Topological Quantum Field Theory: Compact Implies Finite BoundariesAtiyah's definitions of Topological Quantum Field TheoryShowing $S^n$ is compactQuestion about the definition of compact topological spaceA theorem on compactnessIs this the right way to prove finite cartesian product of compact sets is compactTopological Quantum Field TheoryFinite intersection of compact topological subspaces is compact.Understanding compactness and how it relates to finitenessLet $qcolon Eto X$ be a covering map. If $E$ is compact then $X$ is compact and $q$ is a finite-sheeted covering.Does a “flexibly fine” open-interval cover of a compact nowhere dense set admit a disjoint finite subcover?
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Topological Quantum Field Theory: Compact Implies Finite Boundaries
Atiyah's definitions of Topological Quantum Field TheoryShowing $S^n$ is compactQuestion about the definition of compact topological spaceA theorem on compactnessIs this the right way to prove finite cartesian product of compact sets is compactTopological Quantum Field TheoryFinite intersection of compact topological subspaces is compact.Understanding compactness and how it relates to finitenessLet $qcolon Eto X$ be a covering map. If $E$ is compact then $X$ is compact and $q$ is a finite-sheeted covering.Does a “flexibly fine” open-interval cover of a compact nowhere dense set admit a disjoint finite subcover?
$begingroup$
The following is from this paper.
A bordism $B:coprod_iin I S^1 to coprod_jin J S^1$ is just a
“generalized pair of pants” - a pair of pants with one waist hole for
each $iin I$ and one leg hole for each $jin J$ Since both $I$ and
$J$ are finite sets, as required by compactness....
I assume the author is using the general topology version of compactness, i.e., the existence of a finite subcover. However, I don't see how to apply compactness to get the finiteness of $i$ and $j$ conclusion. What is the open cover and what is the finite subcover in this case? The result seems to imply that the disjoint union of $S^1$ is the finite subcover but I can't seem to make this precise.
EDIT: This is actually fairly simple, not sure what I was thinking. Basically if $i,j$ aren't finite then compactness would be violated.
general-topology mathematical-physics quantum-field-theory cobordism topological-quantum-field-theory
edited Mar 11 at 20:54
Andrews
1,2691421
asked Dec 14 '16 at 21:21
BobBob
684410
$endgroup$
add a comment |
$begingroup$
The following is from this paper.
A bordism $B:coprod_iin I S^1 to coprod_jin J S^1$ is just a
“generalized pair of pants” - a pair of pants with one waist hole for
each $iin I$ and one leg hole for each $jin J$ Since both $I$ and
$J$ are finite sets, as required by compactness....
I assume the author is using the general topology version of compactness, i.e., the existence of a finite subcover. However, I don't see how to apply compactness to get the finiteness of $i$ and $j$ conclusion. What is the open cover and what is the finite subcover in this case? The result seems to imply that the disjoint union of $S^1$ is the finite subcover but I can't seem to make this precise.
EDIT: This is actually fairly simple, not sure what I was thinking. Basically if $i,j$ aren't finite then compactness would be violated.
general-topology mathematical-physics quantum-field-theory cobordism topological-quantum-field-theory
edited Mar 11 at 20:54
Andrews
1,2691421
asked Dec 14 '16 at 21:21
BobBob
684410
$endgroup$
$begingroup$
If a manifold is compact, every infinite subset has a accumulation point. An infinite disjoint union of circles is not compact , since you can choose one point from each circle, and that set has no accumulation points. You want to start with something less heady. Try this: mat.uab.es/~kock/TQFT/FS.pdf
$endgroup$
– Charlie Frohman
Dec 14 '16 at 21:36
$begingroup$
May I ask you why you reject my edit and roll back? Thanks
$endgroup$
– Andrews
Mar 4 at 19:35
$begingroup$
@Andrews So if I understand correctly the edit was to remove the general topology tag. IMHO this question, although fairly basic, had a lot to do with general topology.
$endgroup$
– Bob
Mar 4 at 19:50
add a comment |
$begingroup$
The following is from this paper.
A bordism $B:coprod_iin I S^1 to coprod_jin J S^1$ is just a
“generalized pair of pants” - a pair of pants with one waist hole for
each $iin I$ and one leg hole for each $jin J$ Since both $I$ and
$J$ are finite sets, as required by compactness....
I assume the author is using the general topology version of compactness, i.e., the existence of a finite subcover. However, I don't see how to apply compactness to get the finiteness of $i$ and $j$ conclusion. What is the open cover and what is the finite subcover in this case? The result seems to imply that the disjoint union of $S^1$ is the finite subcover but I can't seem to make this precise.
EDIT: This is actually fairly simple, not sure what I was thinking. Basically if $i,j$ aren't finite then compactness would be violated.
general-topology mathematical-physics quantum-field-theory cobordism topological-quantum-field-theory
edited Mar 11 at 20:54
Andrews
1,2691421
asked Dec 14 '16 at 21:21
BobBob
684410
$endgroup$
The following is from this paper.
A bordism $B:coprod_iin I S^1 to coprod_jin J S^1$ is just a
“generalized pair of pants” - a pair of pants with one waist hole for
each $iin I$ and one leg hole for each $jin J$ Since both $I$ and
$J$ are finite sets, as required by compactness....
I assume the author is using the general topology version of compactness, i.e., the existence of a finite subcover. However, I don't see how to apply compactness to get the finiteness of $i$ and $j$ conclusion. What is the open cover and what is the finite subcover in this case? The result seems to imply that the disjoint union of $S^1$ is the finite subcover but I can't seem to make this precise.
EDIT: This is actually fairly simple, not sure what I was thinking. Basically if $i,j$ aren't finite then compactness would be violated.
general-topology mathematical-physics quantum-field-theory cobordism topological-quantum-field-theory
general-topology mathematical-physics quantum-field-theory cobordism topological-quantum-field-theory
edited Mar 11 at 20:54
Andrews
1,2691421
asked Dec 14 '16 at 21:21
BobBob
684410
edited Mar 11 at 20:54
Andrews
1,2691421
asked Dec 14 '16 at 21:21
BobBob
684410
edited Mar 11 at 20:54
Andrews
1,2691421
edited Mar 11 at 20:54
Andrews
1,2691421
edited Mar 11 at 20:54
Andrews
1,2691421
1,2691421
asked Dec 14 '16 at 21:21
BobBob
684410
asked Dec 14 '16 at 21:21
BobBob
684410
asked Dec 14 '16 at 21:21
BobBob
684410
684410
$begingroup$
If a manifold is compact, every infinite subset has a accumulation point. An infinite disjoint union of circles is not compact , since you can choose one point from each circle, and that set has no accumulation points. You want to start with something less heady. Try this: mat.uab.es/~kock/TQFT/FS.pdf
$endgroup$
– Charlie Frohman
Dec 14 '16 at 21:36
$begingroup$
May I ask you why you reject my edit and roll back? Thanks
$endgroup$
– Andrews
Mar 4 at 19:35
$begingroup$
@Andrews So if I understand correctly the edit was to remove the general topology tag. IMHO this question, although fairly basic, had a lot to do with general topology.
$endgroup$
– Bob
Mar 4 at 19:50
add a comment |
$begingroup$
If a manifold is compact, every infinite subset has a accumulation point. An infinite disjoint union of circles is not compact , since you can choose one point from each circle, and that set has no accumulation points. You want to start with something less heady. Try this: mat.uab.es/~kock/TQFT/FS.pdf
$endgroup$
– Charlie Frohman
Dec 14 '16 at 21:36
$begingroup$
May I ask you why you reject my edit and roll back? Thanks
$endgroup$
– Andrews
Mar 4 at 19:35
$begingroup$
@Andrews So if I understand correctly the edit was to remove the general topology tag. IMHO this question, although fairly basic, had a lot to do with general topology.
$endgroup$
– Bob
Mar 4 at 19:50
$begingroup$
If a manifold is compact, every infinite subset has a accumulation point. An infinite disjoint union of circles is not compact , since you can choose one point from each circle, and that set has no accumulation points. You want to start with something less heady. Try this: mat.uab.es/~kock/TQFT/FS.pdf
$endgroup$
– Charlie Frohman
Dec 14 '16 at 21:36
$begingroup$
If a manifold is compact, every infinite subset has a accumulation point. An infinite disjoint union of circles is not compact , since you can choose one point from each circle, and that set has no accumulation points. You want to start with something less heady. Try this: mat.uab.es/~kock/TQFT/FS.pdf
$endgroup$
– Charlie Frohman
Dec 14 '16 at 21:36
$begingroup$
May I ask you why you reject my edit and roll back? Thanks
$endgroup$
– Andrews
Mar 4 at 19:35
$begingroup$
May I ask you why you reject my edit and roll back? Thanks
$endgroup$
– Andrews
Mar 4 at 19:35
$begingroup$
@Andrews So if I understand correctly the edit was to remove the general topology tag. IMHO this question, although fairly basic, had a lot to do with general topology.
$endgroup$
– Bob
Mar 4 at 19:50
$begingroup$
@Andrews So if I understand correctly the edit was to remove the general topology tag. IMHO this question, although fairly basic, had a lot to do with general topology.
$endgroup$
– Bob
Mar 4 at 19:50
add a comment |
1 Answer
1
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oldest
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$begingroup$
Take $X=coprod_iin I S_i$ (where each $S_i$ is homeomorphic to $S^1$) with $I$ infinite. An open cover of $X$ is $C=S_i_iin I$. For any finite subset $Dsubset C$ there is some $iin I$ such that $S_inotin D$ and thus $D$ can't be a subcover. We have found a cover $C$ without finite subcover, and thus $X$ isn't compact.
answered Dec 14 '16 at 21:39
coconutcoconut
596211
$endgroup$
add a comment |
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1 Answer
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$begingroup$
Take $X=coprod_iin I S_i$ (where each $S_i$ is homeomorphic to $S^1$) with $I$ infinite. An open cover of $X$ is $C=S_i_iin I$. For any finite subset $Dsubset C$ there is some $iin I$ such that $S_inotin D$ and thus $D$ can't be a subcover. We have found a cover $C$ without finite subcover, and thus $X$ isn't compact.
answered Dec 14 '16 at 21:39
coconutcoconut
596211
$endgroup$
add a comment |
$begingroup$
Take $X=coprod_iin I S_i$ (where each $S_i$ is homeomorphic to $S^1$) with $I$ infinite. An open cover of $X$ is $C=S_i_iin I$. For any finite subset $Dsubset C$ there is some $iin I$ such that $S_inotin D$ and thus $D$ can't be a subcover. We have found a cover $C$ without finite subcover, and thus $X$ isn't compact.
answered Dec 14 '16 at 21:39
coconutcoconut
596211
$endgroup$
add a comment |
$begingroup$
Take $X=coprod_iin I S_i$ (where each $S_i$ is homeomorphic to $S^1$) with $I$ infinite. An open cover of $X$ is $C=S_i_iin I$. For any finite subset $Dsubset C$ there is some $iin I$ such that $S_inotin D$ and thus $D$ can't be a subcover. We have found a cover $C$ without finite subcover, and thus $X$ isn't compact.
answered Dec 14 '16 at 21:39
coconutcoconut
596211
$endgroup$
Take $X=coprod_iin I S_i$ (where each $S_i$ is homeomorphic to $S^1$) with $I$ infinite. An open cover of $X$ is $C=S_i_iin I$. For any finite subset $Dsubset C$ there is some $iin I$ such that $S_inotin D$ and thus $D$ can't be a subcover. We have found a cover $C$ without finite subcover, and thus $X$ isn't compact.
answered Dec 14 '16 at 21:39
coconutcoconut
596211
answered Dec 14 '16 at 21:39
coconutcoconut
596211
answered Dec 14 '16 at 21:39
coconutcoconut
596211
answered Dec 14 '16 at 21:39
coconutcoconut
596211
596211
add a comment |
add a comment |
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If a manifold is compact, every infinite subset has a accumulation point. An infinite disjoint union of circles is not compact , since you can choose one point from each circle, and that set has no accumulation points. You want to start with something less heady. Try this: mat.uab.es/~kock/TQFT/FS.pdf
$endgroup$
– Charlie Frohman
Dec 14 '16 at 21:36
$begingroup$
May I ask you why you reject my edit and roll back? Thanks
$endgroup$
– Andrews
Mar 4 at 19:35
$begingroup$
@Andrews So if I understand correctly the edit was to remove the general topology tag. IMHO this question, although fairly basic, had a lot to do with general topology.
$endgroup$
– Bob
Mar 4 at 19:50