Sequences that converge to the same pointCauchy Sequences and Analysistopology - triangle...
Brexit - No Deal Rejection
Describing a chess game in a novel
Why one should not leave fingerprints on bulbs and plugs?
How difficult is it to simply disable/disengage the MCAS on Boeing 737 Max 8 & 9 Aircraft?
Why does overlay work only on the first tcolorbox?
Is there a hypothetical scenario that would make Earth uninhabitable for humans, but not for (the majority of) other animals?
Why does a Star of David appear at a rally with Francisco Franco?
How do you talk to someone whose loved one is dying?
Non-trivial topology where only open sets are closed
What is a ^ b and (a & b) << 1?
What is the Japanese sound word for the clinking of money?
Why is the President allowed to veto a cancellation of emergency powers?
Is there a place to find the pricing for things not mentioned in the PHB? (non-magical)
Examples of transfinite towers
Why do tuner card drivers fail to build after kernel update to 4.4.0-143-generic?
What is "focus distance lower/upper" and how is it different from depth of field?
If I am holding an item before I cast Blink, will it move with me through the Ethereal Plane?
How to terminate ping <dest> &
Do the common programs (for example: "ls", "cat") in Linux and BSD come from the same source code?
Why is a white electrical wire connected to 2 black wires?
A single argument pattern definition applies to multiple-argument patterns?
ERC721: How to get the owned tokens of an address
How to pronounce "I ♥ Huckabees"?
Have the tides ever turned twice on any open problem?
Sequences that converge to the same point
Cauchy Sequences and Analysistopology - triangle inequalityConvergence and metric - Proof?Complete metric on the space of sequencesConverge in mean square of sum of random variablesProve that if two sequences converge to the same value but their image sequences converge to different values, then the limit DNEProve that either both sequences converge to the same limit or both diverge.Completion of metric space (via formal limits of Cauchy sequences)Proving completeness of space of converging sequencesFind non convergent Cauchy sequence
$begingroup$
In $(X,d)$ metric space, If ${x_n}_{ninBbb N}$ and ${y_n}_{ninBbb N}$ are two sequences which converge to the same point, then prove that
$$lim_{ntoinfty} d(x_n,y_n)=0$$
This question is found in the chapter of completeness, but I could not figure out where to use completeness for this question.
convergence metric-spaces complete-spaces
asked May 11 '17 at 13:14
Jonathan S.Jonathan S.
82
$endgroup$
add a comment |
$begingroup$
In $(X,d)$ metric space, If ${x_n}_{ninBbb N}$ and ${y_n}_{ninBbb N}$ are two sequences which converge to the same point, then prove that
$$lim_{ntoinfty} d(x_n,y_n)=0$$
This question is found in the chapter of completeness, but I could not figure out where to use completeness for this question.
convergence metric-spaces complete-spaces
asked May 11 '17 at 13:14
Jonathan S.Jonathan S.
82
$endgroup$
$begingroup$
You do not need to use completeness in any shape or form. The result is true regardless of even mentioning it.
$endgroup$
– user228113
May 11 '17 at 13:27
add a comment |
$begingroup$
In $(X,d)$ metric space, If ${x_n}_{ninBbb N}$ and ${y_n}_{ninBbb N}$ are two sequences which converge to the same point, then prove that
$$lim_{ntoinfty} d(x_n,y_n)=0$$
This question is found in the chapter of completeness, but I could not figure out where to use completeness for this question.
convergence metric-spaces complete-spaces
asked May 11 '17 at 13:14
Jonathan S.Jonathan S.
82
$endgroup$
In $(X,d)$ metric space, If ${x_n}_{ninBbb N}$ and ${y_n}_{ninBbb N}$ are two sequences which converge to the same point, then prove that
$$lim_{ntoinfty} d(x_n,y_n)=0$$
This question is found in the chapter of completeness, but I could not figure out where to use completeness for this question.
convergence metric-spaces complete-spaces
convergence metric-spaces complete-spaces
asked May 11 '17 at 13:14
Jonathan S.Jonathan S.
82
asked May 11 '17 at 13:14
Jonathan S.Jonathan S.
82
edited May 11 '17 at 13:25
user228113
asked May 11 '17 at 13:14
Jonathan S.Jonathan S.
82
asked May 11 '17 at 13:14
Jonathan S.Jonathan S.
82
asked May 11 '17 at 13:14
Jonathan S.Jonathan S.
82
82
$begingroup$
You do not need to use completeness in any shape or form. The result is true regardless of even mentioning it.
$endgroup$
– user228113
May 11 '17 at 13:27
add a comment |
$begingroup$
You do not need to use completeness in any shape or form. The result is true regardless of even mentioning it.
$endgroup$
– user228113
May 11 '17 at 13:27
$begingroup$
You do not need to use completeness in any shape or form. The result is true regardless of even mentioning it.
$endgroup$
– user228113
May 11 '17 at 13:27
$begingroup$
You do not need to use completeness in any shape or form. The result is true regardless of even mentioning it.
$endgroup$
– user228113
May 11 '17 at 13:27
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
let $X_nrightarrow X$ thus $Y_n$ converges to Y as well. Given $epsilon>0$, $exists min N$ such that $forall n>m$ $d(X_n,X)<epsilon /2$ and $d(Y_n,X)<epsilon /2$ Then $forall n>m $ we have $d(X_n,Y_n)<d(X-n,X)+d(Y_n,X)<epsilon$. Thus $d(X_n,Y_n)rightarrow 0$
$endgroup$
add a comment |
$begingroup$
Hint:
If $x$ is the limit of $x_n$ and $y$ is the limit of $y_n$, then
$$d(x_n, y_n) leq d(x_n, x) + d(x, y) + d(y, y_n)$$
You can use this to prove (using the $epsilon$ -definition) that $$lim_{ntoinfty} d(x_n, y_n) = 0$$
answered May 11 '17 at 13:16
5xum5xum
91.4k394161
$endgroup$
add a comment |
$begingroup$
Let $l$ be the limit of the sequences $(x_{n}), (y_{n})$ by definition Note that $d(x_{n},y_{n}) leq d(x_{n},l) + d(l, y_{n})$ for all $n$. If $varepsilon > 0$, then there are $N_{1},N_{2}$, by assumption, giving $d(x_{n},l), d(l,y_{n}) < varepsilon/2$ for all $n geq N_{1}$ and $geq N_{2}$. So $d(x_{n},l) + d(l,y_{n}) < varepsilon$ for all $n geq max { N_{1},N_{2} }$; so $d(x_{n},y_{n}) < varepsilon$ for all $n geq max { N_{1},N_{2} }$.
answered May 11 '17 at 13:37
Gary MooreGary Moore
17.3k21546
$endgroup$
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%2f2276339%2fsequences-that-converge-to-the-same-point%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
let $X_nrightarrow X$ thus $Y_n$ converges to Y as well. Given $epsilon>0$, $exists min N$ such that $forall n>m$ $d(X_n,X)<epsilon /2$ and $d(Y_n,X)<epsilon /2$ Then $forall n>m $ we have $d(X_n,Y_n)<d(X-n,X)+d(Y_n,X)<epsilon$. Thus $d(X_n,Y_n)rightarrow 0$
$endgroup$
add a comment |
$begingroup$
let $X_nrightarrow X$ thus $Y_n$ converges to Y as well. Given $epsilon>0$, $exists min N$ such that $forall n>m$ $d(X_n,X)<epsilon /2$ and $d(Y_n,X)<epsilon /2$ Then $forall n>m $ we have $d(X_n,Y_n)<d(X-n,X)+d(Y_n,X)<epsilon$. Thus $d(X_n,Y_n)rightarrow 0$
$endgroup$
add a comment |
$begingroup$
let $X_nrightarrow X$ thus $Y_n$ converges to Y as well. Given $epsilon>0$, $exists min N$ such that $forall n>m$ $d(X_n,X)<epsilon /2$ and $d(Y_n,X)<epsilon /2$ Then $forall n>m $ we have $d(X_n,Y_n)<d(X-n,X)+d(Y_n,X)<epsilon$. Thus $d(X_n,Y_n)rightarrow 0$
$endgroup$
let $X_nrightarrow X$ thus $Y_n$ converges to Y as well. Given $epsilon>0$, $exists min N$ such that $forall n>m$ $d(X_n,X)<epsilon /2$ and $d(Y_n,X)<epsilon /2$ Then $forall n>m $ we have $d(X_n,Y_n)<d(X-n,X)+d(Y_n,X)<epsilon$. Thus $d(X_n,Y_n)rightarrow 0$
answered May 11 '17 at 13:20
user379195
add a comment |
add a comment |
$begingroup$
Hint:
If $x$ is the limit of $x_n$ and $y$ is the limit of $y_n$, then
$$d(x_n, y_n) leq d(x_n, x) + d(x, y) + d(y, y_n)$$
You can use this to prove (using the $epsilon$ -definition) that $$lim_{ntoinfty} d(x_n, y_n) = 0$$
answered May 11 '17 at 13:16
5xum5xum
91.4k394161
$endgroup$
add a comment |
$begingroup$
Hint:
If $x$ is the limit of $x_n$ and $y$ is the limit of $y_n$, then
$$d(x_n, y_n) leq d(x_n, x) + d(x, y) + d(y, y_n)$$
You can use this to prove (using the $epsilon$ -definition) that $$lim_{ntoinfty} d(x_n, y_n) = 0$$
answered May 11 '17 at 13:16
5xum5xum
91.4k394161
$endgroup$
add a comment |
$begingroup$
Hint:
If $x$ is the limit of $x_n$ and $y$ is the limit of $y_n$, then
$$d(x_n, y_n) leq d(x_n, x) + d(x, y) + d(y, y_n)$$
You can use this to prove (using the $epsilon$ -definition) that $$lim_{ntoinfty} d(x_n, y_n) = 0$$
answered May 11 '17 at 13:16
5xum5xum
91.4k394161
$endgroup$
Hint:
If $x$ is the limit of $x_n$ and $y$ is the limit of $y_n$, then
$$d(x_n, y_n) leq d(x_n, x) + d(x, y) + d(y, y_n)$$
You can use this to prove (using the $epsilon$ -definition) that $$lim_{ntoinfty} d(x_n, y_n) = 0$$
answered May 11 '17 at 13:16
5xum5xum
91.4k394161
answered May 11 '17 at 13:16
5xum5xum
91.4k394161
answered May 11 '17 at 13:16
5xum5xum
91.4k394161
answered May 11 '17 at 13:16
5xum5xum
91.4k394161
91.4k394161
add a comment |
add a comment |
$begingroup$
Let $l$ be the limit of the sequences $(x_{n}), (y_{n})$ by definition Note that $d(x_{n},y_{n}) leq d(x_{n},l) + d(l, y_{n})$ for all $n$. If $varepsilon > 0$, then there are $N_{1},N_{2}$, by assumption, giving $d(x_{n},l), d(l,y_{n}) < varepsilon/2$ for all $n geq N_{1}$ and $geq N_{2}$. So $d(x_{n},l) + d(l,y_{n}) < varepsilon$ for all $n geq max { N_{1},N_{2} }$; so $d(x_{n},y_{n}) < varepsilon$ for all $n geq max { N_{1},N_{2} }$.
answered May 11 '17 at 13:37
Gary MooreGary Moore
17.3k21546
$endgroup$
add a comment |
$begingroup$
Let $l$ be the limit of the sequences $(x_{n}), (y_{n})$ by definition Note that $d(x_{n},y_{n}) leq d(x_{n},l) + d(l, y_{n})$ for all $n$. If $varepsilon > 0$, then there are $N_{1},N_{2}$, by assumption, giving $d(x_{n},l), d(l,y_{n}) < varepsilon/2$ for all $n geq N_{1}$ and $geq N_{2}$. So $d(x_{n},l) + d(l,y_{n}) < varepsilon$ for all $n geq max { N_{1},N_{2} }$; so $d(x_{n},y_{n}) < varepsilon$ for all $n geq max { N_{1},N_{2} }$.
answered May 11 '17 at 13:37
Gary MooreGary Moore
17.3k21546
$endgroup$
add a comment |
$begingroup$
Let $l$ be the limit of the sequences $(x_{n}), (y_{n})$ by definition Note that $d(x_{n},y_{n}) leq d(x_{n},l) + d(l, y_{n})$ for all $n$. If $varepsilon > 0$, then there are $N_{1},N_{2}$, by assumption, giving $d(x_{n},l), d(l,y_{n}) < varepsilon/2$ for all $n geq N_{1}$ and $geq N_{2}$. So $d(x_{n},l) + d(l,y_{n}) < varepsilon$ for all $n geq max { N_{1},N_{2} }$; so $d(x_{n},y_{n}) < varepsilon$ for all $n geq max { N_{1},N_{2} }$.
answered May 11 '17 at 13:37
Gary MooreGary Moore
17.3k21546
$endgroup$
Let $l$ be the limit of the sequences $(x_{n}), (y_{n})$ by definition Note that $d(x_{n},y_{n}) leq d(x_{n},l) + d(l, y_{n})$ for all $n$. If $varepsilon > 0$, then there are $N_{1},N_{2}$, by assumption, giving $d(x_{n},l), d(l,y_{n}) < varepsilon/2$ for all $n geq N_{1}$ and $geq N_{2}$. So $d(x_{n},l) + d(l,y_{n}) < varepsilon$ for all $n geq max { N_{1},N_{2} }$; so $d(x_{n},y_{n}) < varepsilon$ for all $n geq max { N_{1},N_{2} }$.
answered May 11 '17 at 13:37
Gary MooreGary Moore
17.3k21546
answered May 11 '17 at 13:37
Gary MooreGary Moore
17.3k21546
answered May 11 '17 at 13:37
Gary MooreGary Moore
17.3k21546
answered May 11 '17 at 13:37
Gary MooreGary Moore
17.3k21546
17.3k21546
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%2f2276339%2fsequences-that-converge-to-the-same-point%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
$begingroup$
You do not need to use completeness in any shape or form. The result is true regardless of even mentioning it.
$endgroup$
– user228113
May 11 '17 at 13:27