Is every real valued continuous function on $(0,1)$ is uniformly continuous?
$begingroup$
Is every real valued continuous function on the interval $(0,1)$ is uniformly continuous?
I think the answer is no, and to reject the statement, we need to come up a continuous function probably $f(x)=frac{1}{x}$ and follow the following link:
Coming up with an example, a function that is continuous but not uniformly continuous
But it does not work because $delta=min(x,1)$ cannot be applied because $x$ cannot attain $1$.
continuity
asked Dec 9 '18 at 6:16
SepideSepide
3038
$endgroup$
add a comment |
$begingroup$
Is every real valued continuous function on the interval $(0,1)$ is uniformly continuous?
I think the answer is no, and to reject the statement, we need to come up a continuous function probably $f(x)=frac{1}{x}$ and follow the following link:
Coming up with an example, a function that is continuous but not uniformly continuous
But it does not work because $delta=min(x,1)$ cannot be applied because $x$ cannot attain $1$.
continuity
asked Dec 9 '18 at 6:16
SepideSepide
3038
$endgroup$
4
$begingroup$
Possible duplicate of Coming up with an example, a function that is continuous but not uniformly continuous
$endgroup$
– Boshu
Dec 9 '18 at 6:20
$begingroup$
@Boshu: That does not work as I explained in the question.
$endgroup$
– Sepide
Dec 9 '18 at 6:46
1
$begingroup$
Take a good look at the proof, and the definition of uniform continuity; just because you cannot copy a proof word for word does not mean that it does not hold. You need to understand what cause $frac{1}{x}$ to be not uniformly continuous, and whether changing the right hand side of the interval actually affects it.
$endgroup$
– Boshu
Dec 9 '18 at 6:52
$begingroup$
@Boshu: Sorry I cannot understand, that's why I asked. Could you clarify it for me?
$endgroup$
– Sepide
Dec 9 '18 at 6:57
$begingroup$
Will write a brief answer.
$endgroup$
– Boshu
Dec 9 '18 at 7:05
add a comment |
$begingroup$
Is every real valued continuous function on the interval $(0,1)$ is uniformly continuous?
I think the answer is no, and to reject the statement, we need to come up a continuous function probably $f(x)=frac{1}{x}$ and follow the following link:
Coming up with an example, a function that is continuous but not uniformly continuous
But it does not work because $delta=min(x,1)$ cannot be applied because $x$ cannot attain $1$.
continuity
asked Dec 9 '18 at 6:16
SepideSepide
3038
$endgroup$
Is every real valued continuous function on the interval $(0,1)$ is uniformly continuous?
I think the answer is no, and to reject the statement, we need to come up a continuous function probably $f(x)=frac{1}{x}$ and follow the following link:
Coming up with an example, a function that is continuous but not uniformly continuous
But it does not work because $delta=min(x,1)$ cannot be applied because $x$ cannot attain $1$.
continuity
continuity
asked Dec 9 '18 at 6:16
SepideSepide
3038
asked Dec 9 '18 at 6:16
SepideSepide
3038
edited Dec 9 '18 at 6:44
Sepide
asked Dec 9 '18 at 6:16
SepideSepide
3038
asked Dec 9 '18 at 6:16
SepideSepide
3038
asked Dec 9 '18 at 6:16
SepideSepide
3038
3038
4
$begingroup$
Possible duplicate of Coming up with an example, a function that is continuous but not uniformly continuous
$endgroup$
– Boshu
Dec 9 '18 at 6:20
$begingroup$
@Boshu: That does not work as I explained in the question.
$endgroup$
– Sepide
Dec 9 '18 at 6:46
1
$begingroup$
Take a good look at the proof, and the definition of uniform continuity; just because you cannot copy a proof word for word does not mean that it does not hold. You need to understand what cause $frac{1}{x}$ to be not uniformly continuous, and whether changing the right hand side of the interval actually affects it.
$endgroup$
– Boshu
Dec 9 '18 at 6:52
$begingroup$
@Boshu: Sorry I cannot understand, that's why I asked. Could you clarify it for me?
$endgroup$
– Sepide
Dec 9 '18 at 6:57
$begingroup$
Will write a brief answer.
$endgroup$
– Boshu
Dec 9 '18 at 7:05
add a comment |
4
$begingroup$
Possible duplicate of Coming up with an example, a function that is continuous but not uniformly continuous
$endgroup$
– Boshu
Dec 9 '18 at 6:20
$begingroup$
@Boshu: That does not work as I explained in the question.
$endgroup$
– Sepide
Dec 9 '18 at 6:46
1
$begingroup$
Take a good look at the proof, and the definition of uniform continuity; just because you cannot copy a proof word for word does not mean that it does not hold. You need to understand what cause $frac{1}{x}$ to be not uniformly continuous, and whether changing the right hand side of the interval actually affects it.
$endgroup$
– Boshu
Dec 9 '18 at 6:52
$begingroup$
@Boshu: Sorry I cannot understand, that's why I asked. Could you clarify it for me?
$endgroup$
– Sepide
Dec 9 '18 at 6:57
$begingroup$
Will write a brief answer.
$endgroup$
– Boshu
Dec 9 '18 at 7:05
4
4
$begingroup$
Possible duplicate of Coming up with an example, a function that is continuous but not uniformly continuous
$endgroup$
– Boshu
Dec 9 '18 at 6:20
$begingroup$
Possible duplicate of Coming up with an example, a function that is continuous but not uniformly continuous
$endgroup$
– Boshu
Dec 9 '18 at 6:20
$begingroup$
@Boshu: That does not work as I explained in the question.
$endgroup$
– Sepide
Dec 9 '18 at 6:46
$begingroup$
@Boshu: That does not work as I explained in the question.
$endgroup$
– Sepide
Dec 9 '18 at 6:46
1
1
$begingroup$
Take a good look at the proof, and the definition of uniform continuity; just because you cannot copy a proof word for word does not mean that it does not hold. You need to understand what cause $frac{1}{x}$ to be not uniformly continuous, and whether changing the right hand side of the interval actually affects it.
$endgroup$
– Boshu
Dec 9 '18 at 6:52
$begingroup$
Take a good look at the proof, and the definition of uniform continuity; just because you cannot copy a proof word for word does not mean that it does not hold. You need to understand what cause $frac{1}{x}$ to be not uniformly continuous, and whether changing the right hand side of the interval actually affects it.
$endgroup$
– Boshu
Dec 9 '18 at 6:52
$begingroup$
@Boshu: Sorry I cannot understand, that's why I asked. Could you clarify it for me?
$endgroup$
– Sepide
Dec 9 '18 at 6:57
$begingroup$
@Boshu: Sorry I cannot understand, that's why I asked. Could you clarify it for me?
$endgroup$
– Sepide
Dec 9 '18 at 6:57
$begingroup$
Will write a brief answer.
$endgroup$
– Boshu
Dec 9 '18 at 7:05
$begingroup$
Will write a brief answer.
$endgroup$
– Boshu
Dec 9 '18 at 7:05
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The link I provided above is fairly comprehensive. What I think will help you is understanding uniform continuity better.
Uniform continuity means that if two points are within some fixed distance of each other, the values taken at those points can only be so far apart. However, because $dfrac{1}{x}$ goes to infinity within that small interval, it means that the difference between the values taken by a function at two points at some distance, gets larger as we move closer to $0$. Consequently, $dfrac{1}{x}$ is not uniformly continuous. You can now attempt to formally write this down following from the answer I've linked above.
answered Dec 9 '18 at 7:08
BoshuBoshu
705315
$endgroup$
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%2f3032073%2fis-every-real-valued-continuous-function-on-0-1-is-uniformly-continuous%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
$begingroup$
The link I provided above is fairly comprehensive. What I think will help you is understanding uniform continuity better.
Uniform continuity means that if two points are within some fixed distance of each other, the values taken at those points can only be so far apart. However, because $dfrac{1}{x}$ goes to infinity within that small interval, it means that the difference between the values taken by a function at two points at some distance, gets larger as we move closer to $0$. Consequently, $dfrac{1}{x}$ is not uniformly continuous. You can now attempt to formally write this down following from the answer I've linked above.
answered Dec 9 '18 at 7:08
BoshuBoshu
705315
$endgroup$
add a comment |
$begingroup$
The link I provided above is fairly comprehensive. What I think will help you is understanding uniform continuity better.
Uniform continuity means that if two points are within some fixed distance of each other, the values taken at those points can only be so far apart. However, because $dfrac{1}{x}$ goes to infinity within that small interval, it means that the difference between the values taken by a function at two points at some distance, gets larger as we move closer to $0$. Consequently, $dfrac{1}{x}$ is not uniformly continuous. You can now attempt to formally write this down following from the answer I've linked above.
answered Dec 9 '18 at 7:08
BoshuBoshu
705315
$endgroup$
add a comment |
$begingroup$
The link I provided above is fairly comprehensive. What I think will help you is understanding uniform continuity better.
Uniform continuity means that if two points are within some fixed distance of each other, the values taken at those points can only be so far apart. However, because $dfrac{1}{x}$ goes to infinity within that small interval, it means that the difference between the values taken by a function at two points at some distance, gets larger as we move closer to $0$. Consequently, $dfrac{1}{x}$ is not uniformly continuous. You can now attempt to formally write this down following from the answer I've linked above.
answered Dec 9 '18 at 7:08
BoshuBoshu
705315
$endgroup$
The link I provided above is fairly comprehensive. What I think will help you is understanding uniform continuity better.
Uniform continuity means that if two points are within some fixed distance of each other, the values taken at those points can only be so far apart. However, because $dfrac{1}{x}$ goes to infinity within that small interval, it means that the difference between the values taken by a function at two points at some distance, gets larger as we move closer to $0$. Consequently, $dfrac{1}{x}$ is not uniformly continuous. You can now attempt to formally write this down following from the answer I've linked above.
answered Dec 9 '18 at 7:08
BoshuBoshu
705315
answered Dec 9 '18 at 7:08
BoshuBoshu
705315
answered Dec 9 '18 at 7:08
BoshuBoshu
705315
answered Dec 9 '18 at 7:08
BoshuBoshu
705315
705315
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%2f3032073%2fis-every-real-valued-continuous-function-on-0-1-is-uniformly-continuous%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
4
$begingroup$
Possible duplicate of Coming up with an example, a function that is continuous but not uniformly continuous
$endgroup$
– Boshu
Dec 9 '18 at 6:20
$begingroup$
@Boshu: That does not work as I explained in the question.
$endgroup$
– Sepide
Dec 9 '18 at 6:46
1
$begingroup$
Take a good look at the proof, and the definition of uniform continuity; just because you cannot copy a proof word for word does not mean that it does not hold. You need to understand what cause $frac{1}{x}$ to be not uniformly continuous, and whether changing the right hand side of the interval actually affects it.
$endgroup$
– Boshu
Dec 9 '18 at 6:52
$begingroup$
@Boshu: Sorry I cannot understand, that's why I asked. Could you clarify it for me?
$endgroup$
– Sepide
Dec 9 '18 at 6:57
$begingroup$
Will write a brief answer.
$endgroup$
– Boshu
Dec 9 '18 at 7:05