Mixing
Nerve Theorem: Is the finite union of closed convex sets triangulable?
My Question: Let $A_1, ldots, A_k subseteq mathbb{R}^n$ be closed convex sets.
Is the union $bigcup_{i=1}^k A_i$ triangulable$^1$? If so, why?
Background:
I'm trying to better understand the Nerve Theorem from Topology. In his Book Computational Topology (p.71) Edelsbrunner presents the following Nerve theorem without giving a proof:
Let F be a finite collection of closed, convex sets in Euclidean Space. Then the nerve$^2$ of $F$ is homotopy equivalent to $bigcup F$.
He also mentions the following classical nerve theorem, which one can find in Topological methods (p.1850), and which is supposed to be more general:
Let $X$ be a triangulable space and let
$mathcal{A} = {A_1, ldots, A_k}$ be a finite closed cover$^3$ of $X$ such that
every non-empty intersection of the ${A_i}'s$ is contractible. Then the nerve of $mathcal{A}$ is homotopy equivalent to $X$.
So does the former follow from the later?
$^1$ A space is called triangulable if it is homeomorphic to some simplicial complex.
$^2$ The nerve of a collection $F$ of sets is the abstract simplicial complex ${Y subseteq F : bigcap Y neq emptyset}$.
$^3$ A closed cover is a covering by closed subset of a topological space.
general-topology convex-analysis computational-mathematics
asked Jun 6 '14 at 5:27
kleenstar
966
add a comment |
My Question: Let $A_1, ldots, A_k subseteq mathbb{R}^n$ be closed convex sets.
Is the union $bigcup_{i=1}^k A_i$ triangulable$^1$? If so, why?
Background:
I'm trying to better understand the Nerve Theorem from Topology. In his Book Computational Topology (p.71) Edelsbrunner presents the following Nerve theorem without giving a proof:
Let F be a finite collection of closed, convex sets in Euclidean Space. Then the nerve$^2$ of $F$ is homotopy equivalent to $bigcup F$.
He also mentions the following classical nerve theorem, which one can find in Topological methods (p.1850), and which is supposed to be more general:
Let $X$ be a triangulable space and let
$mathcal{A} = {A_1, ldots, A_k}$ be a finite closed cover$^3$ of $X$ such that
every non-empty intersection of the ${A_i}'s$ is contractible. Then the nerve of $mathcal{A}$ is homotopy equivalent to $X$.
So does the former follow from the later?
$^1$ A space is called triangulable if it is homeomorphic to some simplicial complex.
$^2$ The nerve of a collection $F$ of sets is the abstract simplicial complex ${Y subseteq F : bigcap Y neq emptyset}$.
$^3$ A closed cover is a covering by closed subset of a topological space.
general-topology convex-analysis computational-mathematics
asked Jun 6 '14 at 5:27
kleenstar
966
The definition of the nerve of F would make this question more self-contained.
– coffeemath
Jun 6 '14 at 5:31
1
Thanks for suggesting this. I added the definition.
– kleenstar
Jun 6 '14 at 5:36
add a comment |
My Question: Let $A_1, ldots, A_k subseteq mathbb{R}^n$ be closed convex sets.
Is the union $bigcup_{i=1}^k A_i$ triangulable$^1$? If so, why?
Background:
I'm trying to better understand the Nerve Theorem from Topology. In his Book Computational Topology (p.71) Edelsbrunner presents the following Nerve theorem without giving a proof:
Let F be a finite collection of closed, convex sets in Euclidean Space. Then the nerve$^2$ of $F$ is homotopy equivalent to $bigcup F$.
He also mentions the following classical nerve theorem, which one can find in Topological methods (p.1850), and which is supposed to be more general:
Let $X$ be a triangulable space and let
$mathcal{A} = {A_1, ldots, A_k}$ be a finite closed cover$^3$ of $X$ such that
every non-empty intersection of the ${A_i}'s$ is contractible. Then the nerve of $mathcal{A}$ is homotopy equivalent to $X$.
So does the former follow from the later?
$^1$ A space is called triangulable if it is homeomorphic to some simplicial complex.
$^2$ The nerve of a collection $F$ of sets is the abstract simplicial complex ${Y subseteq F : bigcap Y neq emptyset}$.
$^3$ A closed cover is a covering by closed subset of a topological space.
general-topology convex-analysis computational-mathematics
asked Jun 6 '14 at 5:27
kleenstar
966
My Question: Let $A_1, ldots, A_k subseteq mathbb{R}^n$ be closed convex sets.
Is the union $bigcup_{i=1}^k A_i$ triangulable$^1$? If so, why?
Background:
I'm trying to better understand the Nerve Theorem from Topology. In his Book Computational Topology (p.71) Edelsbrunner presents the following Nerve theorem without giving a proof:
Let F be a finite collection of closed, convex sets in Euclidean Space. Then the nerve$^2$ of $F$ is homotopy equivalent to $bigcup F$.
He also mentions the following classical nerve theorem, which one can find in Topological methods (p.1850), and which is supposed to be more general:
Let $X$ be a triangulable space and let
$mathcal{A} = {A_1, ldots, A_k}$ be a finite closed cover$^3$ of $X$ such that
every non-empty intersection of the ${A_i}'s$ is contractible. Then the nerve of $mathcal{A}$ is homotopy equivalent to $X$.
So does the former follow from the later?
$^1$ A space is called triangulable if it is homeomorphic to some simplicial complex.
$^2$ The nerve of a collection $F$ of sets is the abstract simplicial complex ${Y subseteq F : bigcap Y neq emptyset}$.
$^3$ A closed cover is a covering by closed subset of a topological space.
general-topology convex-analysis computational-mathematics
general-topology convex-analysis computational-mathematics
asked Jun 6 '14 at 5:27
kleenstar
966
asked Jun 6 '14 at 5:27
kleenstar
966
edited Jun 24 '14 at 10:38
asked Jun 6 '14 at 5:27
kleenstar
966
asked Jun 6 '14 at 5:27
kleenstar
966
asked Jun 6 '14 at 5:27
kleenstar
966
966
The definition of the nerve of F would make this question more self-contained.
– coffeemath
Jun 6 '14 at 5:31
1
Thanks for suggesting this. I added the definition.
– kleenstar
Jun 6 '14 at 5:36
add a comment |
The definition of the nerve of F would make this question more self-contained.
– coffeemath
Jun 6 '14 at 5:31
1
Thanks for suggesting this. I added the definition.
– kleenstar
Jun 6 '14 at 5:36
The definition of the nerve of F would make this question more self-contained.
– coffeemath
Jun 6 '14 at 5:31
The definition of the nerve of F would make this question more self-contained.
– coffeemath
Jun 6 '14 at 5:31
1
1
Thanks for suggesting this. I added the definition.
– kleenstar
Jun 6 '14 at 5:36
Thanks for suggesting this. I added the definition.
– kleenstar
Jun 6 '14 at 5:36
add a comment |
1 Answer
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It seems like this is a few years too late, but yes, the former follows for the latter. To see this, we need the following two things:
- convex sets are contractible
- an intersection of convex sets is convex.
From here, the former statement follows from the latter, since a collection of closed convex sets in Euclidean space is a closed cover of their union such that each non-empty intersection of the sets is contractible.
answered May 18 '17 at 23:02
Caitlin
11
The question isn't about the cover or the contractibility conditions, but rather the part "let $X$ be a triangulable space". I shouldn't have voted "Looks OK".
– epimorphic
May 19 '17 at 1:41
Oh, wow. I completely misread the question. I was wondering why it was still unanswered.
– Caitlin
Jun 6 '17 at 16:15
I believe the answer is yes. First you need to show that a single closed convex set can be triangulated with simplices that are straight with respect to the standard metric on $mathbb R^n$. Once you have this, you can triangulate a union by separately triangulating the pieces. Since both triangulations are straight, you can find a common subdivision.
– Cheerful Parsnip
Mar 18 '18 at 6:15
add a comment |
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1 Answer
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It seems like this is a few years too late, but yes, the former follows for the latter. To see this, we need the following two things:
- convex sets are contractible
- an intersection of convex sets is convex.
From here, the former statement follows from the latter, since a collection of closed convex sets in Euclidean space is a closed cover of their union such that each non-empty intersection of the sets is contractible.
answered May 18 '17 at 23:02
Caitlin
11
The question isn't about the cover or the contractibility conditions, but rather the part "let $X$ be a triangulable space". I shouldn't have voted "Looks OK".
– epimorphic
May 19 '17 at 1:41
Oh, wow. I completely misread the question. I was wondering why it was still unanswered.
– Caitlin
Jun 6 '17 at 16:15
I believe the answer is yes. First you need to show that a single closed convex set can be triangulated with simplices that are straight with respect to the standard metric on $mathbb R^n$. Once you have this, you can triangulate a union by separately triangulating the pieces. Since both triangulations are straight, you can find a common subdivision.
– Cheerful Parsnip
Mar 18 '18 at 6:15
add a comment |
It seems like this is a few years too late, but yes, the former follows for the latter. To see this, we need the following two things:
- convex sets are contractible
- an intersection of convex sets is convex.
From here, the former statement follows from the latter, since a collection of closed convex sets in Euclidean space is a closed cover of their union such that each non-empty intersection of the sets is contractible.
answered May 18 '17 at 23:02
Caitlin
11
The question isn't about the cover or the contractibility conditions, but rather the part "let $X$ be a triangulable space". I shouldn't have voted "Looks OK".
– epimorphic
May 19 '17 at 1:41
Oh, wow. I completely misread the question. I was wondering why it was still unanswered.
– Caitlin
Jun 6 '17 at 16:15
I believe the answer is yes. First you need to show that a single closed convex set can be triangulated with simplices that are straight with respect to the standard metric on $mathbb R^n$. Once you have this, you can triangulate a union by separately triangulating the pieces. Since both triangulations are straight, you can find a common subdivision.
– Cheerful Parsnip
Mar 18 '18 at 6:15
add a comment |
It seems like this is a few years too late, but yes, the former follows for the latter. To see this, we need the following two things:
- convex sets are contractible
- an intersection of convex sets is convex.
From here, the former statement follows from the latter, since a collection of closed convex sets in Euclidean space is a closed cover of their union such that each non-empty intersection of the sets is contractible.
answered May 18 '17 at 23:02
Caitlin
11
It seems like this is a few years too late, but yes, the former follows for the latter. To see this, we need the following two things:
- convex sets are contractible
- an intersection of convex sets is convex.
From here, the former statement follows from the latter, since a collection of closed convex sets in Euclidean space is a closed cover of their union such that each non-empty intersection of the sets is contractible.
answered May 18 '17 at 23:02
Caitlin
11
answered May 18 '17 at 23:02
Caitlin
11
answered May 18 '17 at 23:02
Caitlin
11
answered May 18 '17 at 23:02
Caitlin
11
11
The question isn't about the cover or the contractibility conditions, but rather the part "let $X$ be a triangulable space". I shouldn't have voted "Looks OK".
– epimorphic
May 19 '17 at 1:41
Oh, wow. I completely misread the question. I was wondering why it was still unanswered.
– Caitlin
Jun 6 '17 at 16:15
I believe the answer is yes. First you need to show that a single closed convex set can be triangulated with simplices that are straight with respect to the standard metric on $mathbb R^n$. Once you have this, you can triangulate a union by separately triangulating the pieces. Since both triangulations are straight, you can find a common subdivision.
– Cheerful Parsnip
Mar 18 '18 at 6:15
add a comment |
The question isn't about the cover or the contractibility conditions, but rather the part "let $X$ be a triangulable space". I shouldn't have voted "Looks OK".
– epimorphic
May 19 '17 at 1:41
Oh, wow. I completely misread the question. I was wondering why it was still unanswered.
– Caitlin
Jun 6 '17 at 16:15
I believe the answer is yes. First you need to show that a single closed convex set can be triangulated with simplices that are straight with respect to the standard metric on $mathbb R^n$. Once you have this, you can triangulate a union by separately triangulating the pieces. Since both triangulations are straight, you can find a common subdivision.
– Cheerful Parsnip
Mar 18 '18 at 6:15
The question isn't about the cover or the contractibility conditions, but rather the part "let $X$ be a triangulable space". I shouldn't have voted "Looks OK".
– epimorphic
May 19 '17 at 1:41
The question isn't about the cover or the contractibility conditions, but rather the part "let $X$ be a triangulable space". I shouldn't have voted "Looks OK".
– epimorphic
May 19 '17 at 1:41
Oh, wow. I completely misread the question. I was wondering why it was still unanswered.
– Caitlin
Jun 6 '17 at 16:15
Oh, wow. I completely misread the question. I was wondering why it was still unanswered.
– Caitlin
Jun 6 '17 at 16:15
I believe the answer is yes. First you need to show that a single closed convex set can be triangulated with simplices that are straight with respect to the standard metric on $mathbb R^n$. Once you have this, you can triangulate a union by separately triangulating the pieces. Since both triangulations are straight, you can find a common subdivision.
– Cheerful Parsnip
Mar 18 '18 at 6:15
I believe the answer is yes. First you need to show that a single closed convex set can be triangulated with simplices that are straight with respect to the standard metric on $mathbb R^n$. Once you have this, you can triangulate a union by separately triangulating the pieces. Since both triangulations are straight, you can find a common subdivision.
– Cheerful Parsnip
Mar 18 '18 at 6:15
add a comment |
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The definition of the nerve of F would make this question more self-contained.
– coffeemath
Jun 6 '14 at 5:31
1
Thanks for suggesting this. I added the definition.
– kleenstar
Jun 6 '14 at 5:36