A $delta$- chain is $delta$ shadowed by identity map
$begingroup$
Let $f$ be the identity map on the space $X={x_i}_{i=1}^{infty} $, where $x_i=sum_{n=1}^ifrac{1}{n}$, given the metric inherited from $mathbb{R}$. Also, let for finite sequence ${y_n}_{n=0}^msubseteq X$, we have $|y_0-y_m|<delta$ and $|y_n-y_{n+1}|<delta$ for $n=0, 1, ldots m-1$.
Question. Is it true there is $zin{y_n}_{n=0}^m$ such that $|z-y_n|<delta$ for all $n=0, 1, ldots, m$?
real-analysis calculus analysis
$endgroup$
add a comment |
$begingroup$
Let $f$ be the identity map on the space $X={x_i}_{i=1}^{infty} $, where $x_i=sum_{n=1}^ifrac{1}{n}$, given the metric inherited from $mathbb{R}$. Also, let for finite sequence ${y_n}_{n=0}^msubseteq X$, we have $|y_0-y_m|<delta$ and $|y_n-y_{n+1}|<delta$ for $n=0, 1, ldots m-1$.
Question. Is it true there is $zin{y_n}_{n=0}^m$ such that $|z-y_n|<delta$ for all $n=0, 1, ldots, m$?
real-analysis calculus analysis
$endgroup$
$begingroup$
No, it is not true. You can go as far away as you want and then come back. Why do you think this should be true? Maybe you want to impose further conditions?
$endgroup$
– Severin Schraven
Dec 2 '18 at 18:44
$begingroup$
@S.Schraven. There is no further conditions. give an example please , if $delta>0$ is sufficiently small
$endgroup$
– user479859
Dec 3 '18 at 0:00
$begingroup$
Only I can choice $zin X$, hence
$endgroup$
– user479859
Dec 3 '18 at 0:34
$begingroup$
Is it true there is $zin X$ such that $|z-y_n|<delta$ for all $n=0, 1, ldots, m$?
$endgroup$
– user479859
Dec 3 '18 at 0:34
add a comment |
$begingroup$
Let $f$ be the identity map on the space $X={x_i}_{i=1}^{infty} $, where $x_i=sum_{n=1}^ifrac{1}{n}$, given the metric inherited from $mathbb{R}$. Also, let for finite sequence ${y_n}_{n=0}^msubseteq X$, we have $|y_0-y_m|<delta$ and $|y_n-y_{n+1}|<delta$ for $n=0, 1, ldots m-1$.
Question. Is it true there is $zin{y_n}_{n=0}^m$ such that $|z-y_n|<delta$ for all $n=0, 1, ldots, m$?
real-analysis calculus analysis
$endgroup$
Let $f$ be the identity map on the space $X={x_i}_{i=1}^{infty} $, where $x_i=sum_{n=1}^ifrac{1}{n}$, given the metric inherited from $mathbb{R}$. Also, let for finite sequence ${y_n}_{n=0}^msubseteq X$, we have $|y_0-y_m|<delta$ and $|y_n-y_{n+1}|<delta$ for $n=0, 1, ldots m-1$.
Question. Is it true there is $zin{y_n}_{n=0}^m$ such that $|z-y_n|<delta$ for all $n=0, 1, ldots, m$?
real-analysis calculus analysis
real-analysis calculus analysis
asked Dec 2 '18 at 18:27
user479859user479859
756
756
$begingroup$
No, it is not true. You can go as far away as you want and then come back. Why do you think this should be true? Maybe you want to impose further conditions?
$endgroup$
– Severin Schraven
Dec 2 '18 at 18:44
$begingroup$
@S.Schraven. There is no further conditions. give an example please , if $delta>0$ is sufficiently small
$endgroup$
– user479859
Dec 3 '18 at 0:00
$begingroup$
Only I can choice $zin X$, hence
$endgroup$
– user479859
Dec 3 '18 at 0:34
$begingroup$
Is it true there is $zin X$ such that $|z-y_n|<delta$ for all $n=0, 1, ldots, m$?
$endgroup$
– user479859
Dec 3 '18 at 0:34
add a comment |
$begingroup$
No, it is not true. You can go as far away as you want and then come back. Why do you think this should be true? Maybe you want to impose further conditions?
$endgroup$
– Severin Schraven
Dec 2 '18 at 18:44
$begingroup$
@S.Schraven. There is no further conditions. give an example please , if $delta>0$ is sufficiently small
$endgroup$
– user479859
Dec 3 '18 at 0:00
$begingroup$
Only I can choice $zin X$, hence
$endgroup$
– user479859
Dec 3 '18 at 0:34
$begingroup$
Is it true there is $zin X$ such that $|z-y_n|<delta$ for all $n=0, 1, ldots, m$?
$endgroup$
– user479859
Dec 3 '18 at 0:34
$begingroup$
No, it is not true. You can go as far away as you want and then come back. Why do you think this should be true? Maybe you want to impose further conditions?
$endgroup$
– Severin Schraven
Dec 2 '18 at 18:44
$begingroup$
No, it is not true. You can go as far away as you want and then come back. Why do you think this should be true? Maybe you want to impose further conditions?
$endgroup$
– Severin Schraven
Dec 2 '18 at 18:44
$begingroup$
@S.Schraven. There is no further conditions. give an example please , if $delta>0$ is sufficiently small
$endgroup$
– user479859
Dec 3 '18 at 0:00
$begingroup$
@S.Schraven. There is no further conditions. give an example please , if $delta>0$ is sufficiently small
$endgroup$
– user479859
Dec 3 '18 at 0:00
$begingroup$
Only I can choice $zin X$, hence
$endgroup$
– user479859
Dec 3 '18 at 0:34
$begingroup$
Only I can choice $zin X$, hence
$endgroup$
– user479859
Dec 3 '18 at 0:34
$begingroup$
Is it true there is $zin X$ such that $|z-y_n|<delta$ for all $n=0, 1, ldots, m$?
$endgroup$
– user479859
Dec 3 '18 at 0:34
$begingroup$
Is it true there is $zin X$ such that $|z-y_n|<delta$ for all $n=0, 1, ldots, m$?
$endgroup$
– user479859
Dec 3 '18 at 0:34
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Choose $delta >0$ in any way you like. Pick $N>0$ sufficiently large s.t. $1/N<delta$. Then consider the sequence $y_0=x_N=y_{2l}, y_1=x_{N+1}=y_{2l-1}, dots, y_{l}=x_{N+l}$. Clearly $vert y_0 -y_{2l} vert =0<delta$ and $vert y_n - y_{n+1} vert leq 1/N <delta $, but $vert y_0 - y_l vert = sum_{k=N+1}^{N+l} 1/k$ which we can make as large as we want by making $l$ large.
Note that in fact there does not even exist $zin X$ (i.e. if we are not forced to pick $zin (y_m)_{m=0}^{2l})$ such that $vert z - y_m vert<delta$ for all $m=0, dots , 2l$.
$endgroup$
add a comment |
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
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3023014%2fa-delta-chain-is-delta-shadowed-by-identity-map%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
$begingroup$
Choose $delta >0$ in any way you like. Pick $N>0$ sufficiently large s.t. $1/N<delta$. Then consider the sequence $y_0=x_N=y_{2l}, y_1=x_{N+1}=y_{2l-1}, dots, y_{l}=x_{N+l}$. Clearly $vert y_0 -y_{2l} vert =0<delta$ and $vert y_n - y_{n+1} vert leq 1/N <delta $, but $vert y_0 - y_l vert = sum_{k=N+1}^{N+l} 1/k$ which we can make as large as we want by making $l$ large.
Note that in fact there does not even exist $zin X$ (i.e. if we are not forced to pick $zin (y_m)_{m=0}^{2l})$ such that $vert z - y_m vert<delta$ for all $m=0, dots , 2l$.
$endgroup$
add a comment |
$begingroup$
Choose $delta >0$ in any way you like. Pick $N>0$ sufficiently large s.t. $1/N<delta$. Then consider the sequence $y_0=x_N=y_{2l}, y_1=x_{N+1}=y_{2l-1}, dots, y_{l}=x_{N+l}$. Clearly $vert y_0 -y_{2l} vert =0<delta$ and $vert y_n - y_{n+1} vert leq 1/N <delta $, but $vert y_0 - y_l vert = sum_{k=N+1}^{N+l} 1/k$ which we can make as large as we want by making $l$ large.
Note that in fact there does not even exist $zin X$ (i.e. if we are not forced to pick $zin (y_m)_{m=0}^{2l})$ such that $vert z - y_m vert<delta$ for all $m=0, dots , 2l$.
$endgroup$
add a comment |
$begingroup$
Choose $delta >0$ in any way you like. Pick $N>0$ sufficiently large s.t. $1/N<delta$. Then consider the sequence $y_0=x_N=y_{2l}, y_1=x_{N+1}=y_{2l-1}, dots, y_{l}=x_{N+l}$. Clearly $vert y_0 -y_{2l} vert =0<delta$ and $vert y_n - y_{n+1} vert leq 1/N <delta $, but $vert y_0 - y_l vert = sum_{k=N+1}^{N+l} 1/k$ which we can make as large as we want by making $l$ large.
Note that in fact there does not even exist $zin X$ (i.e. if we are not forced to pick $zin (y_m)_{m=0}^{2l})$ such that $vert z - y_m vert<delta$ for all $m=0, dots , 2l$.
$endgroup$
Choose $delta >0$ in any way you like. Pick $N>0$ sufficiently large s.t. $1/N<delta$. Then consider the sequence $y_0=x_N=y_{2l}, y_1=x_{N+1}=y_{2l-1}, dots, y_{l}=x_{N+l}$. Clearly $vert y_0 -y_{2l} vert =0<delta$ and $vert y_n - y_{n+1} vert leq 1/N <delta $, but $vert y_0 - y_l vert = sum_{k=N+1}^{N+l} 1/k$ which we can make as large as we want by making $l$ large.
Note that in fact there does not even exist $zin X$ (i.e. if we are not forced to pick $zin (y_m)_{m=0}^{2l})$ such that $vert z - y_m vert<delta$ for all $m=0, dots , 2l$.
edited Dec 4 '18 at 7:20
answered Dec 3 '18 at 14:23
Severin SchravenSeverin Schraven
6,0081934
6,0081934
add a comment |
add a comment |
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.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3023014%2fa-delta-chain-is-delta-shadowed-by-identity-map%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
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
$begingroup$
No, it is not true. You can go as far away as you want and then come back. Why do you think this should be true? Maybe you want to impose further conditions?
$endgroup$
– Severin Schraven
Dec 2 '18 at 18:44
$begingroup$
@S.Schraven. There is no further conditions. give an example please , if $delta>0$ is sufficiently small
$endgroup$
– user479859
Dec 3 '18 at 0:00
$begingroup$
Only I can choice $zin X$, hence
$endgroup$
– user479859
Dec 3 '18 at 0:34
$begingroup$
Is it true there is $zin X$ such that $|z-y_n|<delta$ for all $n=0, 1, ldots, m$?
$endgroup$
– user479859
Dec 3 '18 at 0:34