For $x_nto x$, is there homeomorphisms $f_n$ with $d(f_n(x_n), f_n(x))>epsilon$ ? ($forall ninmathbb{N}$)












-1












$begingroup$


Let $(X, d)$ be a compact metric space and $x_nto x$ as $nto infty$.



Let $epsilon>0$ be given. Is there a sequence of homeomorphisms $f_n:Xto X$ such that $d(f_n(x_n), f_n(x))>epsilon$ for some $epsilon>0$? (for all $ninmathbb{N}$)






functional-analysis analysis







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$endgroup$












  • $begingroup$
    Obviously false if $x_n=x$ for all $n$.
    $endgroup$
    – Kavi Rama Murthy
    Jan 8 at 23:25
















-1












$begingroup$


Let $(X, d)$ be a compact metric space and $x_nto x$ as $nto infty$.



Let $epsilon>0$ be given. Is there a sequence of homeomorphisms $f_n:Xto X$ such that $d(f_n(x_n), f_n(x))>epsilon$ for some $epsilon>0$? (for all $ninmathbb{N}$)






functional-analysis analysis







share|cite|improve this question









$endgroup$












  • $begingroup$
    Obviously false if $x_n=x$ for all $n$.
    $endgroup$
    – Kavi Rama Murthy
    Jan 8 at 23:25














-1












-1








-1





$begingroup$


Let $(X, d)$ be a compact metric space and $x_nto x$ as $nto infty$.



Let $epsilon>0$ be given. Is there a sequence of homeomorphisms $f_n:Xto X$ such that $d(f_n(x_n), f_n(x))>epsilon$ for some $epsilon>0$? (for all $ninmathbb{N}$)






functional-analysis analysis







share|cite|improve this question









$endgroup$




Let $(X, d)$ be a compact metric space and $x_nto x$ as $nto infty$.



Let $epsilon>0$ be given. Is there a sequence of homeomorphisms $f_n:Xto X$ such that $d(f_n(x_n), f_n(x))>epsilon$ for some $epsilon>0$? (for all $ninmathbb{N}$)






functional-analysis analysis




functional-analysis analysis






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 8 at 22:50









user479859user479859

806




806












  • $begingroup$
    Obviously false if $x_n=x$ for all $n$.
    $endgroup$
    – Kavi Rama Murthy
    Jan 8 at 23:25


















  • $begingroup$
    Obviously false if $x_n=x$ for all $n$.
    $endgroup$
    – Kavi Rama Murthy
    Jan 8 at 23:25
















$begingroup$
Obviously false if $x_n=x$ for all $n$.
$endgroup$
– Kavi Rama Murthy
Jan 8 at 23:25




$begingroup$
Obviously false if $x_n=x$ for all $n$.
$endgroup$
– Kavi Rama Murthy
Jan 8 at 23:25










1 Answer
1






active

oldest

votes


















1












$begingroup$

Not necessarily. Take $X=bigcup_{n geq 0}{S^n}$ where the union is disjoint and $S^0$ is a singleton ${x^0}$.



The distance between any $x,y in S^n$ when $n geq 1$ is $frac{1}{10n^3}frac{1}{1+|x-y|_{mathbb{R}^{n+1}}}$. The distance between $x in S^n$, $y in S^m$, $m > n > 0$, is $frac{1}{n}-frac{1}{m}$, and the distance between $x in S^n$ and $x^0$ is $1/n$.



Take for every $n geq 1$ $x_n in S^n$. Then $x_n$ converges to $x^0$, and, for every homeomorphism $f$ of $X$ into itself, $f(x_n) in S^n$ so is at distance $1/n$ from $x^0$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks. In my research, I obtain a sequence homeomorphisms $f_n:Xto X$ such that for all $nin mathbb{N}$, $d(f_n(x_n), f_n(x))>epsilon$. Can I say that it is a contradiction?($X$ is a compact metric space)
    $endgroup$
    – user479859
    Jan 9 at 4:26










  • $begingroup$
    As is, no. Maybe though you can have some more information about $X$, the $f_n$’s qnd the $x_n$’s?
    $endgroup$
    – Mindlack
    Jan 9 at 9:02










  • $begingroup$
    In my research, there is a sequence ${g_k:Xto X}$, such that $d_0(g_k, f)<frac{1}{k}$ and for every $kin mathbb{N}$ there is $n(k)inmathbb{N}$ such that $d(g^{n(k)}_k(x_k), g^{n(k)}_k(x))>epsilon$. Note that $x_kto x$ as $kto infty$, $d_0(g, f)= sup_{xin X} d(f(x), g(x))$ and $f^n=fcirc fcircldots circ f$ ($n$- times).
    $endgroup$
    – user479859
    Jan 9 at 9:39











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1 Answer
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active

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1 Answer
1






active

oldest

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active

oldest

votes






active

oldest

votes









1












$begingroup$

Not necessarily. Take $X=bigcup_{n geq 0}{S^n}$ where the union is disjoint and $S^0$ is a singleton ${x^0}$.



The distance between any $x,y in S^n$ when $n geq 1$ is $frac{1}{10n^3}frac{1}{1+|x-y|_{mathbb{R}^{n+1}}}$. The distance between $x in S^n$, $y in S^m$, $m > n > 0$, is $frac{1}{n}-frac{1}{m}$, and the distance between $x in S^n$ and $x^0$ is $1/n$.



Take for every $n geq 1$ $x_n in S^n$. Then $x_n$ converges to $x^0$, and, for every homeomorphism $f$ of $X$ into itself, $f(x_n) in S^n$ so is at distance $1/n$ from $x^0$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks. In my research, I obtain a sequence homeomorphisms $f_n:Xto X$ such that for all $nin mathbb{N}$, $d(f_n(x_n), f_n(x))>epsilon$. Can I say that it is a contradiction?($X$ is a compact metric space)
    $endgroup$
    – user479859
    Jan 9 at 4:26










  • $begingroup$
    As is, no. Maybe though you can have some more information about $X$, the $f_n$’s qnd the $x_n$’s?
    $endgroup$
    – Mindlack
    Jan 9 at 9:02










  • $begingroup$
    In my research, there is a sequence ${g_k:Xto X}$, such that $d_0(g_k, f)<frac{1}{k}$ and for every $kin mathbb{N}$ there is $n(k)inmathbb{N}$ such that $d(g^{n(k)}_k(x_k), g^{n(k)}_k(x))>epsilon$. Note that $x_kto x$ as $kto infty$, $d_0(g, f)= sup_{xin X} d(f(x), g(x))$ and $f^n=fcirc fcircldots circ f$ ($n$- times).
    $endgroup$
    – user479859
    Jan 9 at 9:39
















1












$begingroup$

Not necessarily. Take $X=bigcup_{n geq 0}{S^n}$ where the union is disjoint and $S^0$ is a singleton ${x^0}$.



The distance between any $x,y in S^n$ when $n geq 1$ is $frac{1}{10n^3}frac{1}{1+|x-y|_{mathbb{R}^{n+1}}}$. The distance between $x in S^n$, $y in S^m$, $m > n > 0$, is $frac{1}{n}-frac{1}{m}$, and the distance between $x in S^n$ and $x^0$ is $1/n$.



Take for every $n geq 1$ $x_n in S^n$. Then $x_n$ converges to $x^0$, and, for every homeomorphism $f$ of $X$ into itself, $f(x_n) in S^n$ so is at distance $1/n$ from $x^0$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks. In my research, I obtain a sequence homeomorphisms $f_n:Xto X$ such that for all $nin mathbb{N}$, $d(f_n(x_n), f_n(x))>epsilon$. Can I say that it is a contradiction?($X$ is a compact metric space)
    $endgroup$
    – user479859
    Jan 9 at 4:26










  • $begingroup$
    As is, no. Maybe though you can have some more information about $X$, the $f_n$’s qnd the $x_n$’s?
    $endgroup$
    – Mindlack
    Jan 9 at 9:02










  • $begingroup$
    In my research, there is a sequence ${g_k:Xto X}$, such that $d_0(g_k, f)<frac{1}{k}$ and for every $kin mathbb{N}$ there is $n(k)inmathbb{N}$ such that $d(g^{n(k)}_k(x_k), g^{n(k)}_k(x))>epsilon$. Note that $x_kto x$ as $kto infty$, $d_0(g, f)= sup_{xin X} d(f(x), g(x))$ and $f^n=fcirc fcircldots circ f$ ($n$- times).
    $endgroup$
    – user479859
    Jan 9 at 9:39














1












1








1





$begingroup$

Not necessarily. Take $X=bigcup_{n geq 0}{S^n}$ where the union is disjoint and $S^0$ is a singleton ${x^0}$.



The distance between any $x,y in S^n$ when $n geq 1$ is $frac{1}{10n^3}frac{1}{1+|x-y|_{mathbb{R}^{n+1}}}$. The distance between $x in S^n$, $y in S^m$, $m > n > 0$, is $frac{1}{n}-frac{1}{m}$, and the distance between $x in S^n$ and $x^0$ is $1/n$.



Take for every $n geq 1$ $x_n in S^n$. Then $x_n$ converges to $x^0$, and, for every homeomorphism $f$ of $X$ into itself, $f(x_n) in S^n$ so is at distance $1/n$ from $x^0$.






share|cite|improve this answer











$endgroup$



Not necessarily. Take $X=bigcup_{n geq 0}{S^n}$ where the union is disjoint and $S^0$ is a singleton ${x^0}$.



The distance between any $x,y in S^n$ when $n geq 1$ is $frac{1}{10n^3}frac{1}{1+|x-y|_{mathbb{R}^{n+1}}}$. The distance between $x in S^n$, $y in S^m$, $m > n > 0$, is $frac{1}{n}-frac{1}{m}$, and the distance between $x in S^n$ and $x^0$ is $1/n$.



Take for every $n geq 1$ $x_n in S^n$. Then $x_n$ converges to $x^0$, and, for every homeomorphism $f$ of $X$ into itself, $f(x_n) in S^n$ so is at distance $1/n$ from $x^0$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 8 at 23:20

























answered Jan 8 at 23:10









MindlackMindlack

4,740210




4,740210












  • $begingroup$
    Thanks. In my research, I obtain a sequence homeomorphisms $f_n:Xto X$ such that for all $nin mathbb{N}$, $d(f_n(x_n), f_n(x))>epsilon$. Can I say that it is a contradiction?($X$ is a compact metric space)
    $endgroup$
    – user479859
    Jan 9 at 4:26










  • $begingroup$
    As is, no. Maybe though you can have some more information about $X$, the $f_n$’s qnd the $x_n$’s?
    $endgroup$
    – Mindlack
    Jan 9 at 9:02










  • $begingroup$
    In my research, there is a sequence ${g_k:Xto X}$, such that $d_0(g_k, f)<frac{1}{k}$ and for every $kin mathbb{N}$ there is $n(k)inmathbb{N}$ such that $d(g^{n(k)}_k(x_k), g^{n(k)}_k(x))>epsilon$. Note that $x_kto x$ as $kto infty$, $d_0(g, f)= sup_{xin X} d(f(x), g(x))$ and $f^n=fcirc fcircldots circ f$ ($n$- times).
    $endgroup$
    – user479859
    Jan 9 at 9:39


















  • $begingroup$
    Thanks. In my research, I obtain a sequence homeomorphisms $f_n:Xto X$ such that for all $nin mathbb{N}$, $d(f_n(x_n), f_n(x))>epsilon$. Can I say that it is a contradiction?($X$ is a compact metric space)
    $endgroup$
    – user479859
    Jan 9 at 4:26










  • $begingroup$
    As is, no. Maybe though you can have some more information about $X$, the $f_n$’s qnd the $x_n$’s?
    $endgroup$
    – Mindlack
    Jan 9 at 9:02










  • $begingroup$
    In my research, there is a sequence ${g_k:Xto X}$, such that $d_0(g_k, f)<frac{1}{k}$ and for every $kin mathbb{N}$ there is $n(k)inmathbb{N}$ such that $d(g^{n(k)}_k(x_k), g^{n(k)}_k(x))>epsilon$. Note that $x_kto x$ as $kto infty$, $d_0(g, f)= sup_{xin X} d(f(x), g(x))$ and $f^n=fcirc fcircldots circ f$ ($n$- times).
    $endgroup$
    – user479859
    Jan 9 at 9:39
















$begingroup$
Thanks. In my research, I obtain a sequence homeomorphisms $f_n:Xto X$ such that for all $nin mathbb{N}$, $d(f_n(x_n), f_n(x))>epsilon$. Can I say that it is a contradiction?($X$ is a compact metric space)
$endgroup$
– user479859
Jan 9 at 4:26




$begingroup$
Thanks. In my research, I obtain a sequence homeomorphisms $f_n:Xto X$ such that for all $nin mathbb{N}$, $d(f_n(x_n), f_n(x))>epsilon$. Can I say that it is a contradiction?($X$ is a compact metric space)
$endgroup$
– user479859
Jan 9 at 4:26












$begingroup$
As is, no. Maybe though you can have some more information about $X$, the $f_n$’s qnd the $x_n$’s?
$endgroup$
– Mindlack
Jan 9 at 9:02




$begingroup$
As is, no. Maybe though you can have some more information about $X$, the $f_n$’s qnd the $x_n$’s?
$endgroup$
– Mindlack
Jan 9 at 9:02












$begingroup$
In my research, there is a sequence ${g_k:Xto X}$, such that $d_0(g_k, f)<frac{1}{k}$ and for every $kin mathbb{N}$ there is $n(k)inmathbb{N}$ such that $d(g^{n(k)}_k(x_k), g^{n(k)}_k(x))>epsilon$. Note that $x_kto x$ as $kto infty$, $d_0(g, f)= sup_{xin X} d(f(x), g(x))$ and $f^n=fcirc fcircldots circ f$ ($n$- times).
$endgroup$
– user479859
Jan 9 at 9:39




$begingroup$
In my research, there is a sequence ${g_k:Xto X}$, such that $d_0(g_k, f)<frac{1}{k}$ and for every $kin mathbb{N}$ there is $n(k)inmathbb{N}$ such that $d(g^{n(k)}_k(x_k), g^{n(k)}_k(x))>epsilon$. Note that $x_kto x$ as $kto infty$, $d_0(g, f)= sup_{xin X} d(f(x), g(x))$ and $f^n=fcirc fcircldots circ f$ ($n$- times).
$endgroup$
– user479859
Jan 9 at 9:39


















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