Nerve Theorem: Is the finite union of closed convex sets triangulable?
Clash Royale CLAN TAG#URR8PPP
up vote
10
down vote
favorite
My Question: Let $A_1, ldots, A_k subseteq mathbbR^n$ 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
$mathcalA = A_1, ldots, A_k$ 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 $mathcalA$ 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
816
add a comment |Â
up vote
10
down vote
favorite
My Question: Let $A_1, ldots, A_k subseteq mathbbR^n$ 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
$mathcalA = A_1, ldots, A_k$ 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 $mathcalA$ 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
816
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 |Â
up vote
10
down vote
favorite
up vote
10
down vote
favorite
My Question: Let $A_1, ldots, A_k subseteq mathbbR^n$ 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
$mathcalA = A_1, ldots, A_k$ 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 $mathcalA$ 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
816
My Question: Let $A_1, ldots, A_k subseteq mathbbR^n$ 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
$mathcalA = A_1, ldots, A_k$ 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 $mathcalA$ 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
816
edited Jun 24 '14 at 10:38
asked Jun 6 '14 at 5:27
kleenstar
816
asked Jun 6 '14 at 5:27
kleenstar
816
asked Jun 6 '14 at 5:27
kleenstar
816
816
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 |Â
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 |Â
1 Answer
1
active
oldest
votes
up vote
0
down vote
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 at 6:15
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
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 at 6:15
add a comment |Â
up vote
0
down vote
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 at 6:15
add a comment |Â
up vote
0
down vote
up vote
0
down vote
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 at 6:15
add a comment |Â
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 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 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 at 6:15
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f822535%2fnerve-theorem-is-the-finite-union-of-closed-convex-sets-triangulable%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
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