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







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












$begingroup$


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

987




987












  • $begingroup$
    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












Your Answer





StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3066832%2ffor-x-n-to-x-is-there-homeomorphisms-f-n-with-df-nx-n-f-nx-epsilon%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









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

4,900211




4,900211












  • $begingroup$
    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


















draft saved

draft discarded




















































Thanks for contributing an answer to Mathematics Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid …



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3066832%2ffor-x-n-to-x-is-there-homeomorphisms-f-n-with-df-nx-n-f-nx-epsilon%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







-

Popular posts from this blog

Bressuire

Image
.mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid ...
Read more

Cabo Verde

Image
  Nota: Cabo-verdiano redireciona para este artigo. Para a língua falada em Cabo Verde, veja crioulo cabo-verdiano. Para outros significados, veja Cabo Verde (desambiguação). Coordenadas: 14° 55' N 23° 31' O Cabo Verde República de Cabo Verde Bandeira Brasão de Armas Lema: Unidade, Trabalho, Progresso Hino nacional: Cântico da Liberdade Gentílico: cabo-verdiano [ 1 ] Capital Praia 14° 55' N 23° 31' O Cidade mais populosa Praia Língua oficial Português Governo República unitária semipresidencialista [ 2 ]  - Presidente Jorge Carlos Fonseca  - Primeiro-ministro Ulisses Correia e Silva História    - Descoberta 1460   - Independência 00 • de Portugal 5 de Julho de 1975   - Multipartidarismo 13 de Janeiro de 1990  Área    - Total 4 033 km² (146.º)  - Água (%) 0,08 População    - Estimativa para 2017 560 899 [ 3 ] hab. (173.º) ...
Read more

Gyllenstierna

Image
.mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid ...
Read more