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What is a compact set?
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According to one of the definition I found:
A set $S subset R^n$ is compact if every sequence in $S$ has a
convergent subsequence, whose limit lies in $S$.
Specifically, I am not clear what does convergent subsequence mean but it would be really helpful if someone can explain what's a compact set by breaking down the above definition into sub parts. Also, an example and a counter-example will be really helpful.
This definition has been used in the context of game theory to proof existence of Nash equilibrium.
general-topology
asked Jul 18 at 8:07
Rishabh
397
add a comment |Â
up vote
-3
down vote
favorite
According to one of the definition I found:
A set $S subset R^n$ is compact if every sequence in $S$ has a
convergent subsequence, whose limit lies in $S$.
Specifically, I am not clear what does convergent subsequence mean but it would be really helpful if someone can explain what's a compact set by breaking down the above definition into sub parts. Also, an example and a counter-example will be really helpful.
This definition has been used in the context of game theory to proof existence of Nash equilibrium.
general-topology
asked Jul 18 at 8:07
Rishabh
397
1
In $Bbb R^n$ exactly the bounded closed sets are the compact ones.
– Berci
Jul 18 at 8:11
What's the reason for down vote? Is the question not clear?
– Rishabh
Jul 18 at 9:32
2
What you have described is actually sequential compactness. Thankfully, for metric spaces such as $mathbbR^n$, sequential compactness implies compactness (and vice-versa).
– feynhat
Jul 18 at 9:38
@moderator Could you please explain what details is the question missing?
– Rishabh
Jul 19 at 19:20
add a comment |Â
up vote
-3
down vote
favorite
up vote
-3
down vote
favorite
According to one of the definition I found:
A set $S subset R^n$ is compact if every sequence in $S$ has a
convergent subsequence, whose limit lies in $S$.
Specifically, I am not clear what does convergent subsequence mean but it would be really helpful if someone can explain what's a compact set by breaking down the above definition into sub parts. Also, an example and a counter-example will be really helpful.
This definition has been used in the context of game theory to proof existence of Nash equilibrium.
general-topology
asked Jul 18 at 8:07
Rishabh
397
According to one of the definition I found:
A set $S subset R^n$ is compact if every sequence in $S$ has a
convergent subsequence, whose limit lies in $S$.
Specifically, I am not clear what does convergent subsequence mean but it would be really helpful if someone can explain what's a compact set by breaking down the above definition into sub parts. Also, an example and a counter-example will be really helpful.
This definition has been used in the context of game theory to proof existence of Nash equilibrium.
general-topology
asked Jul 18 at 8:07
Rishabh
397
edited Jul 19 at 19:19
asked Jul 18 at 8:07
Rishabh
397
asked Jul 18 at 8:07
Rishabh
397
asked Jul 18 at 8:07
Rishabh
397
397
1
In $Bbb R^n$ exactly the bounded closed sets are the compact ones.
– Berci
Jul 18 at 8:11
What's the reason for down vote? Is the question not clear?
– Rishabh
Jul 18 at 9:32
2
What you have described is actually sequential compactness. Thankfully, for metric spaces such as $mathbbR^n$, sequential compactness implies compactness (and vice-versa).
– feynhat
Jul 18 at 9:38
@moderator Could you please explain what details is the question missing?
– Rishabh
Jul 19 at 19:20
add a comment |Â
1
In $Bbb R^n$ exactly the bounded closed sets are the compact ones.
– Berci
Jul 18 at 8:11
What's the reason for down vote? Is the question not clear?
– Rishabh
Jul 18 at 9:32
2
What you have described is actually sequential compactness. Thankfully, for metric spaces such as $mathbbR^n$, sequential compactness implies compactness (and vice-versa).
– feynhat
Jul 18 at 9:38
@moderator Could you please explain what details is the question missing?
– Rishabh
Jul 19 at 19:20
1
1
In $Bbb R^n$ exactly the bounded closed sets are the compact ones.
– Berci
Jul 18 at 8:11
In $Bbb R^n$ exactly the bounded closed sets are the compact ones.
– Berci
Jul 18 at 8:11
What's the reason for down vote? Is the question not clear?
– Rishabh
Jul 18 at 9:32
What's the reason for down vote? Is the question not clear?
– Rishabh
Jul 18 at 9:32
2
2
What you have described is actually sequential compactness. Thankfully, for metric spaces such as $mathbbR^n$, sequential compactness implies compactness (and vice-versa).
– feynhat
Jul 18 at 9:38
What you have described is actually sequential compactness. Thankfully, for metric spaces such as $mathbbR^n$, sequential compactness implies compactness (and vice-versa).
– feynhat
Jul 18 at 9:38
@moderator Could you please explain what details is the question missing?
– Rishabh
Jul 19 at 19:20
@moderator Could you please explain what details is the question missing?
– Rishabh
Jul 19 at 19:20
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
7
down vote
accepted
Let's take $[0,1]subseteq Bbb R$ as an example of a compact set. There are many sequences contained in that set. Some converge, others do not. For instance, $1, frac12, frac13,ldots$ does converge, while $0,1,0,1,0,ldots$ does not.
However, if you look at that second sequence, we can "remove" every other term, and be left with $0,0,0,ldots$, which is a converging sequence. That's what a "convergent subsequence" is. "Convergent" because it, well, converges. And "subsequence" because it's a sequence made of terms from the original sequence. (Technical note: in a subsequence, the terms must appear in the same order as they did in the original sequence, even though not all terms are present. You are allowed to remove terms, but not shuffle them or duplicate them when you make a subsequence.)
If a sequence is convergent, then all (infinite) subsequences will necessarily converge as well, to the same limit.
So $[0,1]$ being compact means that for any sequence $a_n$ with $a_nin [0,1]$, it is possible to remove terms from that sequence to make a subsequence which is convergent, and where the limit is contained in $[0,1]$. That last part is important, because $(0,1)$ isn't compact (or, rather, we don't want it to be, so we make a definition which excludes it). This is illustrated by the sequence $1, frac12, frac13,ldots$ which converges to $0$ (and so does any subsequence). But $0$ is not contained in $(0,1)$, so in this case we have found a sequence in which all convergent subsequences converge to a limit outside $(0,1)$.
answered Jul 18 at 8:23
Arthur
98.8k793175
Must the terms of a subsequence be taken at regular intervals in the supersequence as per your example, or can they be taken at any sequence of intervals?
– Robert Frost
Aug 1 at 7:10
1
@RobertFrost Good point. No, it can be any sequence of intervals, as long as you always move ahead in the sequence.
– Arthur
Aug 1 at 7:24
add a comment |Â
up vote
2
down vote
In other terms, the definition says that any infinite sequence of points has an accumulation point (points getting arbitrarily close to a given one), and that the accumulation point belongs to the set.
This property could be invalidated in two ways:
by having points escaping to infinity (then no accumulation point),
by having an accumulation point in the closure of the set but not in the set itself.
Compactness, i.e. being bounded and closed ensures that this does not happen. This is useful when a proof requires that a limit point lies in the set.
answered Jul 18 at 8:25
Yves Daoust
111k665204
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
7
down vote
accepted
Let's take $[0,1]subseteq Bbb R$ as an example of a compact set. There are many sequences contained in that set. Some converge, others do not. For instance, $1, frac12, frac13,ldots$ does converge, while $0,1,0,1,0,ldots$ does not.
However, if you look at that second sequence, we can "remove" every other term, and be left with $0,0,0,ldots$, which is a converging sequence. That's what a "convergent subsequence" is. "Convergent" because it, well, converges. And "subsequence" because it's a sequence made of terms from the original sequence. (Technical note: in a subsequence, the terms must appear in the same order as they did in the original sequence, even though not all terms are present. You are allowed to remove terms, but not shuffle them or duplicate them when you make a subsequence.)
If a sequence is convergent, then all (infinite) subsequences will necessarily converge as well, to the same limit.
So $[0,1]$ being compact means that for any sequence $a_n$ with $a_nin [0,1]$, it is possible to remove terms from that sequence to make a subsequence which is convergent, and where the limit is contained in $[0,1]$. That last part is important, because $(0,1)$ isn't compact (or, rather, we don't want it to be, so we make a definition which excludes it). This is illustrated by the sequence $1, frac12, frac13,ldots$ which converges to $0$ (and so does any subsequence). But $0$ is not contained in $(0,1)$, so in this case we have found a sequence in which all convergent subsequences converge to a limit outside $(0,1)$.
answered Jul 18 at 8:23
Arthur
98.8k793175
Must the terms of a subsequence be taken at regular intervals in the supersequence as per your example, or can they be taken at any sequence of intervals?
– Robert Frost
Aug 1 at 7:10
1
@RobertFrost Good point. No, it can be any sequence of intervals, as long as you always move ahead in the sequence.
– Arthur
Aug 1 at 7:24
add a comment |Â
up vote
7
down vote
accepted
Let's take $[0,1]subseteq Bbb R$ as an example of a compact set. There are many sequences contained in that set. Some converge, others do not. For instance, $1, frac12, frac13,ldots$ does converge, while $0,1,0,1,0,ldots$ does not.
However, if you look at that second sequence, we can "remove" every other term, and be left with $0,0,0,ldots$, which is a converging sequence. That's what a "convergent subsequence" is. "Convergent" because it, well, converges. And "subsequence" because it's a sequence made of terms from the original sequence. (Technical note: in a subsequence, the terms must appear in the same order as they did in the original sequence, even though not all terms are present. You are allowed to remove terms, but not shuffle them or duplicate them when you make a subsequence.)
If a sequence is convergent, then all (infinite) subsequences will necessarily converge as well, to the same limit.
So $[0,1]$ being compact means that for any sequence $a_n$ with $a_nin [0,1]$, it is possible to remove terms from that sequence to make a subsequence which is convergent, and where the limit is contained in $[0,1]$. That last part is important, because $(0,1)$ isn't compact (or, rather, we don't want it to be, so we make a definition which excludes it). This is illustrated by the sequence $1, frac12, frac13,ldots$ which converges to $0$ (and so does any subsequence). But $0$ is not contained in $(0,1)$, so in this case we have found a sequence in which all convergent subsequences converge to a limit outside $(0,1)$.
answered Jul 18 at 8:23
Arthur
98.8k793175
Must the terms of a subsequence be taken at regular intervals in the supersequence as per your example, or can they be taken at any sequence of intervals?
– Robert Frost
Aug 1 at 7:10
1
@RobertFrost Good point. No, it can be any sequence of intervals, as long as you always move ahead in the sequence.
– Arthur
Aug 1 at 7:24
add a comment |Â
up vote
7
down vote
accepted
up vote
7
down vote
accepted
Let's take $[0,1]subseteq Bbb R$ as an example of a compact set. There are many sequences contained in that set. Some converge, others do not. For instance, $1, frac12, frac13,ldots$ does converge, while $0,1,0,1,0,ldots$ does not.
However, if you look at that second sequence, we can "remove" every other term, and be left with $0,0,0,ldots$, which is a converging sequence. That's what a "convergent subsequence" is. "Convergent" because it, well, converges. And "subsequence" because it's a sequence made of terms from the original sequence. (Technical note: in a subsequence, the terms must appear in the same order as they did in the original sequence, even though not all terms are present. You are allowed to remove terms, but not shuffle them or duplicate them when you make a subsequence.)
If a sequence is convergent, then all (infinite) subsequences will necessarily converge as well, to the same limit.
So $[0,1]$ being compact means that for any sequence $a_n$ with $a_nin [0,1]$, it is possible to remove terms from that sequence to make a subsequence which is convergent, and where the limit is contained in $[0,1]$. That last part is important, because $(0,1)$ isn't compact (or, rather, we don't want it to be, so we make a definition which excludes it). This is illustrated by the sequence $1, frac12, frac13,ldots$ which converges to $0$ (and so does any subsequence). But $0$ is not contained in $(0,1)$, so in this case we have found a sequence in which all convergent subsequences converge to a limit outside $(0,1)$.
answered Jul 18 at 8:23
Arthur
98.8k793175
Let's take $[0,1]subseteq Bbb R$ as an example of a compact set. There are many sequences contained in that set. Some converge, others do not. For instance, $1, frac12, frac13,ldots$ does converge, while $0,1,0,1,0,ldots$ does not.
However, if you look at that second sequence, we can "remove" every other term, and be left with $0,0,0,ldots$, which is a converging sequence. That's what a "convergent subsequence" is. "Convergent" because it, well, converges. And "subsequence" because it's a sequence made of terms from the original sequence. (Technical note: in a subsequence, the terms must appear in the same order as they did in the original sequence, even though not all terms are present. You are allowed to remove terms, but not shuffle them or duplicate them when you make a subsequence.)
If a sequence is convergent, then all (infinite) subsequences will necessarily converge as well, to the same limit.
So $[0,1]$ being compact means that for any sequence $a_n$ with $a_nin [0,1]$, it is possible to remove terms from that sequence to make a subsequence which is convergent, and where the limit is contained in $[0,1]$. That last part is important, because $(0,1)$ isn't compact (or, rather, we don't want it to be, so we make a definition which excludes it). This is illustrated by the sequence $1, frac12, frac13,ldots$ which converges to $0$ (and so does any subsequence). But $0$ is not contained in $(0,1)$, so in this case we have found a sequence in which all convergent subsequences converge to a limit outside $(0,1)$.
answered Jul 18 at 8:23
Arthur
98.8k793175
answered Jul 18 at 8:23
Arthur
98.8k793175
answered Jul 18 at 8:23
Arthur
98.8k793175
answered Jul 18 at 8:23
Arthur
98.8k793175
98.8k793175
Must the terms of a subsequence be taken at regular intervals in the supersequence as per your example, or can they be taken at any sequence of intervals?
– Robert Frost
Aug 1 at 7:10
1
@RobertFrost Good point. No, it can be any sequence of intervals, as long as you always move ahead in the sequence.
– Arthur
Aug 1 at 7:24
add a comment |Â
Must the terms of a subsequence be taken at regular intervals in the supersequence as per your example, or can they be taken at any sequence of intervals?
– Robert Frost
Aug 1 at 7:10
1
@RobertFrost Good point. No, it can be any sequence of intervals, as long as you always move ahead in the sequence.
– Arthur
Aug 1 at 7:24
Must the terms of a subsequence be taken at regular intervals in the supersequence as per your example, or can they be taken at any sequence of intervals?
– Robert Frost
Aug 1 at 7:10
Must the terms of a subsequence be taken at regular intervals in the supersequence as per your example, or can they be taken at any sequence of intervals?
– Robert Frost
Aug 1 at 7:10
1
1
@RobertFrost Good point. No, it can be any sequence of intervals, as long as you always move ahead in the sequence.
– Arthur
Aug 1 at 7:24
@RobertFrost Good point. No, it can be any sequence of intervals, as long as you always move ahead in the sequence.
– Arthur
Aug 1 at 7:24
add a comment |Â
up vote
2
down vote
In other terms, the definition says that any infinite sequence of points has an accumulation point (points getting arbitrarily close to a given one), and that the accumulation point belongs to the set.
This property could be invalidated in two ways:
by having points escaping to infinity (then no accumulation point),
by having an accumulation point in the closure of the set but not in the set itself.
Compactness, i.e. being bounded and closed ensures that this does not happen. This is useful when a proof requires that a limit point lies in the set.
answered Jul 18 at 8:25
Yves Daoust
111k665204
add a comment |Â
up vote
2
down vote
In other terms, the definition says that any infinite sequence of points has an accumulation point (points getting arbitrarily close to a given one), and that the accumulation point belongs to the set.
This property could be invalidated in two ways:
by having points escaping to infinity (then no accumulation point),
by having an accumulation point in the closure of the set but not in the set itself.
Compactness, i.e. being bounded and closed ensures that this does not happen. This is useful when a proof requires that a limit point lies in the set.
answered Jul 18 at 8:25
Yves Daoust
111k665204
add a comment |Â
up vote
2
down vote
up vote
2
down vote
In other terms, the definition says that any infinite sequence of points has an accumulation point (points getting arbitrarily close to a given one), and that the accumulation point belongs to the set.
This property could be invalidated in two ways:
by having points escaping to infinity (then no accumulation point),
by having an accumulation point in the closure of the set but not in the set itself.
Compactness, i.e. being bounded and closed ensures that this does not happen. This is useful when a proof requires that a limit point lies in the set.
answered Jul 18 at 8:25
Yves Daoust
111k665204
In other terms, the definition says that any infinite sequence of points has an accumulation point (points getting arbitrarily close to a given one), and that the accumulation point belongs to the set.
This property could be invalidated in two ways:
by having points escaping to infinity (then no accumulation point),
by having an accumulation point in the closure of the set but not in the set itself.
Compactness, i.e. being bounded and closed ensures that this does not happen. This is useful when a proof requires that a limit point lies in the set.
answered Jul 18 at 8:25
Yves Daoust
111k665204
answered Jul 18 at 8:25
Yves Daoust
111k665204
answered Jul 18 at 8:25
Yves Daoust
111k665204
answered Jul 18 at 8:25
Yves Daoust
111k665204
111k665204
add a comment |Â
add a comment |Â
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1
In $Bbb R^n$ exactly the bounded closed sets are the compact ones.
– Berci
Jul 18 at 8:11
What's the reason for down vote? Is the question not clear?
– Rishabh
Jul 18 at 9:32
2
What you have described is actually sequential compactness. Thankfully, for metric spaces such as $mathbbR^n$, sequential compactness implies compactness (and vice-versa).
– feynhat
Jul 18 at 9:38
@moderator Could you please explain what details is the question missing?
– Rishabh
Jul 19 at 19:20