Real Analysis - Continuity
up vote
3
down vote
favorite
a. Give an example of a function defined everywhere on the interval $[0,1]$, which does not achieve its maximum.
b. Give an example of a function defined on $mathbb{R}$, that is nowhere continuous.
c. Give an example of a continuous function defined on a bounded set, which is not uniformly continuous.
My answers are
- $f(x) = x^2$
$f(x) = {1$, if $x$ is rational; $-1$, if $x$ is irrational$}$
- $f(x) = 1/x$
My workings are attached. Please help verify if the working is correct.Solutions
real-analysis continuity maxima-minima uniform-continuity
asked Dec 1 at 22:36
Ty Johnson
283
add a comment |
up vote
3
down vote
favorite
a. Give an example of a function defined everywhere on the interval $[0,1]$, which does not achieve its maximum.
b. Give an example of a function defined on $mathbb{R}$, that is nowhere continuous.
c. Give an example of a continuous function defined on a bounded set, which is not uniformly continuous.
My answers are
- $f(x) = x^2$
$f(x) = {1$, if $x$ is rational; $-1$, if $x$ is irrational$}$
- $f(x) = 1/x$
My workings are attached. Please help verify if the working is correct.Solutions
real-analysis continuity maxima-minima uniform-continuity
asked Dec 1 at 22:36
Ty Johnson
283
4
2. is correct. For 3, you need to specify the bounded domain. Your answer to 1. is incorrect, as the function certainly achieves its maximum on the interval $[0,1]$.
– Ted Shifrin
Dec 1 at 22:41
add a comment |
up vote
3
down vote
favorite
up vote
3
down vote
favorite
a. Give an example of a function defined everywhere on the interval $[0,1]$, which does not achieve its maximum.
b. Give an example of a function defined on $mathbb{R}$, that is nowhere continuous.
c. Give an example of a continuous function defined on a bounded set, which is not uniformly continuous.
My answers are
- $f(x) = x^2$
$f(x) = {1$, if $x$ is rational; $-1$, if $x$ is irrational$}$
- $f(x) = 1/x$
My workings are attached. Please help verify if the working is correct.Solutions
real-analysis continuity maxima-minima uniform-continuity
asked Dec 1 at 22:36
Ty Johnson
283
a. Give an example of a function defined everywhere on the interval $[0,1]$, which does not achieve its maximum.
b. Give an example of a function defined on $mathbb{R}$, that is nowhere continuous.
c. Give an example of a continuous function defined on a bounded set, which is not uniformly continuous.
My answers are
- $f(x) = x^2$
$f(x) = {1$, if $x$ is rational; $-1$, if $x$ is irrational$}$
- $f(x) = 1/x$
My workings are attached. Please help verify if the working is correct.Solutions
real-analysis continuity maxima-minima uniform-continuity
real-analysis continuity maxima-minima uniform-continuity
asked Dec 1 at 22:36
Ty Johnson
283
asked Dec 1 at 22:36
Ty Johnson
283
asked Dec 1 at 22:36
Ty Johnson
283
asked Dec 1 at 22:36
Ty Johnson
283
asked Dec 1 at 22:36
Ty Johnson
283
283
4
2. is correct. For 3, you need to specify the bounded domain. Your answer to 1. is incorrect, as the function certainly achieves its maximum on the interval $[0,1]$.
– Ted Shifrin
Dec 1 at 22:41
add a comment |
4
2. is correct. For 3, you need to specify the bounded domain. Your answer to 1. is incorrect, as the function certainly achieves its maximum on the interval $[0,1]$.
– Ted Shifrin
Dec 1 at 22:41
4
4
2. is correct. For 3, you need to specify the bounded domain. Your answer to 1. is incorrect, as the function certainly achieves its maximum on the interval $[0,1]$.
– Ted Shifrin
Dec 1 at 22:41
2. is correct. For 3, you need to specify the bounded domain. Your answer to 1. is incorrect, as the function certainly achieves its maximum on the interval $[0,1]$.
– Ted Shifrin
Dec 1 at 22:41
add a comment |
3 Answers
3
active
oldest
votes
up vote
6
down vote
accepted
$a)$ Consider $f(x)$ on $[0,1]$ such that $f(0) = 0, f(1) = 0, f(x) = dfrac{1}{x}, 0 <x < 1$.
$b)$ The function you had is a good one.
$c)$ Consider $f(x)$ on $(0,1)$ and $f(x) = dfrac{1}{x}$.
answered Dec 1 at 22:50
DeepSea
70.8k54487
add a comment |
up vote
4
down vote
- Wrong: the maximum is $1$, which is $f(1)$.
- Right.
- Your answer is incomplete, at best, since you did not state what is the domain of your function.
answered Dec 1 at 22:44
José Carlos Santos
147k22117217
add a comment |
up vote
2
down vote
1)
Extreme value theorem says that every continuous function over a closed interval has a maximum.
You need a function that is not continuous.
$f(x) = begin{cases} f(x) = x & x<1\f(x) = 0 & x=1 end{cases}$
answered Dec 1 at 23:27
Doug M
43.5k31854
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',
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%2f3021931%2freal-analysis-continuity%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
up vote
6
down vote
accepted
$a)$ Consider $f(x)$ on $[0,1]$ such that $f(0) = 0, f(1) = 0, f(x) = dfrac{1}{x}, 0 <x < 1$.
$b)$ The function you had is a good one.
$c)$ Consider $f(x)$ on $(0,1)$ and $f(x) = dfrac{1}{x}$.
answered Dec 1 at 22:50
DeepSea
70.8k54487
add a comment |
up vote
6
down vote
accepted
$a)$ Consider $f(x)$ on $[0,1]$ such that $f(0) = 0, f(1) = 0, f(x) = dfrac{1}{x}, 0 <x < 1$.
$b)$ The function you had is a good one.
$c)$ Consider $f(x)$ on $(0,1)$ and $f(x) = dfrac{1}{x}$.
answered Dec 1 at 22:50
DeepSea
70.8k54487
add a comment |
up vote
6
down vote
accepted
up vote
6
down vote
accepted
$a)$ Consider $f(x)$ on $[0,1]$ such that $f(0) = 0, f(1) = 0, f(x) = dfrac{1}{x}, 0 <x < 1$.
$b)$ The function you had is a good one.
$c)$ Consider $f(x)$ on $(0,1)$ and $f(x) = dfrac{1}{x}$.
answered Dec 1 at 22:50
DeepSea
70.8k54487
$a)$ Consider $f(x)$ on $[0,1]$ such that $f(0) = 0, f(1) = 0, f(x) = dfrac{1}{x}, 0 <x < 1$.
$b)$ The function you had is a good one.
$c)$ Consider $f(x)$ on $(0,1)$ and $f(x) = dfrac{1}{x}$.
answered Dec 1 at 22:50
DeepSea
70.8k54487
answered Dec 1 at 22:50
DeepSea
70.8k54487
answered Dec 1 at 22:50
DeepSea
70.8k54487
answered Dec 1 at 22:50
DeepSea
70.8k54487
70.8k54487
add a comment |
add a comment |
up vote
4
down vote
- Wrong: the maximum is $1$, which is $f(1)$.
- Right.
- Your answer is incomplete, at best, since you did not state what is the domain of your function.
answered Dec 1 at 22:44
José Carlos Santos
147k22117217
add a comment |
up vote
4
down vote
- Wrong: the maximum is $1$, which is $f(1)$.
- Right.
- Your answer is incomplete, at best, since you did not state what is the domain of your function.
answered Dec 1 at 22:44
José Carlos Santos
147k22117217
add a comment |
up vote
4
down vote
up vote
4
down vote
- Wrong: the maximum is $1$, which is $f(1)$.
- Right.
- Your answer is incomplete, at best, since you did not state what is the domain of your function.
answered Dec 1 at 22:44
José Carlos Santos
147k22117217
- Wrong: the maximum is $1$, which is $f(1)$.
- Right.
- Your answer is incomplete, at best, since you did not state what is the domain of your function.
answered Dec 1 at 22:44
José Carlos Santos
147k22117217
answered Dec 1 at 22:44
José Carlos Santos
147k22117217
answered Dec 1 at 22:44
José Carlos Santos
147k22117217
answered Dec 1 at 22:44
José Carlos Santos
147k22117217
147k22117217
add a comment |
add a comment |
up vote
2
down vote
1)
Extreme value theorem says that every continuous function over a closed interval has a maximum.
You need a function that is not continuous.
$f(x) = begin{cases} f(x) = x & x<1\f(x) = 0 & x=1 end{cases}$
answered Dec 1 at 23:27
Doug M
43.5k31854
add a comment |
up vote
2
down vote
1)
Extreme value theorem says that every continuous function over a closed interval has a maximum.
You need a function that is not continuous.
$f(x) = begin{cases} f(x) = x & x<1\f(x) = 0 & x=1 end{cases}$
answered Dec 1 at 23:27
Doug M
43.5k31854
add a comment |
up vote
2
down vote
up vote
2
down vote
1)
Extreme value theorem says that every continuous function over a closed interval has a maximum.
You need a function that is not continuous.
$f(x) = begin{cases} f(x) = x & x<1\f(x) = 0 & x=1 end{cases}$
answered Dec 1 at 23:27
Doug M
43.5k31854
1)
Extreme value theorem says that every continuous function over a closed interval has a maximum.
You need a function that is not continuous.
$f(x) = begin{cases} f(x) = x & x<1\f(x) = 0 & x=1 end{cases}$
answered Dec 1 at 23:27
Doug M
43.5k31854
answered Dec 1 at 23:27
Doug M
43.5k31854
answered Dec 1 at 23:27
Doug M
43.5k31854
answered Dec 1 at 23:27
Doug M
43.5k31854
43.5k31854
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.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- 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.
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%2f3021931%2freal-analysis-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
4
2. is correct. For 3, you need to specify the bounded domain. Your answer to 1. is incorrect, as the function certainly achieves its maximum on the interval $[0,1]$.
– Ted Shifrin
Dec 1 at 22:41