Divergent real sequence for which $forall epsilon >0 exists a in mathbb{R}: |x_n-a|<epsilon$ for almost...
$begingroup$
Looking at the definition of convergence:
$exists a in mathbb{R}forall epsilon >0: |x_n-a|<epsilon$ for almost all n
I switched the order of the first two quantifiers. Is there a sequence for which
$forall epsilon >0 exists b in mathbb{R}: |x_n-b|<epsilon$ for almost all n
holds but it's not convergent?
real-analysis sequences-and-series convergence
edited Dec 18 '18 at 22:47
Bernard
121k740116
asked Dec 18 '18 at 22:20
DDevelopsDDevelops
583
$endgroup$
add a comment |
$begingroup$
Looking at the definition of convergence:
$exists a in mathbb{R}forall epsilon >0: |x_n-a|<epsilon$ for almost all n
I switched the order of the first two quantifiers. Is there a sequence for which
$forall epsilon >0 exists b in mathbb{R}: |x_n-b|<epsilon$ for almost all n
holds but it's not convergent?
real-analysis sequences-and-series convergence
edited Dec 18 '18 at 22:47
Bernard
121k740116
asked Dec 18 '18 at 22:20
DDevelopsDDevelops
583
$endgroup$
add a comment |
$begingroup$
Looking at the definition of convergence:
$exists a in mathbb{R}forall epsilon >0: |x_n-a|<epsilon$ for almost all n
I switched the order of the first two quantifiers. Is there a sequence for which
$forall epsilon >0 exists b in mathbb{R}: |x_n-b|<epsilon$ for almost all n
holds but it's not convergent?
real-analysis sequences-and-series convergence
edited Dec 18 '18 at 22:47
Bernard
121k740116
asked Dec 18 '18 at 22:20
DDevelopsDDevelops
583
$endgroup$
Looking at the definition of convergence:
$exists a in mathbb{R}forall epsilon >0: |x_n-a|<epsilon$ for almost all n
I switched the order of the first two quantifiers. Is there a sequence for which
$forall epsilon >0 exists b in mathbb{R}: |x_n-b|<epsilon$ for almost all n
holds but it's not convergent?
real-analysis sequences-and-series convergence
real-analysis sequences-and-series convergence
edited Dec 18 '18 at 22:47
Bernard
121k740116
asked Dec 18 '18 at 22:20
DDevelopsDDevelops
583
edited Dec 18 '18 at 22:47
Bernard
121k740116
asked Dec 18 '18 at 22:20
DDevelopsDDevelops
583
edited Dec 18 '18 at 22:47
Bernard
121k740116
edited Dec 18 '18 at 22:47
Bernard
121k740116
edited Dec 18 '18 at 22:47
Bernard
121k740116
121k740116
asked Dec 18 '18 at 22:20
DDevelopsDDevelops
583
asked Dec 18 '18 at 22:20
DDevelopsDDevelops
583
asked Dec 18 '18 at 22:20
DDevelopsDDevelops
583
583
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Any sequence satisfying this condition is Cauchy, and therefore convergent to some finite limit by the completeness of $mathbb{R}$. To see this, suppose $epsilon > 0$. Then there exists $b$ such that $|x_n - b| < frac{epsilon}{2}$ for almost all $n$. In order words, for such a $b$, there exists $N$ such that $|x_n - b| < frac{epsilon}{2}$ whenever $n ge N$. But then whenever $m, n ge N$, we have
$$|x_m - x_n| = |(x_m - b) - (x_n - b)| le |x_m - b| + |x_n - b| < frac{epsilon}{2} + frac{epsilon}{2} = epsilon.$$
answered Dec 18 '18 at 22:31
Daniel ScheplerDaniel Schepler
8,9591620
$endgroup$
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%2f3045779%2fdivergent-real-sequence-for-which-forall-epsilon-0-exists-a-in-mathbbr%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
$begingroup$
Any sequence satisfying this condition is Cauchy, and therefore convergent to some finite limit by the completeness of $mathbb{R}$. To see this, suppose $epsilon > 0$. Then there exists $b$ such that $|x_n - b| < frac{epsilon}{2}$ for almost all $n$. In order words, for such a $b$, there exists $N$ such that $|x_n - b| < frac{epsilon}{2}$ whenever $n ge N$. But then whenever $m, n ge N$, we have
$$|x_m - x_n| = |(x_m - b) - (x_n - b)| le |x_m - b| + |x_n - b| < frac{epsilon}{2} + frac{epsilon}{2} = epsilon.$$
answered Dec 18 '18 at 22:31
Daniel ScheplerDaniel Schepler
8,9591620
$endgroup$
add a comment |
$begingroup$
Any sequence satisfying this condition is Cauchy, and therefore convergent to some finite limit by the completeness of $mathbb{R}$. To see this, suppose $epsilon > 0$. Then there exists $b$ such that $|x_n - b| < frac{epsilon}{2}$ for almost all $n$. In order words, for such a $b$, there exists $N$ such that $|x_n - b| < frac{epsilon}{2}$ whenever $n ge N$. But then whenever $m, n ge N$, we have
$$|x_m - x_n| = |(x_m - b) - (x_n - b)| le |x_m - b| + |x_n - b| < frac{epsilon}{2} + frac{epsilon}{2} = epsilon.$$
answered Dec 18 '18 at 22:31
Daniel ScheplerDaniel Schepler
8,9591620
$endgroup$
add a comment |
$begingroup$
Any sequence satisfying this condition is Cauchy, and therefore convergent to some finite limit by the completeness of $mathbb{R}$. To see this, suppose $epsilon > 0$. Then there exists $b$ such that $|x_n - b| < frac{epsilon}{2}$ for almost all $n$. In order words, for such a $b$, there exists $N$ such that $|x_n - b| < frac{epsilon}{2}$ whenever $n ge N$. But then whenever $m, n ge N$, we have
$$|x_m - x_n| = |(x_m - b) - (x_n - b)| le |x_m - b| + |x_n - b| < frac{epsilon}{2} + frac{epsilon}{2} = epsilon.$$
answered Dec 18 '18 at 22:31
Daniel ScheplerDaniel Schepler
8,9591620
$endgroup$
Any sequence satisfying this condition is Cauchy, and therefore convergent to some finite limit by the completeness of $mathbb{R}$. To see this, suppose $epsilon > 0$. Then there exists $b$ such that $|x_n - b| < frac{epsilon}{2}$ for almost all $n$. In order words, for such a $b$, there exists $N$ such that $|x_n - b| < frac{epsilon}{2}$ whenever $n ge N$. But then whenever $m, n ge N$, we have
$$|x_m - x_n| = |(x_m - b) - (x_n - b)| le |x_m - b| + |x_n - b| < frac{epsilon}{2} + frac{epsilon}{2} = epsilon.$$
answered Dec 18 '18 at 22:31
Daniel ScheplerDaniel Schepler
8,9591620
answered Dec 18 '18 at 22:31
Daniel ScheplerDaniel Schepler
8,9591620
answered Dec 18 '18 at 22:31
Daniel ScheplerDaniel Schepler
8,9591620
answered Dec 18 '18 at 22:31
Daniel ScheplerDaniel Schepler
8,9591620
8,9591620
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%2f3045779%2fdivergent-real-sequence-for-which-forall-epsilon-0-exists-a-in-mathbbr%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
