What topological spaces satisfy another property involving relatively compact sets?
$begingroup$
This is a follow-up to my question here. A subset of a topological space is called relatively compact if its closure is compact. My question is, what kind of topological spaces satisfy the following property: there exist countably many relatively compact sets $S_1,S_2,...$ such that every relatively compact set $S$ is a subset of some $S_n$? Or to put it another way, the collection of relatively compact sets has a countable cofinal subcollection.
Is there some category of topological spaces which satisfies this property? Maybe sigma-compact spaces?
My reason for asking this question, by the way, is that relatively compact sets form a bornology for $T_1$ spaces, and this property is one of the conditions for a bornology to be induced by a compatible metric, as I discuss here.
general-topology compactness examples-counterexamples separation-axioms
asked Dec 21 '18 at 3:32
Keshav SrinivasanKeshav Srinivasan
2,33021445
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add a comment |
$begingroup$
This is a follow-up to my question here. A subset of a topological space is called relatively compact if its closure is compact. My question is, what kind of topological spaces satisfy the following property: there exist countably many relatively compact sets $S_1,S_2,...$ such that every relatively compact set $S$ is a subset of some $S_n$? Or to put it another way, the collection of relatively compact sets has a countable cofinal subcollection.
Is there some category of topological spaces which satisfies this property? Maybe sigma-compact spaces?
My reason for asking this question, by the way, is that relatively compact sets form a bornology for $T_1$ spaces, and this property is one of the conditions for a bornology to be induced by a compatible metric, as I discuss here.
general-topology compactness examples-counterexamples separation-axioms
asked Dec 21 '18 at 3:32
Keshav SrinivasanKeshav Srinivasan
2,33021445
$endgroup$
add a comment |
$begingroup$
This is a follow-up to my question here. A subset of a topological space is called relatively compact if its closure is compact. My question is, what kind of topological spaces satisfy the following property: there exist countably many relatively compact sets $S_1,S_2,...$ such that every relatively compact set $S$ is a subset of some $S_n$? Or to put it another way, the collection of relatively compact sets has a countable cofinal subcollection.
Is there some category of topological spaces which satisfies this property? Maybe sigma-compact spaces?
My reason for asking this question, by the way, is that relatively compact sets form a bornology for $T_1$ spaces, and this property is one of the conditions for a bornology to be induced by a compatible metric, as I discuss here.
general-topology compactness examples-counterexamples separation-axioms
asked Dec 21 '18 at 3:32
Keshav SrinivasanKeshav Srinivasan
2,33021445
$endgroup$
This is a follow-up to my question here. A subset of a topological space is called relatively compact if its closure is compact. My question is, what kind of topological spaces satisfy the following property: there exist countably many relatively compact sets $S_1,S_2,...$ such that every relatively compact set $S$ is a subset of some $S_n$? Or to put it another way, the collection of relatively compact sets has a countable cofinal subcollection.
Is there some category of topological spaces which satisfies this property? Maybe sigma-compact spaces?
My reason for asking this question, by the way, is that relatively compact sets form a bornology for $T_1$ spaces, and this property is one of the conditions for a bornology to be induced by a compatible metric, as I discuss here.
general-topology compactness examples-counterexamples separation-axioms
general-topology compactness examples-counterexamples separation-axioms
asked Dec 21 '18 at 3:32
Keshav SrinivasanKeshav Srinivasan
2,33021445
asked Dec 21 '18 at 3:32
Keshav SrinivasanKeshav Srinivasan
2,33021445
asked Dec 21 '18 at 3:32
Keshav SrinivasanKeshav Srinivasan
2,33021445
asked Dec 21 '18 at 3:32
Keshav SrinivasanKeshav Srinivasan
2,33021445
asked Dec 21 '18 at 3:32
Keshav SrinivasanKeshav Srinivasan
2,33021445
2,33021445
add a comment |
add a comment |
1 Answer
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I think hemicompactness comes close: this means that there are countably many compact $H_nsubseteq X$, $n in omega$, such that every compact subset of $X$ is a subset of some $H_n$. This condition is often considered when studying the compact-open topology on $X$.
This property implies $sigma$-compactness but is stronger: $mathbb{Q}$ is $sigma$-compact but not hemicompact (see this planetmath page), but for locally compact Hausdorff spaces the notions are equivalent.
It is clear that a hemicompact $X$ satisfies your condition, and a space satisfying your condition is hemicompact (using the closures of the base sets).
answered Dec 21 '18 at 5:39
Henno BrandsmaHenno Brandsma
111k348118
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$begingroup$
Thanks. By the way, I just posted a follow-up question: math.stackexchange.com/q/3048232/71829
$endgroup$
– Keshav Srinivasan
Dec 21 '18 at 6:23
add a comment |
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$begingroup$
I think hemicompactness comes close: this means that there are countably many compact $H_nsubseteq X$, $n in omega$, such that every compact subset of $X$ is a subset of some $H_n$. This condition is often considered when studying the compact-open topology on $X$.
This property implies $sigma$-compactness but is stronger: $mathbb{Q}$ is $sigma$-compact but not hemicompact (see this planetmath page), but for locally compact Hausdorff spaces the notions are equivalent.
It is clear that a hemicompact $X$ satisfies your condition, and a space satisfying your condition is hemicompact (using the closures of the base sets).
answered Dec 21 '18 at 5:39
Henno BrandsmaHenno Brandsma
111k348118
$endgroup$
$begingroup$
Thanks. By the way, I just posted a follow-up question: math.stackexchange.com/q/3048232/71829
$endgroup$
– Keshav Srinivasan
Dec 21 '18 at 6:23
add a comment |
$begingroup$
I think hemicompactness comes close: this means that there are countably many compact $H_nsubseteq X$, $n in omega$, such that every compact subset of $X$ is a subset of some $H_n$. This condition is often considered when studying the compact-open topology on $X$.
This property implies $sigma$-compactness but is stronger: $mathbb{Q}$ is $sigma$-compact but not hemicompact (see this planetmath page), but for locally compact Hausdorff spaces the notions are equivalent.
It is clear that a hemicompact $X$ satisfies your condition, and a space satisfying your condition is hemicompact (using the closures of the base sets).
answered Dec 21 '18 at 5:39
Henno BrandsmaHenno Brandsma
111k348118
$endgroup$
$begingroup$
Thanks. By the way, I just posted a follow-up question: math.stackexchange.com/q/3048232/71829
$endgroup$
– Keshav Srinivasan
Dec 21 '18 at 6:23
add a comment |
$begingroup$
I think hemicompactness comes close: this means that there are countably many compact $H_nsubseteq X$, $n in omega$, such that every compact subset of $X$ is a subset of some $H_n$. This condition is often considered when studying the compact-open topology on $X$.
This property implies $sigma$-compactness but is stronger: $mathbb{Q}$ is $sigma$-compact but not hemicompact (see this planetmath page), but for locally compact Hausdorff spaces the notions are equivalent.
It is clear that a hemicompact $X$ satisfies your condition, and a space satisfying your condition is hemicompact (using the closures of the base sets).
answered Dec 21 '18 at 5:39
Henno BrandsmaHenno Brandsma
111k348118
$endgroup$
I think hemicompactness comes close: this means that there are countably many compact $H_nsubseteq X$, $n in omega$, such that every compact subset of $X$ is a subset of some $H_n$. This condition is often considered when studying the compact-open topology on $X$.
This property implies $sigma$-compactness but is stronger: $mathbb{Q}$ is $sigma$-compact but not hemicompact (see this planetmath page), but for locally compact Hausdorff spaces the notions are equivalent.
It is clear that a hemicompact $X$ satisfies your condition, and a space satisfying your condition is hemicompact (using the closures of the base sets).
answered Dec 21 '18 at 5:39
Henno BrandsmaHenno Brandsma
111k348118
edited Dec 21 '18 at 6:28
answered Dec 21 '18 at 5:39
Henno BrandsmaHenno Brandsma
111k348118
answered Dec 21 '18 at 5:39
Henno BrandsmaHenno Brandsma
111k348118
answered Dec 21 '18 at 5:39
Henno BrandsmaHenno Brandsma
111k348118
111k348118
$begingroup$
Thanks. By the way, I just posted a follow-up question: math.stackexchange.com/q/3048232/71829
$endgroup$
– Keshav Srinivasan
Dec 21 '18 at 6:23
add a comment |
$begingroup$
Thanks. By the way, I just posted a follow-up question: math.stackexchange.com/q/3048232/71829
$endgroup$
– Keshav Srinivasan
Dec 21 '18 at 6:23
$begingroup$
Thanks. By the way, I just posted a follow-up question: math.stackexchange.com/q/3048232/71829
$endgroup$
– Keshav Srinivasan
Dec 21 '18 at 6:23
$begingroup$
Thanks. By the way, I just posted a follow-up question: math.stackexchange.com/q/3048232/71829
$endgroup$
– Keshav Srinivasan
Dec 21 '18 at 6:23
add a comment |
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