Mixing
Convergence of a nested sequence of sets that all contain a common set
$begingroup$
I spent hours looking for answers to my question, but I could not find anything.
I am looking for a proof that a nested sequence of sets $A_{n+1} subseteq A_n subseteq ... subseteq A_0$ that all contain a nonempty set B ($B subseteq A_n, forall n$) is converging towards a non-empty set $A_{infty} $
The sets $A_n$ are linear convex sets of $ mathbb R^p $. So I guess I can say they are closed (?)
The problem is that they are not bounded. So, I cannot use Cantor's intersection theorem.
I know there are seome counter examples of nested closed unbounded sets that do not converge towards a nonempty set, such as $F_n = [n,infty[$.
But in my case, I have the additional condition that $B subseteq A_n, forall n$ which should help.
Anyone knows the theorem saying that my sequence is converging?
Thanks a lot!
Quentin
convergence
asked Jan 19 at 22:52
Quentin PLOUSSARDQuentin PLOUSSARD
31
$endgroup$
add a comment |
$begingroup$
I spent hours looking for answers to my question, but I could not find anything.
I am looking for a proof that a nested sequence of sets $A_{n+1} subseteq A_n subseteq ... subseteq A_0$ that all contain a nonempty set B ($B subseteq A_n, forall n$) is converging towards a non-empty set $A_{infty} $
The sets $A_n$ are linear convex sets of $ mathbb R^p $. So I guess I can say they are closed (?)
The problem is that they are not bounded. So, I cannot use Cantor's intersection theorem.
I know there are seome counter examples of nested closed unbounded sets that do not converge towards a nonempty set, such as $F_n = [n,infty[$.
But in my case, I have the additional condition that $B subseteq A_n, forall n$ which should help.
Anyone knows the theorem saying that my sequence is converging?
Thanks a lot!
Quentin
convergence
asked Jan 19 at 22:52
Quentin PLOUSSARDQuentin PLOUSSARD
31
$endgroup$
$begingroup$
What is a linear convex set? Why do you thing such a a set is closed?
$endgroup$
– Kavi Rama Murthy
Jan 19 at 23:18
$begingroup$
What I mean by "linear convex set" is an intersection of half-spaces. In other words, this is a convex region delimited by linear inequalities.
$endgroup$
– Quentin PLOUSSARD
Jan 19 at 23:59
add a comment |
$begingroup$
I spent hours looking for answers to my question, but I could not find anything.
I am looking for a proof that a nested sequence of sets $A_{n+1} subseteq A_n subseteq ... subseteq A_0$ that all contain a nonempty set B ($B subseteq A_n, forall n$) is converging towards a non-empty set $A_{infty} $
The sets $A_n$ are linear convex sets of $ mathbb R^p $. So I guess I can say they are closed (?)
The problem is that they are not bounded. So, I cannot use Cantor's intersection theorem.
I know there are seome counter examples of nested closed unbounded sets that do not converge towards a nonempty set, such as $F_n = [n,infty[$.
But in my case, I have the additional condition that $B subseteq A_n, forall n$ which should help.
Anyone knows the theorem saying that my sequence is converging?
Thanks a lot!
Quentin
convergence
asked Jan 19 at 22:52
Quentin PLOUSSARDQuentin PLOUSSARD
31
$endgroup$
I spent hours looking for answers to my question, but I could not find anything.
I am looking for a proof that a nested sequence of sets $A_{n+1} subseteq A_n subseteq ... subseteq A_0$ that all contain a nonempty set B ($B subseteq A_n, forall n$) is converging towards a non-empty set $A_{infty} $
The sets $A_n$ are linear convex sets of $ mathbb R^p $. So I guess I can say they are closed (?)
The problem is that they are not bounded. So, I cannot use Cantor's intersection theorem.
I know there are seome counter examples of nested closed unbounded sets that do not converge towards a nonempty set, such as $F_n = [n,infty[$.
But in my case, I have the additional condition that $B subseteq A_n, forall n$ which should help.
Anyone knows the theorem saying that my sequence is converging?
Thanks a lot!
Quentin
convergence
convergence
asked Jan 19 at 22:52
Quentin PLOUSSARDQuentin PLOUSSARD
31
asked Jan 19 at 22:52
Quentin PLOUSSARDQuentin PLOUSSARD
31
asked Jan 19 at 22:52
Quentin PLOUSSARDQuentin PLOUSSARD
31
asked Jan 19 at 22:52
Quentin PLOUSSARDQuentin PLOUSSARD
31
asked Jan 19 at 22:52
Quentin PLOUSSARDQuentin PLOUSSARD
31
31
$begingroup$
What is a linear convex set? Why do you thing such a a set is closed?
$endgroup$
– Kavi Rama Murthy
Jan 19 at 23:18
$begingroup$
What I mean by "linear convex set" is an intersection of half-spaces. In other words, this is a convex region delimited by linear inequalities.
$endgroup$
– Quentin PLOUSSARD
Jan 19 at 23:59
add a comment |
$begingroup$
What is a linear convex set? Why do you thing such a a set is closed?
$endgroup$
– Kavi Rama Murthy
Jan 19 at 23:18
$begingroup$
What I mean by "linear convex set" is an intersection of half-spaces. In other words, this is a convex region delimited by linear inequalities.
$endgroup$
– Quentin PLOUSSARD
Jan 19 at 23:59
$begingroup$
What is a linear convex set? Why do you thing such a a set is closed?
$endgroup$
– Kavi Rama Murthy
Jan 19 at 23:18
$begingroup$
What is a linear convex set? Why do you thing such a a set is closed?
$endgroup$
– Kavi Rama Murthy
Jan 19 at 23:18
$begingroup$
What I mean by "linear convex set" is an intersection of half-spaces. In other words, this is a convex region delimited by linear inequalities.
$endgroup$
– Quentin PLOUSSARD
Jan 19 at 23:59
$begingroup$
What I mean by "linear convex set" is an intersection of half-spaces. In other words, this is a convex region delimited by linear inequalities.
$endgroup$
– Quentin PLOUSSARD
Jan 19 at 23:59
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
So I'm not really sure what you mean by "converging" here, as that's not frequently a term that's used for sets (although you could use it once you define a topology on the space of all subsets of a set). In particular, although the Cantor Intersection Theorem does use topological concepts (i.e. compactness), it doesn't use the idea of "convergence" of sequences of sets but only talks about their intersection.
That said, it's a simple fact about sets that
$$ B subset A_i forall i implies B subset cap_{i=o}^{infty} A_i, $$ and that's true of any sequences of sets without even having to talk about topology. So if you just wanted to show that the intersection of all the $A_i$ is not empty then you could define $A_{infty}$ to be $cap_{i=o}^{infty} A_i$ and then say that the $A_i$ "converge to" it, (and the existence of a set $B$ like you described guarantees that $cap_{i=o}^{infty} A_i$ is not empty). But really that wouldn't be very good use of mathematical language.
If I were writing up a problem and all I cared about is that the intersection isn't empty, I would write "Because all of the $A_i$ contain the (non-empty) set $B$, we can see that $cap_{i=o}^{infty} A_i$ is non-empty." and I think that would satisfy most people. If it's not clear to you why that's true then add a comment and we can dig into more details.
answered Jan 19 at 23:11
JonathanZJonathanZ
2,202613
$endgroup$
$begingroup$
Thank you for your answer. I guess the term "converging" was indeed not appropriate. What I wanted to know are the conditions under which such "infinite intersection" can be defined / exists. Is there a theorem about that. Or is it considered obvious that the infinite intersection of a nested sequence of set will always exist?
$endgroup$
– Quentin PLOUSSARD
Jan 19 at 23:51
$begingroup$
@QuentinPLOUSSARD You don't make much sense. Intersection of any family of sets always exists.
$endgroup$
– Kavi Rama Murthy
Jan 20 at 0:03
$begingroup$
@QuentinPLOUSSARD: It's a very basic assumption (or axiom) of set theory that you can always take the intersection of any family of sets, nested or not. (It consists of the points that are in all of the sets.) However it is always possible that the intersection is the empty set, $emptyset$.
$endgroup$
– JonathanZ
Jan 20 at 0:19
$begingroup$
You can get into really deep set theory and ask whether you can always find the intersection of a sequence of sets, but that's the kind of set theory where you also ask "Does the empty set exist?", or "Given two elements $a$ and $b$ can I always create the set ${a,b}$?" It's pretty abstract stuff and most working mathematicians don't worry about such things. You can just write "$cap A_i$" and assume that this intersection exists.
$endgroup$
– JonathanZ
Jan 20 at 0:27
add a comment |
$begingroup$
So, if I understand well your comments, the infinite intersection of a sequence of sets always exists. Am I right ?
Therefore, for a nested sequence of sets $A_{n+1} ⊆ A_n ⊆...⊆ A_0 $, "convergence", i.e. "$lim_{nto 0} A_n$", always exists ?
answered Jan 20 at 1:08
Quentin PLOUSSARDQuentin PLOUSSARD
31
$endgroup$
$begingroup$
Yes, the intersection always exists. And while I get what you're intending when you say "convergence" here, if you write this to other people you're going get reactions like "what the heck do you mean by that?" and "please define 'convergence'?". Is there something extra you get from using the word "limit" that you don't get from "intersection"?
$endgroup$
– JonathanZ
Jan 20 at 1:35
add a comment |
$begingroup$
What I mean is that, contrary to the nested sequence I am refering to, the sequence of sets $ A_n = [0,1+(-1)^n] $ does not "converge", because it is always "jumping" between the intervals $ { 0 } $ and $[0,2]$.
answered Jan 20 at 1:13
Quentin PLOUSSARDQuentin PLOUSSARD
31
$endgroup$
$begingroup$
You can define a metric on {all compact subspaces of $mathbb{R}$} by $d(X,Y) = text{diameter}(X backslash Y cup Y backslash X)$, and in that metric your sequence does not converge.
$endgroup$
– JonathanZ
Jan 20 at 1:41
add a comment |
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3 Answers
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active
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3 Answers
3
active
oldest
votes
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oldest
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active
oldest
votes
$begingroup$
So I'm not really sure what you mean by "converging" here, as that's not frequently a term that's used for sets (although you could use it once you define a topology on the space of all subsets of a set). In particular, although the Cantor Intersection Theorem does use topological concepts (i.e. compactness), it doesn't use the idea of "convergence" of sequences of sets but only talks about their intersection.
That said, it's a simple fact about sets that
$$ B subset A_i forall i implies B subset cap_{i=o}^{infty} A_i, $$ and that's true of any sequences of sets without even having to talk about topology. So if you just wanted to show that the intersection of all the $A_i$ is not empty then you could define $A_{infty}$ to be $cap_{i=o}^{infty} A_i$ and then say that the $A_i$ "converge to" it, (and the existence of a set $B$ like you described guarantees that $cap_{i=o}^{infty} A_i$ is not empty). But really that wouldn't be very good use of mathematical language.
If I were writing up a problem and all I cared about is that the intersection isn't empty, I would write "Because all of the $A_i$ contain the (non-empty) set $B$, we can see that $cap_{i=o}^{infty} A_i$ is non-empty." and I think that would satisfy most people. If it's not clear to you why that's true then add a comment and we can dig into more details.
answered Jan 19 at 23:11
JonathanZJonathanZ
2,202613
$endgroup$
$begingroup$
Thank you for your answer. I guess the term "converging" was indeed not appropriate. What I wanted to know are the conditions under which such "infinite intersection" can be defined / exists. Is there a theorem about that. Or is it considered obvious that the infinite intersection of a nested sequence of set will always exist?
$endgroup$
– Quentin PLOUSSARD
Jan 19 at 23:51
$begingroup$
@QuentinPLOUSSARD You don't make much sense. Intersection of any family of sets always exists.
$endgroup$
– Kavi Rama Murthy
Jan 20 at 0:03
$begingroup$
@QuentinPLOUSSARD: It's a very basic assumption (or axiom) of set theory that you can always take the intersection of any family of sets, nested or not. (It consists of the points that are in all of the sets.) However it is always possible that the intersection is the empty set, $emptyset$.
$endgroup$
– JonathanZ
Jan 20 at 0:19
$begingroup$
You can get into really deep set theory and ask whether you can always find the intersection of a sequence of sets, but that's the kind of set theory where you also ask "Does the empty set exist?", or "Given two elements $a$ and $b$ can I always create the set ${a,b}$?" It's pretty abstract stuff and most working mathematicians don't worry about such things. You can just write "$cap A_i$" and assume that this intersection exists.
$endgroup$
– JonathanZ
Jan 20 at 0:27
add a comment |
$begingroup$
So I'm not really sure what you mean by "converging" here, as that's not frequently a term that's used for sets (although you could use it once you define a topology on the space of all subsets of a set). In particular, although the Cantor Intersection Theorem does use topological concepts (i.e. compactness), it doesn't use the idea of "convergence" of sequences of sets but only talks about their intersection.
That said, it's a simple fact about sets that
$$ B subset A_i forall i implies B subset cap_{i=o}^{infty} A_i, $$ and that's true of any sequences of sets without even having to talk about topology. So if you just wanted to show that the intersection of all the $A_i$ is not empty then you could define $A_{infty}$ to be $cap_{i=o}^{infty} A_i$ and then say that the $A_i$ "converge to" it, (and the existence of a set $B$ like you described guarantees that $cap_{i=o}^{infty} A_i$ is not empty). But really that wouldn't be very good use of mathematical language.
If I were writing up a problem and all I cared about is that the intersection isn't empty, I would write "Because all of the $A_i$ contain the (non-empty) set $B$, we can see that $cap_{i=o}^{infty} A_i$ is non-empty." and I think that would satisfy most people. If it's not clear to you why that's true then add a comment and we can dig into more details.
answered Jan 19 at 23:11
JonathanZJonathanZ
2,202613
$endgroup$
$begingroup$
Thank you for your answer. I guess the term "converging" was indeed not appropriate. What I wanted to know are the conditions under which such "infinite intersection" can be defined / exists. Is there a theorem about that. Or is it considered obvious that the infinite intersection of a nested sequence of set will always exist?
$endgroup$
– Quentin PLOUSSARD
Jan 19 at 23:51
$begingroup$
@QuentinPLOUSSARD You don't make much sense. Intersection of any family of sets always exists.
$endgroup$
– Kavi Rama Murthy
Jan 20 at 0:03
$begingroup$
@QuentinPLOUSSARD: It's a very basic assumption (or axiom) of set theory that you can always take the intersection of any family of sets, nested or not. (It consists of the points that are in all of the sets.) However it is always possible that the intersection is the empty set, $emptyset$.
$endgroup$
– JonathanZ
Jan 20 at 0:19
$begingroup$
You can get into really deep set theory and ask whether you can always find the intersection of a sequence of sets, but that's the kind of set theory where you also ask "Does the empty set exist?", or "Given two elements $a$ and $b$ can I always create the set ${a,b}$?" It's pretty abstract stuff and most working mathematicians don't worry about such things. You can just write "$cap A_i$" and assume that this intersection exists.
$endgroup$
– JonathanZ
Jan 20 at 0:27
add a comment |
$begingroup$
So I'm not really sure what you mean by "converging" here, as that's not frequently a term that's used for sets (although you could use it once you define a topology on the space of all subsets of a set). In particular, although the Cantor Intersection Theorem does use topological concepts (i.e. compactness), it doesn't use the idea of "convergence" of sequences of sets but only talks about their intersection.
That said, it's a simple fact about sets that
$$ B subset A_i forall i implies B subset cap_{i=o}^{infty} A_i, $$ and that's true of any sequences of sets without even having to talk about topology. So if you just wanted to show that the intersection of all the $A_i$ is not empty then you could define $A_{infty}$ to be $cap_{i=o}^{infty} A_i$ and then say that the $A_i$ "converge to" it, (and the existence of a set $B$ like you described guarantees that $cap_{i=o}^{infty} A_i$ is not empty). But really that wouldn't be very good use of mathematical language.
If I were writing up a problem and all I cared about is that the intersection isn't empty, I would write "Because all of the $A_i$ contain the (non-empty) set $B$, we can see that $cap_{i=o}^{infty} A_i$ is non-empty." and I think that would satisfy most people. If it's not clear to you why that's true then add a comment and we can dig into more details.
answered Jan 19 at 23:11
JonathanZJonathanZ
2,202613
$endgroup$
So I'm not really sure what you mean by "converging" here, as that's not frequently a term that's used for sets (although you could use it once you define a topology on the space of all subsets of a set). In particular, although the Cantor Intersection Theorem does use topological concepts (i.e. compactness), it doesn't use the idea of "convergence" of sequences of sets but only talks about their intersection.
That said, it's a simple fact about sets that
$$ B subset A_i forall i implies B subset cap_{i=o}^{infty} A_i, $$ and that's true of any sequences of sets without even having to talk about topology. So if you just wanted to show that the intersection of all the $A_i$ is not empty then you could define $A_{infty}$ to be $cap_{i=o}^{infty} A_i$ and then say that the $A_i$ "converge to" it, (and the existence of a set $B$ like you described guarantees that $cap_{i=o}^{infty} A_i$ is not empty). But really that wouldn't be very good use of mathematical language.
If I were writing up a problem and all I cared about is that the intersection isn't empty, I would write "Because all of the $A_i$ contain the (non-empty) set $B$, we can see that $cap_{i=o}^{infty} A_i$ is non-empty." and I think that would satisfy most people. If it's not clear to you why that's true then add a comment and we can dig into more details.
answered Jan 19 at 23:11
JonathanZJonathanZ
2,202613
edited Jan 19 at 23:25
answered Jan 19 at 23:11
JonathanZJonathanZ
2,202613
answered Jan 19 at 23:11
JonathanZJonathanZ
2,202613
answered Jan 19 at 23:11
JonathanZJonathanZ
2,202613
2,202613
$begingroup$
Thank you for your answer. I guess the term "converging" was indeed not appropriate. What I wanted to know are the conditions under which such "infinite intersection" can be defined / exists. Is there a theorem about that. Or is it considered obvious that the infinite intersection of a nested sequence of set will always exist?
$endgroup$
– Quentin PLOUSSARD
Jan 19 at 23:51
$begingroup$
@QuentinPLOUSSARD You don't make much sense. Intersection of any family of sets always exists.
$endgroup$
– Kavi Rama Murthy
Jan 20 at 0:03
$begingroup$
@QuentinPLOUSSARD: It's a very basic assumption (or axiom) of set theory that you can always take the intersection of any family of sets, nested or not. (It consists of the points that are in all of the sets.) However it is always possible that the intersection is the empty set, $emptyset$.
$endgroup$
– JonathanZ
Jan 20 at 0:19
$begingroup$
You can get into really deep set theory and ask whether you can always find the intersection of a sequence of sets, but that's the kind of set theory where you also ask "Does the empty set exist?", or "Given two elements $a$ and $b$ can I always create the set ${a,b}$?" It's pretty abstract stuff and most working mathematicians don't worry about such things. You can just write "$cap A_i$" and assume that this intersection exists.
$endgroup$
– JonathanZ
Jan 20 at 0:27
add a comment |
$begingroup$
Thank you for your answer. I guess the term "converging" was indeed not appropriate. What I wanted to know are the conditions under which such "infinite intersection" can be defined / exists. Is there a theorem about that. Or is it considered obvious that the infinite intersection of a nested sequence of set will always exist?
$endgroup$
– Quentin PLOUSSARD
Jan 19 at 23:51
$begingroup$
@QuentinPLOUSSARD You don't make much sense. Intersection of any family of sets always exists.
$endgroup$
– Kavi Rama Murthy
Jan 20 at 0:03
$begingroup$
@QuentinPLOUSSARD: It's a very basic assumption (or axiom) of set theory that you can always take the intersection of any family of sets, nested or not. (It consists of the points that are in all of the sets.) However it is always possible that the intersection is the empty set, $emptyset$.
$endgroup$
– JonathanZ
Jan 20 at 0:19
$begingroup$
You can get into really deep set theory and ask whether you can always find the intersection of a sequence of sets, but that's the kind of set theory where you also ask "Does the empty set exist?", or "Given two elements $a$ and $b$ can I always create the set ${a,b}$?" It's pretty abstract stuff and most working mathematicians don't worry about such things. You can just write "$cap A_i$" and assume that this intersection exists.
$endgroup$
– JonathanZ
Jan 20 at 0:27
$begingroup$
Thank you for your answer. I guess the term "converging" was indeed not appropriate. What I wanted to know are the conditions under which such "infinite intersection" can be defined / exists. Is there a theorem about that. Or is it considered obvious that the infinite intersection of a nested sequence of set will always exist?
$endgroup$
– Quentin PLOUSSARD
Jan 19 at 23:51
$begingroup$
Thank you for your answer. I guess the term "converging" was indeed not appropriate. What I wanted to know are the conditions under which such "infinite intersection" can be defined / exists. Is there a theorem about that. Or is it considered obvious that the infinite intersection of a nested sequence of set will always exist?
$endgroup$
– Quentin PLOUSSARD
Jan 19 at 23:51
$begingroup$
@QuentinPLOUSSARD You don't make much sense. Intersection of any family of sets always exists.
$endgroup$
– Kavi Rama Murthy
Jan 20 at 0:03
$begingroup$
@QuentinPLOUSSARD You don't make much sense. Intersection of any family of sets always exists.
$endgroup$
– Kavi Rama Murthy
Jan 20 at 0:03
$begingroup$
@QuentinPLOUSSARD: It's a very basic assumption (or axiom) of set theory that you can always take the intersection of any family of sets, nested or not. (It consists of the points that are in all of the sets.) However it is always possible that the intersection is the empty set, $emptyset$.
$endgroup$
– JonathanZ
Jan 20 at 0:19
$begingroup$
@QuentinPLOUSSARD: It's a very basic assumption (or axiom) of set theory that you can always take the intersection of any family of sets, nested or not. (It consists of the points that are in all of the sets.) However it is always possible that the intersection is the empty set, $emptyset$.
$endgroup$
– JonathanZ
Jan 20 at 0:19
$begingroup$
You can get into really deep set theory and ask whether you can always find the intersection of a sequence of sets, but that's the kind of set theory where you also ask "Does the empty set exist?", or "Given two elements $a$ and $b$ can I always create the set ${a,b}$?" It's pretty abstract stuff and most working mathematicians don't worry about such things. You can just write "$cap A_i$" and assume that this intersection exists.
$endgroup$
– JonathanZ
Jan 20 at 0:27
$begingroup$
You can get into really deep set theory and ask whether you can always find the intersection of a sequence of sets, but that's the kind of set theory where you also ask "Does the empty set exist?", or "Given two elements $a$ and $b$ can I always create the set ${a,b}$?" It's pretty abstract stuff and most working mathematicians don't worry about such things. You can just write "$cap A_i$" and assume that this intersection exists.
$endgroup$
– JonathanZ
Jan 20 at 0:27
add a comment |
$begingroup$
So, if I understand well your comments, the infinite intersection of a sequence of sets always exists. Am I right ?
Therefore, for a nested sequence of sets $A_{n+1} ⊆ A_n ⊆...⊆ A_0 $, "convergence", i.e. "$lim_{nto 0} A_n$", always exists ?
answered Jan 20 at 1:08
Quentin PLOUSSARDQuentin PLOUSSARD
31
$endgroup$
$begingroup$
Yes, the intersection always exists. And while I get what you're intending when you say "convergence" here, if you write this to other people you're going get reactions like "what the heck do you mean by that?" and "please define 'convergence'?". Is there something extra you get from using the word "limit" that you don't get from "intersection"?
$endgroup$
– JonathanZ
Jan 20 at 1:35
add a comment |
$begingroup$
So, if I understand well your comments, the infinite intersection of a sequence of sets always exists. Am I right ?
Therefore, for a nested sequence of sets $A_{n+1} ⊆ A_n ⊆...⊆ A_0 $, "convergence", i.e. "$lim_{nto 0} A_n$", always exists ?
answered Jan 20 at 1:08
Quentin PLOUSSARDQuentin PLOUSSARD
31
$endgroup$
$begingroup$
Yes, the intersection always exists. And while I get what you're intending when you say "convergence" here, if you write this to other people you're going get reactions like "what the heck do you mean by that?" and "please define 'convergence'?". Is there something extra you get from using the word "limit" that you don't get from "intersection"?
$endgroup$
– JonathanZ
Jan 20 at 1:35
add a comment |
$begingroup$
So, if I understand well your comments, the infinite intersection of a sequence of sets always exists. Am I right ?
Therefore, for a nested sequence of sets $A_{n+1} ⊆ A_n ⊆...⊆ A_0 $, "convergence", i.e. "$lim_{nto 0} A_n$", always exists ?
answered Jan 20 at 1:08
Quentin PLOUSSARDQuentin PLOUSSARD
31
$endgroup$
So, if I understand well your comments, the infinite intersection of a sequence of sets always exists. Am I right ?
Therefore, for a nested sequence of sets $A_{n+1} ⊆ A_n ⊆...⊆ A_0 $, "convergence", i.e. "$lim_{nto 0} A_n$", always exists ?
answered Jan 20 at 1:08
Quentin PLOUSSARDQuentin PLOUSSARD
31
answered Jan 20 at 1:08
Quentin PLOUSSARDQuentin PLOUSSARD
31
answered Jan 20 at 1:08
Quentin PLOUSSARDQuentin PLOUSSARD
31
answered Jan 20 at 1:08
Quentin PLOUSSARDQuentin PLOUSSARD
31
31
$begingroup$
Yes, the intersection always exists. And while I get what you're intending when you say "convergence" here, if you write this to other people you're going get reactions like "what the heck do you mean by that?" and "please define 'convergence'?". Is there something extra you get from using the word "limit" that you don't get from "intersection"?
$endgroup$
– JonathanZ
Jan 20 at 1:35
add a comment |
$begingroup$
Yes, the intersection always exists. And while I get what you're intending when you say "convergence" here, if you write this to other people you're going get reactions like "what the heck do you mean by that?" and "please define 'convergence'?". Is there something extra you get from using the word "limit" that you don't get from "intersection"?
$endgroup$
– JonathanZ
Jan 20 at 1:35
$begingroup$
Yes, the intersection always exists. And while I get what you're intending when you say "convergence" here, if you write this to other people you're going get reactions like "what the heck do you mean by that?" and "please define 'convergence'?". Is there something extra you get from using the word "limit" that you don't get from "intersection"?
$endgroup$
– JonathanZ
Jan 20 at 1:35
$begingroup$
Yes, the intersection always exists. And while I get what you're intending when you say "convergence" here, if you write this to other people you're going get reactions like "what the heck do you mean by that?" and "please define 'convergence'?". Is there something extra you get from using the word "limit" that you don't get from "intersection"?
$endgroup$
– JonathanZ
Jan 20 at 1:35
add a comment |
$begingroup$
What I mean is that, contrary to the nested sequence I am refering to, the sequence of sets $ A_n = [0,1+(-1)^n] $ does not "converge", because it is always "jumping" between the intervals $ { 0 } $ and $[0,2]$.
answered Jan 20 at 1:13
Quentin PLOUSSARDQuentin PLOUSSARD
31
$endgroup$
$begingroup$
You can define a metric on {all compact subspaces of $mathbb{R}$} by $d(X,Y) = text{diameter}(X backslash Y cup Y backslash X)$, and in that metric your sequence does not converge.
$endgroup$
– JonathanZ
Jan 20 at 1:41
add a comment |
$begingroup$
What I mean is that, contrary to the nested sequence I am refering to, the sequence of sets $ A_n = [0,1+(-1)^n] $ does not "converge", because it is always "jumping" between the intervals $ { 0 } $ and $[0,2]$.
answered Jan 20 at 1:13
Quentin PLOUSSARDQuentin PLOUSSARD
31
$endgroup$
$begingroup$
You can define a metric on {all compact subspaces of $mathbb{R}$} by $d(X,Y) = text{diameter}(X backslash Y cup Y backslash X)$, and in that metric your sequence does not converge.
$endgroup$
– JonathanZ
Jan 20 at 1:41
add a comment |
$begingroup$
What I mean is that, contrary to the nested sequence I am refering to, the sequence of sets $ A_n = [0,1+(-1)^n] $ does not "converge", because it is always "jumping" between the intervals $ { 0 } $ and $[0,2]$.
answered Jan 20 at 1:13
Quentin PLOUSSARDQuentin PLOUSSARD
31
$endgroup$
What I mean is that, contrary to the nested sequence I am refering to, the sequence of sets $ A_n = [0,1+(-1)^n] $ does not "converge", because it is always "jumping" between the intervals $ { 0 } $ and $[0,2]$.
answered Jan 20 at 1:13
Quentin PLOUSSARDQuentin PLOUSSARD
31
answered Jan 20 at 1:13
Quentin PLOUSSARDQuentin PLOUSSARD
31
answered Jan 20 at 1:13
Quentin PLOUSSARDQuentin PLOUSSARD
31
answered Jan 20 at 1:13
Quentin PLOUSSARDQuentin PLOUSSARD
31
31
$begingroup$
You can define a metric on {all compact subspaces of $mathbb{R}$} by $d(X,Y) = text{diameter}(X backslash Y cup Y backslash X)$, and in that metric your sequence does not converge.
$endgroup$
– JonathanZ
Jan 20 at 1:41
add a comment |
$begingroup$
You can define a metric on {all compact subspaces of $mathbb{R}$} by $d(X,Y) = text{diameter}(X backslash Y cup Y backslash X)$, and in that metric your sequence does not converge.
$endgroup$
– JonathanZ
Jan 20 at 1:41
$begingroup$
You can define a metric on {all compact subspaces of $mathbb{R}$} by $d(X,Y) = text{diameter}(X backslash Y cup Y backslash X)$, and in that metric your sequence does not converge.
$endgroup$
– JonathanZ
Jan 20 at 1:41
$begingroup$
You can define a metric on {all compact subspaces of $mathbb{R}$} by $d(X,Y) = text{diameter}(X backslash Y cup Y backslash X)$, and in that metric your sequence does not converge.
$endgroup$
– JonathanZ
Jan 20 at 1:41
add a comment |
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$begingroup$
What is a linear convex set? Why do you thing such a a set is closed?
$endgroup$
– Kavi Rama Murthy
Jan 19 at 23:18
$begingroup$
What I mean by "linear convex set" is an intersection of half-spaces. In other words, this is a convex region delimited by linear inequalities.
$endgroup$
– Quentin PLOUSSARD
Jan 19 at 23:59