What topological spaces satisfy a property involving relatively compact sets?
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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: for every relatively compact set $S$, there exists a relatively compact set $T$ such that the closure of $S$ is a subset of the interior of $T$?
Is there some category of topological spaces which satisfies this property? And is there an example of a $T_1$ space which does not satisfy this property?
My reason for asking this, by the way, is that relatively compact subsets of a $T_1$ form a bornology, and this property says that that bornology interacts well with the topology.
general-topology compactness examples-counterexamples separation-axioms
asked Dec 20 '18 at 2:41
Keshav SrinivasanKeshav Srinivasan
2,37221445
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add a comment |
$begingroup$
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: for every relatively compact set $S$, there exists a relatively compact set $T$ such that the closure of $S$ is a subset of the interior of $T$?
Is there some category of topological spaces which satisfies this property? And is there an example of a $T_1$ space which does not satisfy this property?
My reason for asking this, by the way, is that relatively compact subsets of a $T_1$ form a bornology, and this property says that that bornology interacts well with the topology.
general-topology compactness examples-counterexamples separation-axioms
asked Dec 20 '18 at 2:41
Keshav SrinivasanKeshav Srinivasan
2,37221445
$endgroup$
add a comment |
$begingroup$
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: for every relatively compact set $S$, there exists a relatively compact set $T$ such that the closure of $S$ is a subset of the interior of $T$?
Is there some category of topological spaces which satisfies this property? And is there an example of a $T_1$ space which does not satisfy this property?
My reason for asking this, by the way, is that relatively compact subsets of a $T_1$ form a bornology, and this property says that that bornology interacts well with the topology.
general-topology compactness examples-counterexamples separation-axioms
asked Dec 20 '18 at 2:41
Keshav SrinivasanKeshav Srinivasan
2,37221445
$endgroup$
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: for every relatively compact set $S$, there exists a relatively compact set $T$ such that the closure of $S$ is a subset of the interior of $T$?
Is there some category of topological spaces which satisfies this property? And is there an example of a $T_1$ space which does not satisfy this property?
My reason for asking this, by the way, is that relatively compact subsets of a $T_1$ form a bornology, and this property says that that bornology interacts well with the topology.
general-topology compactness examples-counterexamples separation-axioms
general-topology compactness examples-counterexamples separation-axioms
asked Dec 20 '18 at 2:41
Keshav SrinivasanKeshav Srinivasan
2,37221445
asked Dec 20 '18 at 2:41
Keshav SrinivasanKeshav Srinivasan
2,37221445
asked Dec 20 '18 at 2:41
Keshav SrinivasanKeshav Srinivasan
2,37221445
asked Dec 20 '18 at 2:41
Keshav SrinivasanKeshav Srinivasan
2,37221445
asked Dec 20 '18 at 2:41
Keshav SrinivasanKeshav Srinivasan
2,37221445
2,37221445
add a comment |
add a comment |
1 Answer
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oldest
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This is equivalent to local compactness.
If a topological space satisfies your property, then a singleton point set ${x}$ is compact, and hence there must be a relatively compact set $T$ whose interior contains $x$ (in fact, it contains $overline{{x}}$, which may be strictly larger). The closure of $T$ is a compact neighbourhood of $x$.
If, on the other hand, if we have locally compact space and a relatively compact set $S$, then $overline{S}$ is compact. For each point $s in overline{S}$, there is an open neighbourhood $mathcal{N}_s$ of $s$ that is compact. Note that the interiors of these neighbourhoods cover $overline{S}$, hence there must be a finite-subcover of $overline{S}$. Unioning this finite subcover yields an open, relatively compact neighbourhood containing $overline{S}$, establishing equivalence.
answered Dec 20 '18 at 3:14
Theo BenditTheo Bendit
19.8k12354
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$begingroup$
I just posted a follow-up question: math.stackexchange.com/q/3048148/71829
$endgroup$
– Keshav Srinivasan
Dec 21 '18 at 3:37
$begingroup$
And now I posted one more 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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1 Answer
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1 Answer
1
active
oldest
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active
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active
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$begingroup$
This is equivalent to local compactness.
If a topological space satisfies your property, then a singleton point set ${x}$ is compact, and hence there must be a relatively compact set $T$ whose interior contains $x$ (in fact, it contains $overline{{x}}$, which may be strictly larger). The closure of $T$ is a compact neighbourhood of $x$.
If, on the other hand, if we have locally compact space and a relatively compact set $S$, then $overline{S}$ is compact. For each point $s in overline{S}$, there is an open neighbourhood $mathcal{N}_s$ of $s$ that is compact. Note that the interiors of these neighbourhoods cover $overline{S}$, hence there must be a finite-subcover of $overline{S}$. Unioning this finite subcover yields an open, relatively compact neighbourhood containing $overline{S}$, establishing equivalence.
answered Dec 20 '18 at 3:14
Theo BenditTheo Bendit
19.8k12354
$endgroup$
$begingroup$
I just posted a follow-up question: math.stackexchange.com/q/3048148/71829
$endgroup$
– Keshav Srinivasan
Dec 21 '18 at 3:37
$begingroup$
And now I posted one more follow-up question: math.stackexchange.com/q/3048232/71829
$endgroup$
– Keshav Srinivasan
Dec 21 '18 at 6:23
add a comment |
$begingroup$
This is equivalent to local compactness.
If a topological space satisfies your property, then a singleton point set ${x}$ is compact, and hence there must be a relatively compact set $T$ whose interior contains $x$ (in fact, it contains $overline{{x}}$, which may be strictly larger). The closure of $T$ is a compact neighbourhood of $x$.
If, on the other hand, if we have locally compact space and a relatively compact set $S$, then $overline{S}$ is compact. For each point $s in overline{S}$, there is an open neighbourhood $mathcal{N}_s$ of $s$ that is compact. Note that the interiors of these neighbourhoods cover $overline{S}$, hence there must be a finite-subcover of $overline{S}$. Unioning this finite subcover yields an open, relatively compact neighbourhood containing $overline{S}$, establishing equivalence.
answered Dec 20 '18 at 3:14
Theo BenditTheo Bendit
19.8k12354
$endgroup$
$begingroup$
I just posted a follow-up question: math.stackexchange.com/q/3048148/71829
$endgroup$
– Keshav Srinivasan
Dec 21 '18 at 3:37
$begingroup$
And now I posted one more follow-up question: math.stackexchange.com/q/3048232/71829
$endgroup$
– Keshav Srinivasan
Dec 21 '18 at 6:23
add a comment |
$begingroup$
This is equivalent to local compactness.
If a topological space satisfies your property, then a singleton point set ${x}$ is compact, and hence there must be a relatively compact set $T$ whose interior contains $x$ (in fact, it contains $overline{{x}}$, which may be strictly larger). The closure of $T$ is a compact neighbourhood of $x$.
If, on the other hand, if we have locally compact space and a relatively compact set $S$, then $overline{S}$ is compact. For each point $s in overline{S}$, there is an open neighbourhood $mathcal{N}_s$ of $s$ that is compact. Note that the interiors of these neighbourhoods cover $overline{S}$, hence there must be a finite-subcover of $overline{S}$. Unioning this finite subcover yields an open, relatively compact neighbourhood containing $overline{S}$, establishing equivalence.
answered Dec 20 '18 at 3:14
Theo BenditTheo Bendit
19.8k12354
$endgroup$
This is equivalent to local compactness.
If a topological space satisfies your property, then a singleton point set ${x}$ is compact, and hence there must be a relatively compact set $T$ whose interior contains $x$ (in fact, it contains $overline{{x}}$, which may be strictly larger). The closure of $T$ is a compact neighbourhood of $x$.
If, on the other hand, if we have locally compact space and a relatively compact set $S$, then $overline{S}$ is compact. For each point $s in overline{S}$, there is an open neighbourhood $mathcal{N}_s$ of $s$ that is compact. Note that the interiors of these neighbourhoods cover $overline{S}$, hence there must be a finite-subcover of $overline{S}$. Unioning this finite subcover yields an open, relatively compact neighbourhood containing $overline{S}$, establishing equivalence.
answered Dec 20 '18 at 3:14
Theo BenditTheo Bendit
19.8k12354
answered Dec 20 '18 at 3:14
Theo BenditTheo Bendit
19.8k12354
answered Dec 20 '18 at 3:14
Theo BenditTheo Bendit
19.8k12354
answered Dec 20 '18 at 3:14
Theo BenditTheo Bendit
19.8k12354
19.8k12354
$begingroup$
I just posted a follow-up question: math.stackexchange.com/q/3048148/71829
$endgroup$
– Keshav Srinivasan
Dec 21 '18 at 3:37
$begingroup$
And now I posted one more follow-up question: math.stackexchange.com/q/3048232/71829
$endgroup$
– Keshav Srinivasan
Dec 21 '18 at 6:23
add a comment |
$begingroup$
I just posted a follow-up question: math.stackexchange.com/q/3048148/71829
$endgroup$
– Keshav Srinivasan
Dec 21 '18 at 3:37
$begingroup$
And now I posted one more follow-up question: math.stackexchange.com/q/3048232/71829
$endgroup$
– Keshav Srinivasan
Dec 21 '18 at 6:23
$begingroup$
I just posted a follow-up question: math.stackexchange.com/q/3048148/71829
$endgroup$
– Keshav Srinivasan
Dec 21 '18 at 3:37
$begingroup$
I just posted a follow-up question: math.stackexchange.com/q/3048148/71829
$endgroup$
– Keshav Srinivasan
Dec 21 '18 at 3:37
$begingroup$
And now I posted one more follow-up question: math.stackexchange.com/q/3048232/71829
$endgroup$
– Keshav Srinivasan
Dec 21 '18 at 6:23
$begingroup$
And now I posted one more 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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