Sequence with anchoring property is convergent in finite dimension
$begingroup$
Let $(x_k)$ be a sequence in $mathbb{R}^n$. Furthermore, assume that the sequence has the following "anchoring" property: $exists y,z inmathbb{R}^n$ such that $lim|x_k-y|=r_y$ and $lim|x_k-z|=r_z$. Show that the sequence $(x_k)$ is convergent.
I have shown it's bounded:
$|x_k| leq |x_k-y|+|y| < r_y + varepsilon + |y| < infty$.
From here I am not sure where to go. Any hints? I thought maybe I should use the sequential compactness of closed balls in $mathbb{R}^n$ to find a convergent subsequence maybe?
real-analysis sequences-and-series functional-analysis
asked Dec 11 '18 at 10:58
Tony S.F.Tony S.F.
3,29821028
$endgroup$
|
show 2 more comments
$begingroup$
Let $(x_k)$ be a sequence in $mathbb{R}^n$. Furthermore, assume that the sequence has the following "anchoring" property: $exists y,z inmathbb{R}^n$ such that $lim|x_k-y|=r_y$ and $lim|x_k-z|=r_z$. Show that the sequence $(x_k)$ is convergent.
I have shown it's bounded:
$|x_k| leq |x_k-y|+|y| < r_y + varepsilon + |y| < infty$.
From here I am not sure where to go. Any hints? I thought maybe I should use the sequential compactness of closed balls in $mathbb{R}^n$ to find a convergent subsequence maybe?
real-analysis sequences-and-series functional-analysis
asked Dec 11 '18 at 10:58
Tony S.F.Tony S.F.
3,29821028
$endgroup$
1
$begingroup$
are there other conditions such as $y ne z$
$endgroup$
– Siong Thye Goh
Dec 11 '18 at 11:05
$begingroup$
Yeah $yneq z$ of course otherwise it's not true (circle)
$endgroup$
– Tony S.F.
Dec 11 '18 at 11:09
$begingroup$
Given the answer below, it seems to be true only for $n=1$.
$endgroup$
– daw
Dec 11 '18 at 12:15
$begingroup$
Shouldn't be $y,zin Bbb R$?
$endgroup$
– Mostafa Ayaz
Dec 11 '18 at 23:14
$begingroup$
@MostafaAyaz they are in R^n. OP writes |y|, but we understand than he/she means the norm of a vector.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Dec 12 '18 at 13:41
|
show 2 more comments
$begingroup$
Let $(x_k)$ be a sequence in $mathbb{R}^n$. Furthermore, assume that the sequence has the following "anchoring" property: $exists y,z inmathbb{R}^n$ such that $lim|x_k-y|=r_y$ and $lim|x_k-z|=r_z$. Show that the sequence $(x_k)$ is convergent.
I have shown it's bounded:
$|x_k| leq |x_k-y|+|y| < r_y + varepsilon + |y| < infty$.
From here I am not sure where to go. Any hints? I thought maybe I should use the sequential compactness of closed balls in $mathbb{R}^n$ to find a convergent subsequence maybe?
real-analysis sequences-and-series functional-analysis
asked Dec 11 '18 at 10:58
Tony S.F.Tony S.F.
3,29821028
$endgroup$
Let $(x_k)$ be a sequence in $mathbb{R}^n$. Furthermore, assume that the sequence has the following "anchoring" property: $exists y,z inmathbb{R}^n$ such that $lim|x_k-y|=r_y$ and $lim|x_k-z|=r_z$. Show that the sequence $(x_k)$ is convergent.
I have shown it's bounded:
$|x_k| leq |x_k-y|+|y| < r_y + varepsilon + |y| < infty$.
From here I am not sure where to go. Any hints? I thought maybe I should use the sequential compactness of closed balls in $mathbb{R}^n$ to find a convergent subsequence maybe?
real-analysis sequences-and-series functional-analysis
real-analysis sequences-and-series functional-analysis
asked Dec 11 '18 at 10:58
Tony S.F.Tony S.F.
3,29821028
asked Dec 11 '18 at 10:58
Tony S.F.Tony S.F.
3,29821028
asked Dec 11 '18 at 10:58
Tony S.F.Tony S.F.
3,29821028
asked Dec 11 '18 at 10:58
Tony S.F.Tony S.F.
3,29821028
asked Dec 11 '18 at 10:58
Tony S.F.Tony S.F.
3,29821028
3,29821028
1
$begingroup$
are there other conditions such as $y ne z$
$endgroup$
– Siong Thye Goh
Dec 11 '18 at 11:05
$begingroup$
Yeah $yneq z$ of course otherwise it's not true (circle)
$endgroup$
– Tony S.F.
Dec 11 '18 at 11:09
$begingroup$
Given the answer below, it seems to be true only for $n=1$.
$endgroup$
– daw
Dec 11 '18 at 12:15
$begingroup$
Shouldn't be $y,zin Bbb R$?
$endgroup$
– Mostafa Ayaz
Dec 11 '18 at 23:14
$begingroup$
@MostafaAyaz they are in R^n. OP writes |y|, but we understand than he/she means the norm of a vector.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Dec 12 '18 at 13:41
|
show 2 more comments
1
$begingroup$
are there other conditions such as $y ne z$
$endgroup$
– Siong Thye Goh
Dec 11 '18 at 11:05
$begingroup$
Yeah $yneq z$ of course otherwise it's not true (circle)
$endgroup$
– Tony S.F.
Dec 11 '18 at 11:09
$begingroup$
Given the answer below, it seems to be true only for $n=1$.
$endgroup$
– daw
Dec 11 '18 at 12:15
$begingroup$
Shouldn't be $y,zin Bbb R$?
$endgroup$
– Mostafa Ayaz
Dec 11 '18 at 23:14
$begingroup$
@MostafaAyaz they are in R^n. OP writes |y|, but we understand than he/she means the norm of a vector.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Dec 12 '18 at 13:41
1
1
$begingroup$
are there other conditions such as $y ne z$
$endgroup$
– Siong Thye Goh
Dec 11 '18 at 11:05
$begingroup$
are there other conditions such as $y ne z$
$endgroup$
– Siong Thye Goh
Dec 11 '18 at 11:05
$begingroup$
Yeah $yneq z$ of course otherwise it's not true (circle)
$endgroup$
– Tony S.F.
Dec 11 '18 at 11:09
$begingroup$
Yeah $yneq z$ of course otherwise it's not true (circle)
$endgroup$
– Tony S.F.
Dec 11 '18 at 11:09
$begingroup$
Given the answer below, it seems to be true only for $n=1$.
$endgroup$
– daw
Dec 11 '18 at 12:15
$begingroup$
Given the answer below, it seems to be true only for $n=1$.
$endgroup$
– daw
Dec 11 '18 at 12:15
$begingroup$
Shouldn't be $y,zin Bbb R$?
$endgroup$
– Mostafa Ayaz
Dec 11 '18 at 23:14
$begingroup$
Shouldn't be $y,zin Bbb R$?
$endgroup$
– Mostafa Ayaz
Dec 11 '18 at 23:14
$begingroup$
@MostafaAyaz they are in R^n. OP writes |y|, but we understand than he/she means the norm of a vector.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Dec 12 '18 at 13:41
$begingroup$
@MostafaAyaz they are in R^n. OP writes |y|, but we understand than he/she means the norm of a vector.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Dec 12 '18 at 13:41
|
show 2 more comments
1 Answer
1
active
oldest
votes
$begingroup$
Even though you have $y ne z$, the statement is still incorrect. For a counterexample, take $n = 2$, and two intersecting circles centred at $y$ and $z$ with radii $r_y$ and $r_z$ respectively. Take $(x_k)_k$ to be the sequence oscillating between the two points of intersection. From the uniqueness of limit in Hausdorff spaces (e.g. $Bbb{R}^n$), we know that $(x_k)_k$ can't be convergent.
$$
%
defplace#1#2#3{smash{rlap{hskip{#1em}raise{#2em}{#3}}}}
%
hskip 1em
%
place{-3}{-0.5}{Hugebigcirc}
place{-3}{0}{bullet_{large y}}
place{-1}{-0.5}{Hugebigcirc}
place{0.5}{0}{bullet_{large z}}
$$
answered Dec 11 '18 at 11:10
GNUSupporter 8964民主女神 地下教會GNUSupporter 8964民主女神 地下教會
13.3k72549
$endgroup$
1
$begingroup$
agreed, if it converges it does in the intersection of two spheres but if they are not tangent it does not converges (but a sub sequence does)
$endgroup$
– Alexis
Dec 11 '18 at 11:14
$begingroup$
You're right, but what if, instead of just two points $y$ and $z$, we assume the following instead: $exists Ssubset mathbb{R}^n$ such that $forall sin S, lim|x_n-s|=r_s$ and int$Sneqemptyset$.
$endgroup$
– Tony S.F.
Dec 12 '18 at 11:14
$begingroup$
You may try to see if it's Cauchy. I think that's true, but I don't know how to prove that. Why not start a new question, so that others can upvote and answer your question.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Dec 12 '18 at 13:56
$begingroup$
I will, but comments are partially intended for this sort of informal brainstorming iirc. Its possible you could answer decisively with a short reason and then i could revise the question to a more meaningful form before posting again. This would avoid too many similar posts.
$endgroup$
– Tony S.F.
Dec 12 '18 at 19:00
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Even though you have $y ne z$, the statement is still incorrect. For a counterexample, take $n = 2$, and two intersecting circles centred at $y$ and $z$ with radii $r_y$ and $r_z$ respectively. Take $(x_k)_k$ to be the sequence oscillating between the two points of intersection. From the uniqueness of limit in Hausdorff spaces (e.g. $Bbb{R}^n$), we know that $(x_k)_k$ can't be convergent.
$$
%
defplace#1#2#3{smash{rlap{hskip{#1em}raise{#2em}{#3}}}}
%
hskip 1em
%
place{-3}{-0.5}{Hugebigcirc}
place{-3}{0}{bullet_{large y}}
place{-1}{-0.5}{Hugebigcirc}
place{0.5}{0}{bullet_{large z}}
$$
answered Dec 11 '18 at 11:10
GNUSupporter 8964民主女神 地下教會GNUSupporter 8964民主女神 地下教會
13.3k72549
$endgroup$
1
$begingroup$
agreed, if it converges it does in the intersection of two spheres but if they are not tangent it does not converges (but a sub sequence does)
$endgroup$
– Alexis
Dec 11 '18 at 11:14
$begingroup$
You're right, but what if, instead of just two points $y$ and $z$, we assume the following instead: $exists Ssubset mathbb{R}^n$ such that $forall sin S, lim|x_n-s|=r_s$ and int$Sneqemptyset$.
$endgroup$
– Tony S.F.
Dec 12 '18 at 11:14
$begingroup$
You may try to see if it's Cauchy. I think that's true, but I don't know how to prove that. Why not start a new question, so that others can upvote and answer your question.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Dec 12 '18 at 13:56
$begingroup$
I will, but comments are partially intended for this sort of informal brainstorming iirc. Its possible you could answer decisively with a short reason and then i could revise the question to a more meaningful form before posting again. This would avoid too many similar posts.
$endgroup$
– Tony S.F.
Dec 12 '18 at 19:00
add a comment |
$begingroup$
Even though you have $y ne z$, the statement is still incorrect. For a counterexample, take $n = 2$, and two intersecting circles centred at $y$ and $z$ with radii $r_y$ and $r_z$ respectively. Take $(x_k)_k$ to be the sequence oscillating between the two points of intersection. From the uniqueness of limit in Hausdorff spaces (e.g. $Bbb{R}^n$), we know that $(x_k)_k$ can't be convergent.
$$
%
defplace#1#2#3{smash{rlap{hskip{#1em}raise{#2em}{#3}}}}
%
hskip 1em
%
place{-3}{-0.5}{Hugebigcirc}
place{-3}{0}{bullet_{large y}}
place{-1}{-0.5}{Hugebigcirc}
place{0.5}{0}{bullet_{large z}}
$$
answered Dec 11 '18 at 11:10
GNUSupporter 8964民主女神 地下教會GNUSupporter 8964民主女神 地下教會
13.3k72549
$endgroup$
1
$begingroup$
agreed, if it converges it does in the intersection of two spheres but if they are not tangent it does not converges (but a sub sequence does)
$endgroup$
– Alexis
Dec 11 '18 at 11:14
$begingroup$
You're right, but what if, instead of just two points $y$ and $z$, we assume the following instead: $exists Ssubset mathbb{R}^n$ such that $forall sin S, lim|x_n-s|=r_s$ and int$Sneqemptyset$.
$endgroup$
– Tony S.F.
Dec 12 '18 at 11:14
$begingroup$
You may try to see if it's Cauchy. I think that's true, but I don't know how to prove that. Why not start a new question, so that others can upvote and answer your question.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Dec 12 '18 at 13:56
$begingroup$
I will, but comments are partially intended for this sort of informal brainstorming iirc. Its possible you could answer decisively with a short reason and then i could revise the question to a more meaningful form before posting again. This would avoid too many similar posts.
$endgroup$
– Tony S.F.
Dec 12 '18 at 19:00
add a comment |
$begingroup$
Even though you have $y ne z$, the statement is still incorrect. For a counterexample, take $n = 2$, and two intersecting circles centred at $y$ and $z$ with radii $r_y$ and $r_z$ respectively. Take $(x_k)_k$ to be the sequence oscillating between the two points of intersection. From the uniqueness of limit in Hausdorff spaces (e.g. $Bbb{R}^n$), we know that $(x_k)_k$ can't be convergent.
$$
%
defplace#1#2#3{smash{rlap{hskip{#1em}raise{#2em}{#3}}}}
%
hskip 1em
%
place{-3}{-0.5}{Hugebigcirc}
place{-3}{0}{bullet_{large y}}
place{-1}{-0.5}{Hugebigcirc}
place{0.5}{0}{bullet_{large z}}
$$
answered Dec 11 '18 at 11:10
GNUSupporter 8964民主女神 地下教會GNUSupporter 8964民主女神 地下教會
13.3k72549
$endgroup$
Even though you have $y ne z$, the statement is still incorrect. For a counterexample, take $n = 2$, and two intersecting circles centred at $y$ and $z$ with radii $r_y$ and $r_z$ respectively. Take $(x_k)_k$ to be the sequence oscillating between the two points of intersection. From the uniqueness of limit in Hausdorff spaces (e.g. $Bbb{R}^n$), we know that $(x_k)_k$ can't be convergent.
$$
%
defplace#1#2#3{smash{rlap{hskip{#1em}raise{#2em}{#3}}}}
%
hskip 1em
%
place{-3}{-0.5}{Hugebigcirc}
place{-3}{0}{bullet_{large y}}
place{-1}{-0.5}{Hugebigcirc}
place{0.5}{0}{bullet_{large z}}
$$
answered Dec 11 '18 at 11:10
GNUSupporter 8964民主女神 地下教會GNUSupporter 8964民主女神 地下教會
13.3k72549
edited Dec 11 '18 at 11:21
answered Dec 11 '18 at 11:10
GNUSupporter 8964民主女神 地下教會GNUSupporter 8964民主女神 地下教會
13.3k72549
answered Dec 11 '18 at 11:10
GNUSupporter 8964民主女神 地下教會GNUSupporter 8964民主女神 地下教會
13.3k72549
answered Dec 11 '18 at 11:10
GNUSupporter 8964民主女神 地下教會GNUSupporter 8964民主女神 地下教會
13.3k72549
13.3k72549
1
$begingroup$
agreed, if it converges it does in the intersection of two spheres but if they are not tangent it does not converges (but a sub sequence does)
$endgroup$
– Alexis
Dec 11 '18 at 11:14
$begingroup$
You're right, but what if, instead of just two points $y$ and $z$, we assume the following instead: $exists Ssubset mathbb{R}^n$ such that $forall sin S, lim|x_n-s|=r_s$ and int$Sneqemptyset$.
$endgroup$
– Tony S.F.
Dec 12 '18 at 11:14
$begingroup$
You may try to see if it's Cauchy. I think that's true, but I don't know how to prove that. Why not start a new question, so that others can upvote and answer your question.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Dec 12 '18 at 13:56
$begingroup$
I will, but comments are partially intended for this sort of informal brainstorming iirc. Its possible you could answer decisively with a short reason and then i could revise the question to a more meaningful form before posting again. This would avoid too many similar posts.
$endgroup$
– Tony S.F.
Dec 12 '18 at 19:00
add a comment |
1
$begingroup$
agreed, if it converges it does in the intersection of two spheres but if they are not tangent it does not converges (but a sub sequence does)
$endgroup$
– Alexis
Dec 11 '18 at 11:14
$begingroup$
You're right, but what if, instead of just two points $y$ and $z$, we assume the following instead: $exists Ssubset mathbb{R}^n$ such that $forall sin S, lim|x_n-s|=r_s$ and int$Sneqemptyset$.
$endgroup$
– Tony S.F.
Dec 12 '18 at 11:14
$begingroup$
You may try to see if it's Cauchy. I think that's true, but I don't know how to prove that. Why not start a new question, so that others can upvote and answer your question.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Dec 12 '18 at 13:56
$begingroup$
I will, but comments are partially intended for this sort of informal brainstorming iirc. Its possible you could answer decisively with a short reason and then i could revise the question to a more meaningful form before posting again. This would avoid too many similar posts.
$endgroup$
– Tony S.F.
Dec 12 '18 at 19:00
1
1
$begingroup$
agreed, if it converges it does in the intersection of two spheres but if they are not tangent it does not converges (but a sub sequence does)
$endgroup$
– Alexis
Dec 11 '18 at 11:14
$begingroup$
agreed, if it converges it does in the intersection of two spheres but if they are not tangent it does not converges (but a sub sequence does)
$endgroup$
– Alexis
Dec 11 '18 at 11:14
$begingroup$
You're right, but what if, instead of just two points $y$ and $z$, we assume the following instead: $exists Ssubset mathbb{R}^n$ such that $forall sin S, lim|x_n-s|=r_s$ and int$Sneqemptyset$.
$endgroup$
– Tony S.F.
Dec 12 '18 at 11:14
$begingroup$
You're right, but what if, instead of just two points $y$ and $z$, we assume the following instead: $exists Ssubset mathbb{R}^n$ such that $forall sin S, lim|x_n-s|=r_s$ and int$Sneqemptyset$.
$endgroup$
– Tony S.F.
Dec 12 '18 at 11:14
$begingroup$
You may try to see if it's Cauchy. I think that's true, but I don't know how to prove that. Why not start a new question, so that others can upvote and answer your question.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Dec 12 '18 at 13:56
$begingroup$
You may try to see if it's Cauchy. I think that's true, but I don't know how to prove that. Why not start a new question, so that others can upvote and answer your question.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Dec 12 '18 at 13:56
$begingroup$
I will, but comments are partially intended for this sort of informal brainstorming iirc. Its possible you could answer decisively with a short reason and then i could revise the question to a more meaningful form before posting again. This would avoid too many similar posts.
$endgroup$
– Tony S.F.
Dec 12 '18 at 19:00
$begingroup$
I will, but comments are partially intended for this sort of informal brainstorming iirc. Its possible you could answer decisively with a short reason and then i could revise the question to a more meaningful form before posting again. This would avoid too many similar posts.
$endgroup$
– Tony S.F.
Dec 12 '18 at 19:00
add a comment |
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1
$begingroup$
are there other conditions such as $y ne z$
$endgroup$
– Siong Thye Goh
Dec 11 '18 at 11:05
$begingroup$
Yeah $yneq z$ of course otherwise it's not true (circle)
$endgroup$
– Tony S.F.
Dec 11 '18 at 11:09
$begingroup$
Given the answer below, it seems to be true only for $n=1$.
$endgroup$
– daw
Dec 11 '18 at 12:15
$begingroup$
Shouldn't be $y,zin Bbb R$?
$endgroup$
– Mostafa Ayaz
Dec 11 '18 at 23:14
$begingroup$
@MostafaAyaz they are in R^n. OP writes |y|, but we understand than he/she means the norm of a vector.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Dec 12 '18 at 13:41