Mixing
simple question : sequential criterion for continuity
$begingroup$
Let $A$ be a subset of $mathbb{R}$, and let $f:Atomathbb{R}$ be a function. Now, let ${r_{n}}$ be any rational sequence in $A$, and let ${s_{n}}$ be any irrational sequence in $A$.
Suppose that $r_{n},s_{n}to ain A$ as $ntoinfty$, and that $f(r_{n}),f(s_{n})to f(a)inmathbb{R}$ as $ntoinfty$.
Is it enough so that $f$ is continuous at $x=a$?
The above two sequences are not "any" sequence in $A$.
But, it seems to be the density of rational and irrational would give such weak condition to Sequential criterion for continuity.
It just my opinion. Is it possible?
Give some comments or counter-examples! Thank you!
real-analysis continuity irrational-numbers rational-numbers
asked Jan 26 at 10:57
PrimaveraPrimavera
30919
$endgroup$
add a comment |
$begingroup$
Let $A$ be a subset of $mathbb{R}$, and let $f:Atomathbb{R}$ be a function. Now, let ${r_{n}}$ be any rational sequence in $A$, and let ${s_{n}}$ be any irrational sequence in $A$.
Suppose that $r_{n},s_{n}to ain A$ as $ntoinfty$, and that $f(r_{n}),f(s_{n})to f(a)inmathbb{R}$ as $ntoinfty$.
Is it enough so that $f$ is continuous at $x=a$?
The above two sequences are not "any" sequence in $A$.
But, it seems to be the density of rational and irrational would give such weak condition to Sequential criterion for continuity.
It just my opinion. Is it possible?
Give some comments or counter-examples! Thank you!
real-analysis continuity irrational-numbers rational-numbers
asked Jan 26 at 10:57
PrimaveraPrimavera
30919
$endgroup$
add a comment |
$begingroup$
Let $A$ be a subset of $mathbb{R}$, and let $f:Atomathbb{R}$ be a function. Now, let ${r_{n}}$ be any rational sequence in $A$, and let ${s_{n}}$ be any irrational sequence in $A$.
Suppose that $r_{n},s_{n}to ain A$ as $ntoinfty$, and that $f(r_{n}),f(s_{n})to f(a)inmathbb{R}$ as $ntoinfty$.
Is it enough so that $f$ is continuous at $x=a$?
The above two sequences are not "any" sequence in $A$.
But, it seems to be the density of rational and irrational would give such weak condition to Sequential criterion for continuity.
It just my opinion. Is it possible?
Give some comments or counter-examples! Thank you!
real-analysis continuity irrational-numbers rational-numbers
asked Jan 26 at 10:57
PrimaveraPrimavera
30919
$endgroup$
Let $A$ be a subset of $mathbb{R}$, and let $f:Atomathbb{R}$ be a function. Now, let ${r_{n}}$ be any rational sequence in $A$, and let ${s_{n}}$ be any irrational sequence in $A$.
Suppose that $r_{n},s_{n}to ain A$ as $ntoinfty$, and that $f(r_{n}),f(s_{n})to f(a)inmathbb{R}$ as $ntoinfty$.
Is it enough so that $f$ is continuous at $x=a$?
The above two sequences are not "any" sequence in $A$.
But, it seems to be the density of rational and irrational would give such weak condition to Sequential criterion for continuity.
It just my opinion. Is it possible?
Give some comments or counter-examples! Thank you!
real-analysis continuity irrational-numbers rational-numbers
real-analysis continuity irrational-numbers rational-numbers
asked Jan 26 at 10:57
PrimaveraPrimavera
30919
asked Jan 26 at 10:57
PrimaveraPrimavera
30919
edited Jan 26 at 11:24
Primavera
asked Jan 26 at 10:57
PrimaveraPrimavera
30919
asked Jan 26 at 10:57
PrimaveraPrimavera
30919
asked Jan 26 at 10:57
PrimaveraPrimavera
30919
30919
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
What makes you think its enough? For example, think about $f(x)$ such that $f(x)=a$ if $xin{a_n}cup{b_n}$ and $=a+1$ elsewhere.
You can never derive continuity simply by having the limits of any finite number of specific sequences coincide, because they'll never be enough to cover all possibilities. Its possible if you have information about generic sequential limits though. For instance, if you change your premise to "for any rational sequence $a_n$ ... and any irrational sequence $b_n$ ...", then you indeed can imply continuity.
answered Jan 26 at 11:02
VimVim
8,14631348
$endgroup$
$begingroup$
I mean ${r_{n}}$ is "for any rational sequence", ${s_{n}}$ is "for any irrational sequence". Then, as you commented, it is possible to reduce the condition in sequential criterion for continuity instead "for any sequence in domain" of "for any rational/irrational sequence in domain, right?
$endgroup$
– Primavera
Jan 26 at 11:15
$begingroup$
@Primavera you said "let ... be a rational sequence" etc so i assumed specificity. If you meant "any", then yes.
$endgroup$
– Vim
Jan 26 at 11:19
$begingroup$
oops, sorry. I have to try proving it. Thank you for comment.
$endgroup$
– Primavera
Jan 26 at 11:23
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%2f3088117%2fsimple-question-sequential-criterion-for-continuity%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
$begingroup$
What makes you think its enough? For example, think about $f(x)$ such that $f(x)=a$ if $xin{a_n}cup{b_n}$ and $=a+1$ elsewhere.
You can never derive continuity simply by having the limits of any finite number of specific sequences coincide, because they'll never be enough to cover all possibilities. Its possible if you have information about generic sequential limits though. For instance, if you change your premise to "for any rational sequence $a_n$ ... and any irrational sequence $b_n$ ...", then you indeed can imply continuity.
answered Jan 26 at 11:02
VimVim
8,14631348
$endgroup$
$begingroup$
I mean ${r_{n}}$ is "for any rational sequence", ${s_{n}}$ is "for any irrational sequence". Then, as you commented, it is possible to reduce the condition in sequential criterion for continuity instead "for any sequence in domain" of "for any rational/irrational sequence in domain, right?
$endgroup$
– Primavera
Jan 26 at 11:15
$begingroup$
@Primavera you said "let ... be a rational sequence" etc so i assumed specificity. If you meant "any", then yes.
$endgroup$
– Vim
Jan 26 at 11:19
$begingroup$
oops, sorry. I have to try proving it. Thank you for comment.
$endgroup$
– Primavera
Jan 26 at 11:23
add a comment |
$begingroup$
What makes you think its enough? For example, think about $f(x)$ such that $f(x)=a$ if $xin{a_n}cup{b_n}$ and $=a+1$ elsewhere.
You can never derive continuity simply by having the limits of any finite number of specific sequences coincide, because they'll never be enough to cover all possibilities. Its possible if you have information about generic sequential limits though. For instance, if you change your premise to "for any rational sequence $a_n$ ... and any irrational sequence $b_n$ ...", then you indeed can imply continuity.
answered Jan 26 at 11:02
VimVim
8,14631348
$endgroup$
$begingroup$
I mean ${r_{n}}$ is "for any rational sequence", ${s_{n}}$ is "for any irrational sequence". Then, as you commented, it is possible to reduce the condition in sequential criterion for continuity instead "for any sequence in domain" of "for any rational/irrational sequence in domain, right?
$endgroup$
– Primavera
Jan 26 at 11:15
$begingroup$
@Primavera you said "let ... be a rational sequence" etc so i assumed specificity. If you meant "any", then yes.
$endgroup$
– Vim
Jan 26 at 11:19
$begingroup$
oops, sorry. I have to try proving it. Thank you for comment.
$endgroup$
– Primavera
Jan 26 at 11:23
add a comment |
$begingroup$
What makes you think its enough? For example, think about $f(x)$ such that $f(x)=a$ if $xin{a_n}cup{b_n}$ and $=a+1$ elsewhere.
You can never derive continuity simply by having the limits of any finite number of specific sequences coincide, because they'll never be enough to cover all possibilities. Its possible if you have information about generic sequential limits though. For instance, if you change your premise to "for any rational sequence $a_n$ ... and any irrational sequence $b_n$ ...", then you indeed can imply continuity.
answered Jan 26 at 11:02
VimVim
8,14631348
$endgroup$
What makes you think its enough? For example, think about $f(x)$ such that $f(x)=a$ if $xin{a_n}cup{b_n}$ and $=a+1$ elsewhere.
You can never derive continuity simply by having the limits of any finite number of specific sequences coincide, because they'll never be enough to cover all possibilities. Its possible if you have information about generic sequential limits though. For instance, if you change your premise to "for any rational sequence $a_n$ ... and any irrational sequence $b_n$ ...", then you indeed can imply continuity.
answered Jan 26 at 11:02
VimVim
8,14631348
edited Jan 26 at 11:08
answered Jan 26 at 11:02
VimVim
8,14631348
answered Jan 26 at 11:02
VimVim
8,14631348
answered Jan 26 at 11:02
VimVim
8,14631348
8,14631348
$begingroup$
I mean ${r_{n}}$ is "for any rational sequence", ${s_{n}}$ is "for any irrational sequence". Then, as you commented, it is possible to reduce the condition in sequential criterion for continuity instead "for any sequence in domain" of "for any rational/irrational sequence in domain, right?
$endgroup$
– Primavera
Jan 26 at 11:15
$begingroup$
@Primavera you said "let ... be a rational sequence" etc so i assumed specificity. If you meant "any", then yes.
$endgroup$
– Vim
Jan 26 at 11:19
$begingroup$
oops, sorry. I have to try proving it. Thank you for comment.
$endgroup$
– Primavera
Jan 26 at 11:23
add a comment |
$begingroup$
I mean ${r_{n}}$ is "for any rational sequence", ${s_{n}}$ is "for any irrational sequence". Then, as you commented, it is possible to reduce the condition in sequential criterion for continuity instead "for any sequence in domain" of "for any rational/irrational sequence in domain, right?
$endgroup$
– Primavera
Jan 26 at 11:15
$begingroup$
@Primavera you said "let ... be a rational sequence" etc so i assumed specificity. If you meant "any", then yes.
$endgroup$
– Vim
Jan 26 at 11:19
$begingroup$
oops, sorry. I have to try proving it. Thank you for comment.
$endgroup$
– Primavera
Jan 26 at 11:23
$begingroup$
I mean ${r_{n}}$ is "for any rational sequence", ${s_{n}}$ is "for any irrational sequence". Then, as you commented, it is possible to reduce the condition in sequential criterion for continuity instead "for any sequence in domain" of "for any rational/irrational sequence in domain, right?
$endgroup$
– Primavera
Jan 26 at 11:15
$begingroup$
I mean ${r_{n}}$ is "for any rational sequence", ${s_{n}}$ is "for any irrational sequence". Then, as you commented, it is possible to reduce the condition in sequential criterion for continuity instead "for any sequence in domain" of "for any rational/irrational sequence in domain, right?
$endgroup$
– Primavera
Jan 26 at 11:15
$begingroup$
@Primavera you said "let ... be a rational sequence" etc so i assumed specificity. If you meant "any", then yes.
$endgroup$
– Vim
Jan 26 at 11:19
$begingroup$
@Primavera you said "let ... be a rational sequence" etc so i assumed specificity. If you meant "any", then yes.
$endgroup$
– Vim
Jan 26 at 11:19
$begingroup$
oops, sorry. I have to try proving it. Thank you for comment.
$endgroup$
– Primavera
Jan 26 at 11:23
$begingroup$
oops, sorry. I have to try proving it. Thank you for comment.
$endgroup$
– Primavera
Jan 26 at 11:23
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%2f3088117%2fsimple-question-sequential-criterion-for-continuity%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
