k457120
answered Nov 30 '18 at 4:57
bofbof
51k457120
answered Nov 30 '18 at 4:57
bofbof
51k457120
51k457120
$begingroup$
If a set $A$ is $F_sigma$ and is of first category, then it is a countable union of closed nowhere dense sets. In this case, is $A$ closed? The closure of an infinite union may not be equal to the infinite union of closures right?
$endgroup$
– bbw
Nov 30 '18 at 5:53
$begingroup$
Right, an $F_sigma$ set of first category is not necessarily closed. Any countable subset of $mathbb R$ is an $F_sigma$ set of first category, but a countable set of real numbers is not necessarily closed, for example, the set $mathbb Q$ of all rational numbers.
$endgroup$
– bof
Nov 30 '18 at 6:16
add a comment |
$begingroup$
If a set $A$ is $F_sigma$ and is of first category, then it is a countable union of closed nowhere dense sets. In this case, is $A$ closed? The closure of an infinite union may not be equal to the infinite union of closures right?
$endgroup$
– bbw
Nov 30 '18 at 5:53
$begingroup$
Right, an $F_sigma$ set of first category is not necessarily closed. Any countable subset of $mathbb R$ is an $F_sigma$ set of first category, but a countable set of real numbers is not necessarily closed, for example, the set $mathbb Q$ of all rational numbers.
$endgroup$
– bof
Nov 30 '18 at 6:16
$begingroup$
If a set $A$ is $F_sigma$ and is of first category, then it is a countable union of closed nowhere dense sets. In this case, is $A$ closed? The closure of an infinite union may not be equal to the infinite union of closures right?
$endgroup$
– bbw
Nov 30 '18 at 5:53
$begingroup$
If a set $A$ is $F_sigma$ and is of first category, then it is a countable union of closed nowhere dense sets. In this case, is $A$ closed? The closure of an infinite union may not be equal to the infinite union of closures right?
$endgroup$
– bbw
Nov 30 '18 at 5:53
$begingroup$
Right, an $F_sigma$ set of first category is not necessarily closed. Any countable subset of $mathbb R$ is an $F_sigma$ set of first category, but a countable set of real numbers is not necessarily closed, for example, the set $mathbb Q$ of all rational numbers.
$endgroup$
– bof
Nov 30 '18 at 6:16
$begingroup$
Right, an $F_sigma$ set of first category is not necessarily closed. Any countable subset of $mathbb R$ is an $F_sigma$ set of first category, but a countable set of real numbers is not necessarily closed, for example, the set $mathbb Q$ of all rational numbers.
$endgroup$
– bof
Nov 30 '18 at 6:16
add a comment |
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