Why is compactness important to find solution of certain problems?
$begingroup$
I've been recently re-thinking about my knowledge of compacts sets and I've realized that other than the definition and maybe understanding with some struggle some proofs I don't actually see why they're very important. From some questions in this website which I used as sort of "soft discussion" about the subject it is my understanding that compactness might allow to state if certain functions are bounded or not.
However I came across this question, and I was considering the second answer specifically (more specifically the bit below):
Unlike their Holder cousins, most Sobolev spaces are reflexive Banach
spaces. Reflexivity is a highly desirable feature for variational
problems because it gives a little bit of compactness enough so you
can prove the existence of minimizers (or more general critical
points) of various energy functionals.
So this made me wonder... Is compactness in general a very highly desirable property that allow to prove existence of solutions of problems in general? If yes, how? I mean what is the general argument that is brought up.
how does knowing that some sets are compacts actually help in proving existence of solutions?
general-topology functional-analysis compactness calculus-of-variations
asked Jan 16 at 17:37
user8469759user8469759
1,5681618
$endgroup$
|
show 1 more comment
$begingroup$
I've been recently re-thinking about my knowledge of compacts sets and I've realized that other than the definition and maybe understanding with some struggle some proofs I don't actually see why they're very important. From some questions in this website which I used as sort of "soft discussion" about the subject it is my understanding that compactness might allow to state if certain functions are bounded or not.
However I came across this question, and I was considering the second answer specifically (more specifically the bit below):
Unlike their Holder cousins, most Sobolev spaces are reflexive Banach
spaces. Reflexivity is a highly desirable feature for variational
problems because it gives a little bit of compactness enough so you
can prove the existence of minimizers (or more general critical
points) of various energy functionals.
So this made me wonder... Is compactness in general a very highly desirable property that allow to prove existence of solutions of problems in general? If yes, how? I mean what is the general argument that is brought up.
how does knowing that some sets are compacts actually help in proving existence of solutions?
general-topology functional-analysis compactness calculus-of-variations
asked Jan 16 at 17:37
user8469759user8469759
1,5681618
$endgroup$
1
$begingroup$
In topology and analysis, "compact" is often a very good generalisation of "finite".
$endgroup$
– Arthur
Jan 16 at 17:39
1
$begingroup$
Perhaps this could be interesting.
$endgroup$
– Arnaud D.
Jan 16 at 17:41
2
$begingroup$
@Arthur I don't have a clue of what that can possible imply.
$endgroup$
– user8469759
Jan 16 at 17:47
$begingroup$
The most simple answer I can think of off the top of my head is "extreme value theorem" which is a more simple existence solution that tends to lend itself to others.
$endgroup$
– Prototank
Jan 16 at 18:00
$begingroup$
It is probably a good idea to write down some properties of compact sets and compare them to properties of finite sets so see the wisdom of Arthur's suggestion. It is okay if you don't understand at first. That is the beginning of knowledge.
$endgroup$
– John Douma
Jan 16 at 18:06
|
show 1 more comment
$begingroup$
I've been recently re-thinking about my knowledge of compacts sets and I've realized that other than the definition and maybe understanding with some struggle some proofs I don't actually see why they're very important. From some questions in this website which I used as sort of "soft discussion" about the subject it is my understanding that compactness might allow to state if certain functions are bounded or not.
However I came across this question, and I was considering the second answer specifically (more specifically the bit below):
Unlike their Holder cousins, most Sobolev spaces are reflexive Banach
spaces. Reflexivity is a highly desirable feature for variational
problems because it gives a little bit of compactness enough so you
can prove the existence of minimizers (or more general critical
points) of various energy functionals.
So this made me wonder... Is compactness in general a very highly desirable property that allow to prove existence of solutions of problems in general? If yes, how? I mean what is the general argument that is brought up.
how does knowing that some sets are compacts actually help in proving existence of solutions?
general-topology functional-analysis compactness calculus-of-variations
asked Jan 16 at 17:37
user8469759user8469759
1,5681618
$endgroup$
I've been recently re-thinking about my knowledge of compacts sets and I've realized that other than the definition and maybe understanding with some struggle some proofs I don't actually see why they're very important. From some questions in this website which I used as sort of "soft discussion" about the subject it is my understanding that compactness might allow to state if certain functions are bounded or not.
However I came across this question, and I was considering the second answer specifically (more specifically the bit below):
Unlike their Holder cousins, most Sobolev spaces are reflexive Banach
spaces. Reflexivity is a highly desirable feature for variational
problems because it gives a little bit of compactness enough so you
can prove the existence of minimizers (or more general critical
points) of various energy functionals.
So this made me wonder... Is compactness in general a very highly desirable property that allow to prove existence of solutions of problems in general? If yes, how? I mean what is the general argument that is brought up.
how does knowing that some sets are compacts actually help in proving existence of solutions?
general-topology functional-analysis compactness calculus-of-variations
general-topology functional-analysis compactness calculus-of-variations
asked Jan 16 at 17:37
user8469759user8469759
1,5681618
asked Jan 16 at 17:37
user8469759user8469759
1,5681618
edited Jan 16 at 17:49
user8469759
asked Jan 16 at 17:37
user8469759user8469759
1,5681618
asked Jan 16 at 17:37
user8469759user8469759
1,5681618
asked Jan 16 at 17:37
user8469759user8469759
1,5681618
1,5681618
1
$begingroup$
In topology and analysis, "compact" is often a very good generalisation of "finite".
$endgroup$
– Arthur
Jan 16 at 17:39
1
$begingroup$
Perhaps this could be interesting.
$endgroup$
– Arnaud D.
Jan 16 at 17:41
2
$begingroup$
@Arthur I don't have a clue of what that can possible imply.
$endgroup$
– user8469759
Jan 16 at 17:47
$begingroup$
The most simple answer I can think of off the top of my head is "extreme value theorem" which is a more simple existence solution that tends to lend itself to others.
$endgroup$
– Prototank
Jan 16 at 18:00
$begingroup$
It is probably a good idea to write down some properties of compact sets and compare them to properties of finite sets so see the wisdom of Arthur's suggestion. It is okay if you don't understand at first. That is the beginning of knowledge.
$endgroup$
– John Douma
Jan 16 at 18:06
|
show 1 more comment
1
$begingroup$
In topology and analysis, "compact" is often a very good generalisation of "finite".
$endgroup$
– Arthur
Jan 16 at 17:39
1
$begingroup$
Perhaps this could be interesting.
$endgroup$
– Arnaud D.
Jan 16 at 17:41
2
$begingroup$
@Arthur I don't have a clue of what that can possible imply.
$endgroup$
– user8469759
Jan 16 at 17:47
$begingroup$
The most simple answer I can think of off the top of my head is "extreme value theorem" which is a more simple existence solution that tends to lend itself to others.
$endgroup$
– Prototank
Jan 16 at 18:00
$begingroup$
It is probably a good idea to write down some properties of compact sets and compare them to properties of finite sets so see the wisdom of Arthur's suggestion. It is okay if you don't understand at first. That is the beginning of knowledge.
$endgroup$
– John Douma
Jan 16 at 18:06
1
1
$begingroup$
In topology and analysis, "compact" is often a very good generalisation of "finite".
$endgroup$
– Arthur
Jan 16 at 17:39
$begingroup$
In topology and analysis, "compact" is often a very good generalisation of "finite".
$endgroup$
– Arthur
Jan 16 at 17:39
1
1
$begingroup$
Perhaps this could be interesting.
$endgroup$
– Arnaud D.
Jan 16 at 17:41
$begingroup$
Perhaps this could be interesting.
$endgroup$
– Arnaud D.
Jan 16 at 17:41
2
2
$begingroup$
@Arthur I don't have a clue of what that can possible imply.
$endgroup$
– user8469759
Jan 16 at 17:47
$begingroup$
@Arthur I don't have a clue of what that can possible imply.
$endgroup$
– user8469759
Jan 16 at 17:47
$begingroup$
The most simple answer I can think of off the top of my head is "extreme value theorem" which is a more simple existence solution that tends to lend itself to others.
$endgroup$
– Prototank
Jan 16 at 18:00
$begingroup$
The most simple answer I can think of off the top of my head is "extreme value theorem" which is a more simple existence solution that tends to lend itself to others.
$endgroup$
– Prototank
Jan 16 at 18:00
$begingroup$
It is probably a good idea to write down some properties of compact sets and compare them to properties of finite sets so see the wisdom of Arthur's suggestion. It is okay if you don't understand at first. That is the beginning of knowledge.
$endgroup$
– John Douma
Jan 16 at 18:06
$begingroup$
It is probably a good idea to write down some properties of compact sets and compare them to properties of finite sets so see the wisdom of Arthur's suggestion. It is okay if you don't understand at first. That is the beginning of knowledge.
$endgroup$
– John Douma
Jan 16 at 18:06
|
show 1 more comment
1 Answer
1
active
oldest
votes
$begingroup$
Arnaud gave a link to a very nice post and Song already showed how compactness is used in a minimization problem in the comments. I would like to elaborate a bit more, specifically on the usefulness of compactness in minimization problem in Calculus of Variation.
Many important problems in mathematics and mathematical physics can be cast in the form of minimization problems of some integrals functional, i.e. we want to find $bar uin S$ such that
$$
bar u = min{ I(u) : uin S }
$$
where
$$
I(u) := int_Omega L(x,u(x),nabla u(x)) ,dx.
$$
Here $S$ is some function space, e.g. $S={ uin W^{1,1} : uvert_{partial Omega} = f }$ for some fixed $fin W^{1,1}$, and $Omega$ is a bounded domain in a Euclidean space. The integrand $L$ represents the feature of the problem we want to solve. For example $L$ can represent the energy density for the Dirichlet energy functional or the internal energy of an elastic body.
One method to solve this is called the Direct Method. Suppose that the functional $I$ is bounded below on $S$, we can then choose a minimizing sequence $(u_n)_n subset S$ such that $I(u_n)to inf_S I$. The minimization problem is solved if we can establish the following:
1.) $(u_n)_n$ has a subsequence that converges (in appropriate some sense) to a $uin S$.
2.) $I(u) le liminf_{ntoinfty} I(u_n)$.
The appropriate mode of convergence is usually the weak convergence, since weak compactness is easier to establish than strong compactness. Here we see that compactness is crucial in guaranteeing that $(u_n)_n$ has a convergent subsequence.
In this case, we want the functional $I$ to be weakly lower semi-continuous in order that the second condition holds.
answered Jan 16 at 18:43
BigbearZzzBigbearZzz
9,04221653
$endgroup$
$begingroup$
Can you complement your answer by elaboraring (maybe an example) on why weak compactness is easier to establish than regular compactness?
$endgroup$
– user8469759
Jan 16 at 19:37
$begingroup$
@user8469759 Because weak topology is coarser than norm topology, hence there are fewer open sets. This automatically implies that any norm-compact set is weakly compact but not vise versa.
$endgroup$
– BigbearZzz
Jan 16 at 19:48
$begingroup$
@user8469759 Also, in practice there are a lot of theorems that we can play around with regarding weak and weak$^*$ compactness, e.g. Banach-Alaoglu, Eberlein-Smulian, Mazur's lemma and other results for reflexive spaces.
$endgroup$
– BigbearZzz
Jan 16 at 19:51
$begingroup$
@user8469759 For a simple example, it is well known that in an infinite dimensional Banach space $X$ the unit ball is never compact and that closed and bounded sets need not be compact. However, if $X$ is a reflexive Banach space then any bounded closed and convex set is weakly compact.
$endgroup$
– BigbearZzz
Jan 16 at 20:06
add a comment |
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$begingroup$
Arnaud gave a link to a very nice post and Song already showed how compactness is used in a minimization problem in the comments. I would like to elaborate a bit more, specifically on the usefulness of compactness in minimization problem in Calculus of Variation.
Many important problems in mathematics and mathematical physics can be cast in the form of minimization problems of some integrals functional, i.e. we want to find $bar uin S$ such that
$$
bar u = min{ I(u) : uin S }
$$
where
$$
I(u) := int_Omega L(x,u(x),nabla u(x)) ,dx.
$$
Here $S$ is some function space, e.g. $S={ uin W^{1,1} : uvert_{partial Omega} = f }$ for some fixed $fin W^{1,1}$, and $Omega$ is a bounded domain in a Euclidean space. The integrand $L$ represents the feature of the problem we want to solve. For example $L$ can represent the energy density for the Dirichlet energy functional or the internal energy of an elastic body.
One method to solve this is called the Direct Method. Suppose that the functional $I$ is bounded below on $S$, we can then choose a minimizing sequence $(u_n)_n subset S$ such that $I(u_n)to inf_S I$. The minimization problem is solved if we can establish the following:
1.) $(u_n)_n$ has a subsequence that converges (in appropriate some sense) to a $uin S$.
2.) $I(u) le liminf_{ntoinfty} I(u_n)$.
The appropriate mode of convergence is usually the weak convergence, since weak compactness is easier to establish than strong compactness. Here we see that compactness is crucial in guaranteeing that $(u_n)_n$ has a convergent subsequence.
In this case, we want the functional $I$ to be weakly lower semi-continuous in order that the second condition holds.
answered Jan 16 at 18:43
BigbearZzzBigbearZzz
9,04221653
$endgroup$
$begingroup$
Can you complement your answer by elaboraring (maybe an example) on why weak compactness is easier to establish than regular compactness?
$endgroup$
– user8469759
Jan 16 at 19:37
$begingroup$
@user8469759 Because weak topology is coarser than norm topology, hence there are fewer open sets. This automatically implies that any norm-compact set is weakly compact but not vise versa.
$endgroup$
– BigbearZzz
Jan 16 at 19:48
$begingroup$
@user8469759 Also, in practice there are a lot of theorems that we can play around with regarding weak and weak$^*$ compactness, e.g. Banach-Alaoglu, Eberlein-Smulian, Mazur's lemma and other results for reflexive spaces.
$endgroup$
– BigbearZzz
Jan 16 at 19:51
$begingroup$
@user8469759 For a simple example, it is well known that in an infinite dimensional Banach space $X$ the unit ball is never compact and that closed and bounded sets need not be compact. However, if $X$ is a reflexive Banach space then any bounded closed and convex set is weakly compact.
$endgroup$
– BigbearZzz
Jan 16 at 20:06
add a comment |
$begingroup$
Arnaud gave a link to a very nice post and Song already showed how compactness is used in a minimization problem in the comments. I would like to elaborate a bit more, specifically on the usefulness of compactness in minimization problem in Calculus of Variation.
Many important problems in mathematics and mathematical physics can be cast in the form of minimization problems of some integrals functional, i.e. we want to find $bar uin S$ such that
$$
bar u = min{ I(u) : uin S }
$$
where
$$
I(u) := int_Omega L(x,u(x),nabla u(x)) ,dx.
$$
Here $S$ is some function space, e.g. $S={ uin W^{1,1} : uvert_{partial Omega} = f }$ for some fixed $fin W^{1,1}$, and $Omega$ is a bounded domain in a Euclidean space. The integrand $L$ represents the feature of the problem we want to solve. For example $L$ can represent the energy density for the Dirichlet energy functional or the internal energy of an elastic body.
One method to solve this is called the Direct Method. Suppose that the functional $I$ is bounded below on $S$, we can then choose a minimizing sequence $(u_n)_n subset S$ such that $I(u_n)to inf_S I$. The minimization problem is solved if we can establish the following:
1.) $(u_n)_n$ has a subsequence that converges (in appropriate some sense) to a $uin S$.
2.) $I(u) le liminf_{ntoinfty} I(u_n)$.
The appropriate mode of convergence is usually the weak convergence, since weak compactness is easier to establish than strong compactness. Here we see that compactness is crucial in guaranteeing that $(u_n)_n$ has a convergent subsequence.
In this case, we want the functional $I$ to be weakly lower semi-continuous in order that the second condition holds.
answered Jan 16 at 18:43
BigbearZzzBigbearZzz
9,04221653
$endgroup$
$begingroup$
Can you complement your answer by elaboraring (maybe an example) on why weak compactness is easier to establish than regular compactness?
$endgroup$
– user8469759
Jan 16 at 19:37
$begingroup$
@user8469759 Because weak topology is coarser than norm topology, hence there are fewer open sets. This automatically implies that any norm-compact set is weakly compact but not vise versa.
$endgroup$
– BigbearZzz
Jan 16 at 19:48
$begingroup$
@user8469759 Also, in practice there are a lot of theorems that we can play around with regarding weak and weak$^*$ compactness, e.g. Banach-Alaoglu, Eberlein-Smulian, Mazur's lemma and other results for reflexive spaces.
$endgroup$
– BigbearZzz
Jan 16 at 19:51
$begingroup$
@user8469759 For a simple example, it is well known that in an infinite dimensional Banach space $X$ the unit ball is never compact and that closed and bounded sets need not be compact. However, if $X$ is a reflexive Banach space then any bounded closed and convex set is weakly compact.
$endgroup$
– BigbearZzz
Jan 16 at 20:06
add a comment |
$begingroup$
Arnaud gave a link to a very nice post and Song already showed how compactness is used in a minimization problem in the comments. I would like to elaborate a bit more, specifically on the usefulness of compactness in minimization problem in Calculus of Variation.
Many important problems in mathematics and mathematical physics can be cast in the form of minimization problems of some integrals functional, i.e. we want to find $bar uin S$ such that
$$
bar u = min{ I(u) : uin S }
$$
where
$$
I(u) := int_Omega L(x,u(x),nabla u(x)) ,dx.
$$
Here $S$ is some function space, e.g. $S={ uin W^{1,1} : uvert_{partial Omega} = f }$ for some fixed $fin W^{1,1}$, and $Omega$ is a bounded domain in a Euclidean space. The integrand $L$ represents the feature of the problem we want to solve. For example $L$ can represent the energy density for the Dirichlet energy functional or the internal energy of an elastic body.
One method to solve this is called the Direct Method. Suppose that the functional $I$ is bounded below on $S$, we can then choose a minimizing sequence $(u_n)_n subset S$ such that $I(u_n)to inf_S I$. The minimization problem is solved if we can establish the following:
1.) $(u_n)_n$ has a subsequence that converges (in appropriate some sense) to a $uin S$.
2.) $I(u) le liminf_{ntoinfty} I(u_n)$.
The appropriate mode of convergence is usually the weak convergence, since weak compactness is easier to establish than strong compactness. Here we see that compactness is crucial in guaranteeing that $(u_n)_n$ has a convergent subsequence.
In this case, we want the functional $I$ to be weakly lower semi-continuous in order that the second condition holds.
answered Jan 16 at 18:43
BigbearZzzBigbearZzz
9,04221653
$endgroup$
Arnaud gave a link to a very nice post and Song already showed how compactness is used in a minimization problem in the comments. I would like to elaborate a bit more, specifically on the usefulness of compactness in minimization problem in Calculus of Variation.
Many important problems in mathematics and mathematical physics can be cast in the form of minimization problems of some integrals functional, i.e. we want to find $bar uin S$ such that
$$
bar u = min{ I(u) : uin S }
$$
where
$$
I(u) := int_Omega L(x,u(x),nabla u(x)) ,dx.
$$
Here $S$ is some function space, e.g. $S={ uin W^{1,1} : uvert_{partial Omega} = f }$ for some fixed $fin W^{1,1}$, and $Omega$ is a bounded domain in a Euclidean space. The integrand $L$ represents the feature of the problem we want to solve. For example $L$ can represent the energy density for the Dirichlet energy functional or the internal energy of an elastic body.
One method to solve this is called the Direct Method. Suppose that the functional $I$ is bounded below on $S$, we can then choose a minimizing sequence $(u_n)_n subset S$ such that $I(u_n)to inf_S I$. The minimization problem is solved if we can establish the following:
1.) $(u_n)_n$ has a subsequence that converges (in appropriate some sense) to a $uin S$.
2.) $I(u) le liminf_{ntoinfty} I(u_n)$.
The appropriate mode of convergence is usually the weak convergence, since weak compactness is easier to establish than strong compactness. Here we see that compactness is crucial in guaranteeing that $(u_n)_n$ has a convergent subsequence.
In this case, we want the functional $I$ to be weakly lower semi-continuous in order that the second condition holds.
answered Jan 16 at 18:43
BigbearZzzBigbearZzz
9,04221653
answered Jan 16 at 18:43
BigbearZzzBigbearZzz
9,04221653
answered Jan 16 at 18:43
BigbearZzzBigbearZzz
9,04221653
answered Jan 16 at 18:43
BigbearZzzBigbearZzz
9,04221653
9,04221653
$begingroup$
Can you complement your answer by elaboraring (maybe an example) on why weak compactness is easier to establish than regular compactness?
$endgroup$
– user8469759
Jan 16 at 19:37
$begingroup$
@user8469759 Because weak topology is coarser than norm topology, hence there are fewer open sets. This automatically implies that any norm-compact set is weakly compact but not vise versa.
$endgroup$
– BigbearZzz
Jan 16 at 19:48
$begingroup$
@user8469759 Also, in practice there are a lot of theorems that we can play around with regarding weak and weak$^*$ compactness, e.g. Banach-Alaoglu, Eberlein-Smulian, Mazur's lemma and other results for reflexive spaces.
$endgroup$
– BigbearZzz
Jan 16 at 19:51
$begingroup$
@user8469759 For a simple example, it is well known that in an infinite dimensional Banach space $X$ the unit ball is never compact and that closed and bounded sets need not be compact. However, if $X$ is a reflexive Banach space then any bounded closed and convex set is weakly compact.
$endgroup$
– BigbearZzz
Jan 16 at 20:06
add a comment |
$begingroup$
Can you complement your answer by elaboraring (maybe an example) on why weak compactness is easier to establish than regular compactness?
$endgroup$
– user8469759
Jan 16 at 19:37
$begingroup$
@user8469759 Because weak topology is coarser than norm topology, hence there are fewer open sets. This automatically implies that any norm-compact set is weakly compact but not vise versa.
$endgroup$
– BigbearZzz
Jan 16 at 19:48
$begingroup$
@user8469759 Also, in practice there are a lot of theorems that we can play around with regarding weak and weak$^*$ compactness, e.g. Banach-Alaoglu, Eberlein-Smulian, Mazur's lemma and other results for reflexive spaces.
$endgroup$
– BigbearZzz
Jan 16 at 19:51
$begingroup$
@user8469759 For a simple example, it is well known that in an infinite dimensional Banach space $X$ the unit ball is never compact and that closed and bounded sets need not be compact. However, if $X$ is a reflexive Banach space then any bounded closed and convex set is weakly compact.
$endgroup$
– BigbearZzz
Jan 16 at 20:06
$begingroup$
Can you complement your answer by elaboraring (maybe an example) on why weak compactness is easier to establish than regular compactness?
$endgroup$
– user8469759
Jan 16 at 19:37
$begingroup$
Can you complement your answer by elaboraring (maybe an example) on why weak compactness is easier to establish than regular compactness?
$endgroup$
– user8469759
Jan 16 at 19:37
$begingroup$
@user8469759 Because weak topology is coarser than norm topology, hence there are fewer open sets. This automatically implies that any norm-compact set is weakly compact but not vise versa.
$endgroup$
– BigbearZzz
Jan 16 at 19:48
$begingroup$
@user8469759 Because weak topology is coarser than norm topology, hence there are fewer open sets. This automatically implies that any norm-compact set is weakly compact but not vise versa.
$endgroup$
– BigbearZzz
Jan 16 at 19:48
$begingroup$
@user8469759 Also, in practice there are a lot of theorems that we can play around with regarding weak and weak$^*$ compactness, e.g. Banach-Alaoglu, Eberlein-Smulian, Mazur's lemma and other results for reflexive spaces.
$endgroup$
– BigbearZzz
Jan 16 at 19:51
$begingroup$
@user8469759 Also, in practice there are a lot of theorems that we can play around with regarding weak and weak$^*$ compactness, e.g. Banach-Alaoglu, Eberlein-Smulian, Mazur's lemma and other results for reflexive spaces.
$endgroup$
– BigbearZzz
Jan 16 at 19:51
$begingroup$
@user8469759 For a simple example, it is well known that in an infinite dimensional Banach space $X$ the unit ball is never compact and that closed and bounded sets need not be compact. However, if $X$ is a reflexive Banach space then any bounded closed and convex set is weakly compact.
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– BigbearZzz
Jan 16 at 20:06
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@user8469759 For a simple example, it is well known that in an infinite dimensional Banach space $X$ the unit ball is never compact and that closed and bounded sets need not be compact. However, if $X$ is a reflexive Banach space then any bounded closed and convex set is weakly compact.
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– BigbearZzz
Jan 16 at 20:06
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1
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In topology and analysis, "compact" is often a very good generalisation of "finite".
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– Arthur
Jan 16 at 17:39
1
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Perhaps this could be interesting.
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– Arnaud D.
Jan 16 at 17:41
2
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@Arthur I don't have a clue of what that can possible imply.
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– user8469759
Jan 16 at 17:47
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The most simple answer I can think of off the top of my head is "extreme value theorem" which is a more simple existence solution that tends to lend itself to others.
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– Prototank
Jan 16 at 18:00
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It is probably a good idea to write down some properties of compact sets and compare them to properties of finite sets so see the wisdom of Arthur's suggestion. It is okay if you don't understand at first. That is the beginning of knowledge.
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– John Douma
Jan 16 at 18:06