Sequence with anchoring property is convergent in finite dimension












1












$begingroup$


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







share|cite|improve this question









$endgroup$








  • 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


















1












$begingroup$


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







share|cite|improve this question









$endgroup$








  • 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
















1












1








1





$begingroup$


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







share|cite|improve this question









$endgroup$




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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










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
















  • 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












1 Answer
1






active

oldest

votes


















2












$begingroup$

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.



$$
%
defplace#1#2#3{smash{rlap{hskip{#1em}raise{#2em}{#3}}}}
%
hskip 1em
%
place{-3}{-0.5}{Hugebigcirc}
place{-3}{0}{bullet_{large y}}
place{-1}{-0.5}{Hugebigcirc}
place{0.5}{0}{bullet_{large z}}
$$






share|cite|improve this answer











$endgroup$









  • 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













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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

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.



$$
%
defplace#1#2#3{smash{rlap{hskip{#1em}raise{#2em}{#3}}}}
%
hskip 1em
%
place{-3}{-0.5}{Hugebigcirc}
place{-3}{0}{bullet_{large y}}
place{-1}{-0.5}{Hugebigcirc}
place{0.5}{0}{bullet_{large z}}
$$






share|cite|improve this answer











$endgroup$









  • 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


















2












$begingroup$

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.



$$
%
defplace#1#2#3{smash{rlap{hskip{#1em}raise{#2em}{#3}}}}
%
hskip 1em
%
place{-3}{-0.5}{Hugebigcirc}
place{-3}{0}{bullet_{large y}}
place{-1}{-0.5}{Hugebigcirc}
place{0.5}{0}{bullet_{large z}}
$$






share|cite|improve this answer











$endgroup$









  • 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
















2












2








2





$begingroup$

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.



$$
%
defplace#1#2#3{smash{rlap{hskip{#1em}raise{#2em}{#3}}}}
%
hskip 1em
%
place{-3}{-0.5}{Hugebigcirc}
place{-3}{0}{bullet_{large y}}
place{-1}{-0.5}{Hugebigcirc}
place{0.5}{0}{bullet_{large z}}
$$






share|cite|improve this answer











$endgroup$



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.



$$
%
defplace#1#2#3{smash{rlap{hskip{#1em}raise{#2em}{#3}}}}
%
hskip 1em
%
place{-3}{-0.5}{Hugebigcirc}
place{-3}{0}{bullet_{large y}}
place{-1}{-0.5}{Hugebigcirc}
place{0.5}{0}{bullet_{large z}}
$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








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




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
















  • 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




















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Puebla de Zaragoza

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Para otros usos de este término, véase Puebla (desambiguación). Heroica Puebla de Zaragoza Localidad De arriba a abajo de izquierda a derecha: Estrella de Puebla, Volcán Popocatépetl, Catedral de Puebla, Monumento a Ignacio Zaragoza, Estadio Cuauhtémoc, Angelópolis y Vista panorámica de la ciudad. Escudo Otros nombres: El relicario de América/ La ciudad de los Ángeles/ La Angelópolis Lema: Muy noble y muy leal ciudad de Puebla de los Ángeles Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en México Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en Puebla Mapa interactivo Coordenadas 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Coordenadas: 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Idioma oficial Español Entidad Localidad  • País   México  • Estado Puebla  • Municipio Puebla Presidente municipal Claudia Rivera Vivanc...
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Musa

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Para el género de plantas, véase Musa (planta). Apolo y la Nueve Musas En la mitología griega, las musas (en griego antiguo μοῦσαι « mousai» ) son, según los escritores más antiguos, las divinidades inspiradoras de las Artes: cada una de ellas está relacionada con ramas artísticas y del conocimiento. Son hijas de Zeus y de Mnemósine, compañeras del séquito de Apolo, dios olímpico de la música y patrón de las bellas artes, quien tuvo romances con cada una de ellas, dejando descendientes. Bajaban a la tierra a susurrar ideas e inspirar a aquellos mortales que las invocaran. En la época más arcaica eran las ninfas inspiradoras de las fuentes, en las cuales eran adoradas. Finalmente, alrededor de los siglos VIII-VII a. C. [ 1 ] ​ prevaleció en todo el territorio de la Hélade la adoración de las nueve Musas, que son Calíope, Clío, Erato, Euterpe, Melpómene, Polimnia, Talía, Terpsícore y Urania. El culto a las musas era originalmente de Tracia y Beocia, y fueron de vital impo...
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