When teaching someone how to prove a function is uniformly continuous, using epsilon/delta, which example...
$begingroup$
I've taught how to use $epsilon, delta$ to prove that a function is continuous at a point, and I'm about to teach how to prove that a function is uniformly continuous over an open interval.
Usually, the examples I can think of that seem easy enough on the outside, require some algebraic trickery that might make it seem more daunting than it needs to be, and may inspire a "damn, this is too difficult" mentality.
Are there some examples of functions that are almost painfully straightforward to give a soft introduction to these, that I may increase the difficulty more smoothly?
calculus limits
edited Mar 3 at 17:30
user3813
3,1321025
asked Mar 2 at 22:55
AlecAlec
619311
$endgroup$
add a comment |
$begingroup$
I've taught how to use $epsilon, delta$ to prove that a function is continuous at a point, and I'm about to teach how to prove that a function is uniformly continuous over an open interval.
Usually, the examples I can think of that seem easy enough on the outside, require some algebraic trickery that might make it seem more daunting than it needs to be, and may inspire a "damn, this is too difficult" mentality.
Are there some examples of functions that are almost painfully straightforward to give a soft introduction to these, that I may increase the difficulty more smoothly?
calculus limits
edited Mar 3 at 17:30
user3813
3,1321025
asked Mar 2 at 22:55
AlecAlec
619311
$endgroup$
2
$begingroup$
A linear function, perhaps?
$endgroup$
– paw88789
Mar 2 at 23:10
$begingroup$
@paw88789 - Definitely a good idea, yeah. Easy, quick, and no long lines of algebra that draw attention away from the end goal. Thanks for the tip! Any natural steps beyond that?
$endgroup$
– Alec
Mar 2 at 23:18
$begingroup$
$|sin x - sin y| le |x-y|$ makes sine a good candidate.
$endgroup$
– user3813
Mar 3 at 3:29
1
$begingroup$
@Alec In the title question you use the phrase "uniformly continuous", and in the question body you say "prove that a function is continuous over an open interval". These are different things. Which did you intend to ask a question about?
$endgroup$
– Steven Gubkin
Mar 3 at 14:45
$begingroup$
@StevenGubkin - Ah, inaccurate wording on my part. I meant uniformly continuous in both cases.
$endgroup$
– Alec
Mar 3 at 16:27
add a comment |
$begingroup$
I've taught how to use $epsilon, delta$ to prove that a function is continuous at a point, and I'm about to teach how to prove that a function is uniformly continuous over an open interval.
Usually, the examples I can think of that seem easy enough on the outside, require some algebraic trickery that might make it seem more daunting than it needs to be, and may inspire a "damn, this is too difficult" mentality.
Are there some examples of functions that are almost painfully straightforward to give a soft introduction to these, that I may increase the difficulty more smoothly?
calculus limits
edited Mar 3 at 17:30
user3813
3,1321025
asked Mar 2 at 22:55
AlecAlec
619311
$endgroup$
I've taught how to use $epsilon, delta$ to prove that a function is continuous at a point, and I'm about to teach how to prove that a function is uniformly continuous over an open interval.
Usually, the examples I can think of that seem easy enough on the outside, require some algebraic trickery that might make it seem more daunting than it needs to be, and may inspire a "damn, this is too difficult" mentality.
Are there some examples of functions that are almost painfully straightforward to give a soft introduction to these, that I may increase the difficulty more smoothly?
calculus limits
calculus limits
edited Mar 3 at 17:30
user3813
3,1321025
asked Mar 2 at 22:55
AlecAlec
619311
edited Mar 3 at 17:30
user3813
3,1321025
asked Mar 2 at 22:55
AlecAlec
619311
edited Mar 3 at 17:30
user3813
3,1321025
edited Mar 3 at 17:30
user3813
3,1321025
edited Mar 3 at 17:30
user3813
3,1321025
3,1321025
asked Mar 2 at 22:55
AlecAlec
619311
asked Mar 2 at 22:55
AlecAlec
619311
asked Mar 2 at 22:55
AlecAlec
619311
619311
2
$begingroup$
A linear function, perhaps?
$endgroup$
– paw88789
Mar 2 at 23:10
$begingroup$
@paw88789 - Definitely a good idea, yeah. Easy, quick, and no long lines of algebra that draw attention away from the end goal. Thanks for the tip! Any natural steps beyond that?
$endgroup$
– Alec
Mar 2 at 23:18
$begingroup$
$|sin x - sin y| le |x-y|$ makes sine a good candidate.
$endgroup$
– user3813
Mar 3 at 3:29
1
$begingroup$
@Alec In the title question you use the phrase "uniformly continuous", and in the question body you say "prove that a function is continuous over an open interval". These are different things. Which did you intend to ask a question about?
$endgroup$
– Steven Gubkin
Mar 3 at 14:45
$begingroup$
@StevenGubkin - Ah, inaccurate wording on my part. I meant uniformly continuous in both cases.
$endgroup$
– Alec
Mar 3 at 16:27
add a comment |
2
$begingroup$
A linear function, perhaps?
$endgroup$
– paw88789
Mar 2 at 23:10
$begingroup$
@paw88789 - Definitely a good idea, yeah. Easy, quick, and no long lines of algebra that draw attention away from the end goal. Thanks for the tip! Any natural steps beyond that?
$endgroup$
– Alec
Mar 2 at 23:18
$begingroup$
$|sin x - sin y| le |x-y|$ makes sine a good candidate.
$endgroup$
– user3813
Mar 3 at 3:29
1
$begingroup$
@Alec In the title question you use the phrase "uniformly continuous", and in the question body you say "prove that a function is continuous over an open interval". These are different things. Which did you intend to ask a question about?
$endgroup$
– Steven Gubkin
Mar 3 at 14:45
$begingroup$
@StevenGubkin - Ah, inaccurate wording on my part. I meant uniformly continuous in both cases.
$endgroup$
– Alec
Mar 3 at 16:27
2
2
$begingroup$
A linear function, perhaps?
$endgroup$
– paw88789
Mar 2 at 23:10
$begingroup$
A linear function, perhaps?
$endgroup$
– paw88789
Mar 2 at 23:10
$begingroup$
@paw88789 - Definitely a good idea, yeah. Easy, quick, and no long lines of algebra that draw attention away from the end goal. Thanks for the tip! Any natural steps beyond that?
$endgroup$
– Alec
Mar 2 at 23:18
$begingroup$
@paw88789 - Definitely a good idea, yeah. Easy, quick, and no long lines of algebra that draw attention away from the end goal. Thanks for the tip! Any natural steps beyond that?
$endgroup$
– Alec
Mar 2 at 23:18
$begingroup$
$|sin x - sin y| le |x-y|$ makes sine a good candidate.
$endgroup$
– user3813
Mar 3 at 3:29
$begingroup$
$|sin x - sin y| le |x-y|$ makes sine a good candidate.
$endgroup$
– user3813
Mar 3 at 3:29
1
1
$begingroup$
@Alec In the title question you use the phrase "uniformly continuous", and in the question body you say "prove that a function is continuous over an open interval". These are different things. Which did you intend to ask a question about?
$endgroup$
– Steven Gubkin
Mar 3 at 14:45
$begingroup$
@Alec In the title question you use the phrase "uniformly continuous", and in the question body you say "prove that a function is continuous over an open interval". These are different things. Which did you intend to ask a question about?
$endgroup$
– Steven Gubkin
Mar 3 at 14:45
$begingroup$
@StevenGubkin - Ah, inaccurate wording on my part. I meant uniformly continuous in both cases.
$endgroup$
– Alec
Mar 3 at 16:27
$begingroup$
@StevenGubkin - Ah, inaccurate wording on my part. I meant uniformly continuous in both cases.
$endgroup$
– Alec
Mar 3 at 16:27
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
I think this cannot be understood without a contrasting example where it fails.
So perhaps, in addition to a linear function as suggested by @paw88789, consider $f(x) = frac{1}{x}$ over the open interval $(0,1)$.
It is continuous over that interval, but not uniformly continuous.
Fix an $epsilon > 0$; then for any $delta > 0$ one can
arrange the difference in $f$-values to exceed $epsilon$ by getting
close enough to $x=0$.
answered Mar 3 at 0:38
Joseph O'RourkeJoseph O'Rourke
15k33280
$endgroup$
1
$begingroup$
Also show $1/x$ is uniformly continuous on $[1,infty)$, or $[a,infty)$ for $a>0$.
$endgroup$
– KCd
Mar 7 at 3:32
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: "548"
};
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: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
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%2fmatheducators.stackexchange.com%2fquestions%2f15310%2fwhen-teaching-someone-how-to-prove-a-function-is-uniformly-continuous-using-eps%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
$begingroup$
I think this cannot be understood without a contrasting example where it fails.
So perhaps, in addition to a linear function as suggested by @paw88789, consider $f(x) = frac{1}{x}$ over the open interval $(0,1)$.
It is continuous over that interval, but not uniformly continuous.
Fix an $epsilon > 0$; then for any $delta > 0$ one can
arrange the difference in $f$-values to exceed $epsilon$ by getting
close enough to $x=0$.
answered Mar 3 at 0:38
Joseph O'RourkeJoseph O'Rourke
15k33280
$endgroup$
1
$begingroup$
Also show $1/x$ is uniformly continuous on $[1,infty)$, or $[a,infty)$ for $a>0$.
$endgroup$
– KCd
Mar 7 at 3:32
add a comment |
$begingroup$
I think this cannot be understood without a contrasting example where it fails.
So perhaps, in addition to a linear function as suggested by @paw88789, consider $f(x) = frac{1}{x}$ over the open interval $(0,1)$.
It is continuous over that interval, but not uniformly continuous.
Fix an $epsilon > 0$; then for any $delta > 0$ one can
arrange the difference in $f$-values to exceed $epsilon$ by getting
close enough to $x=0$.
answered Mar 3 at 0:38
Joseph O'RourkeJoseph O'Rourke
15k33280
$endgroup$
1
$begingroup$
Also show $1/x$ is uniformly continuous on $[1,infty)$, or $[a,infty)$ for $a>0$.
$endgroup$
– KCd
Mar 7 at 3:32
add a comment |
$begingroup$
I think this cannot be understood without a contrasting example where it fails.
So perhaps, in addition to a linear function as suggested by @paw88789, consider $f(x) = frac{1}{x}$ over the open interval $(0,1)$.
It is continuous over that interval, but not uniformly continuous.
Fix an $epsilon > 0$; then for any $delta > 0$ one can
arrange the difference in $f$-values to exceed $epsilon$ by getting
close enough to $x=0$.
answered Mar 3 at 0:38
Joseph O'RourkeJoseph O'Rourke
15k33280
$endgroup$
I think this cannot be understood without a contrasting example where it fails.
So perhaps, in addition to a linear function as suggested by @paw88789, consider $f(x) = frac{1}{x}$ over the open interval $(0,1)$.
It is continuous over that interval, but not uniformly continuous.
Fix an $epsilon > 0$; then for any $delta > 0$ one can
arrange the difference in $f$-values to exceed $epsilon$ by getting
close enough to $x=0$.
answered Mar 3 at 0:38
Joseph O'RourkeJoseph O'Rourke
15k33280
answered Mar 3 at 0:38
Joseph O'RourkeJoseph O'Rourke
15k33280
answered Mar 3 at 0:38
Joseph O'RourkeJoseph O'Rourke
15k33280
answered Mar 3 at 0:38
Joseph O'RourkeJoseph O'Rourke
15k33280
15k33280
1
$begingroup$
Also show $1/x$ is uniformly continuous on $[1,infty)$, or $[a,infty)$ for $a>0$.
$endgroup$
– KCd
Mar 7 at 3:32
add a comment |
1
$begingroup$
Also show $1/x$ is uniformly continuous on $[1,infty)$, or $[a,infty)$ for $a>0$.
$endgroup$
– KCd
Mar 7 at 3:32
1
1
$begingroup$
Also show $1/x$ is uniformly continuous on $[1,infty)$, or $[a,infty)$ for $a>0$.
$endgroup$
– KCd
Mar 7 at 3:32
$begingroup$
Also show $1/x$ is uniformly continuous on $[1,infty)$, or $[a,infty)$ for $a>0$.
$endgroup$
– KCd
Mar 7 at 3:32
add a comment |
Thanks for contributing an answer to Mathematics Educators 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%2fmatheducators.stackexchange.com%2fquestions%2f15310%2fwhen-teaching-someone-how-to-prove-a-function-is-uniformly-continuous-using-eps%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
2
$begingroup$
A linear function, perhaps?
$endgroup$
– paw88789
Mar 2 at 23:10
$begingroup$
@paw88789 - Definitely a good idea, yeah. Easy, quick, and no long lines of algebra that draw attention away from the end goal. Thanks for the tip! Any natural steps beyond that?
$endgroup$
– Alec
Mar 2 at 23:18
$begingroup$
$|sin x - sin y| le |x-y|$ makes sine a good candidate.
$endgroup$
– user3813
Mar 3 at 3:29
1
$begingroup$
@Alec In the title question you use the phrase "uniformly continuous", and in the question body you say "prove that a function is continuous over an open interval". These are different things. Which did you intend to ask a question about?
$endgroup$
– Steven Gubkin
Mar 3 at 14:45
$begingroup$
@StevenGubkin - Ah, inaccurate wording on my part. I meant uniformly continuous in both cases.
$endgroup$
– Alec
Mar 3 at 16:27