Rudin: Chapter 4, Problem 9. Rephrasing the definition of uniform continuity. Understanding a solution.
$begingroup$
This is from Rudin Chapter 4, problem 9:
Show that the requirement in the definition of uniform continuity can
be rephrased as follows, in terms of diameters of sets: To every $epsilon > 0$
there exists a $delta > 0$ such that diam $f(E)<epsilon$ for all $Esubset X$ with diam $E<delta$.
Here is the solution from Baby Rudin that I'm following:
Suppose $f$ is uniformly continuous on a metric space $X$. By the definition, we know that for any $epsilon >0$, there is $delta>0$ such that $d(f(x), f(y))<frac{epsilon}{2}$ for any $x, y in X$ satisfying $d(x, y)<delta$. Thus, for all $E$ with $text{diam}(E)<delta$, if $x, y in E$,then $d(x, y)≤text{diam}(E)<delta$, so that $d(f(x), f(y))<frac{epsilon}{2}$. The last inequality implies $text{diam}(f(E))leq frac{epsilon}{2}<epsilon$.
I don't get the last implication. Can smb explain it? Thanks.
real-analysis metric-spaces proof-explanation
edited Nov 29 '18 at 0:41
Shaun
8,832113681
asked Nov 29 '18 at 0:15
dxdydzdxdydz
1869
$endgroup$
add a comment |
$begingroup$
This is from Rudin Chapter 4, problem 9:
Show that the requirement in the definition of uniform continuity can
be rephrased as follows, in terms of diameters of sets: To every $epsilon > 0$
there exists a $delta > 0$ such that diam $f(E)<epsilon$ for all $Esubset X$ with diam $E<delta$.
Here is the solution from Baby Rudin that I'm following:
Suppose $f$ is uniformly continuous on a metric space $X$. By the definition, we know that for any $epsilon >0$, there is $delta>0$ such that $d(f(x), f(y))<frac{epsilon}{2}$ for any $x, y in X$ satisfying $d(x, y)<delta$. Thus, for all $E$ with $text{diam}(E)<delta$, if $x, y in E$,then $d(x, y)≤text{diam}(E)<delta$, so that $d(f(x), f(y))<frac{epsilon}{2}$. The last inequality implies $text{diam}(f(E))leq frac{epsilon}{2}<epsilon$.
I don't get the last implication. Can smb explain it? Thanks.
real-analysis metric-spaces proof-explanation
edited Nov 29 '18 at 0:41
Shaun
8,832113681
asked Nov 29 '18 at 0:15
dxdydzdxdydz
1869
$endgroup$
$begingroup$
I would place the emphasis on what comes beforehand (i.e., "for all $E$ with $text{diam}(E)<delta$, if $x, y in E$, then $d(x, y)≤text{diam}(E)<delta$"), especially this bit: "if $x, y in E$, then". I don't know if that helps. Recall the definition of the diameter.
$endgroup$
– Shaun
Nov 29 '18 at 0:40
add a comment |
$begingroup$
This is from Rudin Chapter 4, problem 9:
Show that the requirement in the definition of uniform continuity can
be rephrased as follows, in terms of diameters of sets: To every $epsilon > 0$
there exists a $delta > 0$ such that diam $f(E)<epsilon$ for all $Esubset X$ with diam $E<delta$.
Here is the solution from Baby Rudin that I'm following:
Suppose $f$ is uniformly continuous on a metric space $X$. By the definition, we know that for any $epsilon >0$, there is $delta>0$ such that $d(f(x), f(y))<frac{epsilon}{2}$ for any $x, y in X$ satisfying $d(x, y)<delta$. Thus, for all $E$ with $text{diam}(E)<delta$, if $x, y in E$,then $d(x, y)≤text{diam}(E)<delta$, so that $d(f(x), f(y))<frac{epsilon}{2}$. The last inequality implies $text{diam}(f(E))leq frac{epsilon}{2}<epsilon$.
I don't get the last implication. Can smb explain it? Thanks.
real-analysis metric-spaces proof-explanation
edited Nov 29 '18 at 0:41
Shaun
8,832113681
asked Nov 29 '18 at 0:15
dxdydzdxdydz
1869
$endgroup$
This is from Rudin Chapter 4, problem 9:
Show that the requirement in the definition of uniform continuity can
be rephrased as follows, in terms of diameters of sets: To every $epsilon > 0$
there exists a $delta > 0$ such that diam $f(E)<epsilon$ for all $Esubset X$ with diam $E<delta$.
Here is the solution from Baby Rudin that I'm following:
Suppose $f$ is uniformly continuous on a metric space $X$. By the definition, we know that for any $epsilon >0$, there is $delta>0$ such that $d(f(x), f(y))<frac{epsilon}{2}$ for any $x, y in X$ satisfying $d(x, y)<delta$. Thus, for all $E$ with $text{diam}(E)<delta$, if $x, y in E$,then $d(x, y)≤text{diam}(E)<delta$, so that $d(f(x), f(y))<frac{epsilon}{2}$. The last inequality implies $text{diam}(f(E))leq frac{epsilon}{2}<epsilon$.
I don't get the last implication. Can smb explain it? Thanks.
real-analysis metric-spaces proof-explanation
real-analysis metric-spaces proof-explanation
edited Nov 29 '18 at 0:41
Shaun
8,832113681
asked Nov 29 '18 at 0:15
dxdydzdxdydz
1869
edited Nov 29 '18 at 0:41
Shaun
8,832113681
asked Nov 29 '18 at 0:15
dxdydzdxdydz
1869
edited Nov 29 '18 at 0:41
Shaun
8,832113681
edited Nov 29 '18 at 0:41
Shaun
8,832113681
edited Nov 29 '18 at 0:41
Shaun
8,832113681
8,832113681
asked Nov 29 '18 at 0:15
dxdydzdxdydz
1869
asked Nov 29 '18 at 0:15
dxdydzdxdydz
1869
asked Nov 29 '18 at 0:15
dxdydzdxdydz
1869
1869
$begingroup$
I would place the emphasis on what comes beforehand (i.e., "for all $E$ with $text{diam}(E)<delta$, if $x, y in E$, then $d(x, y)≤text{diam}(E)<delta$"), especially this bit: "if $x, y in E$, then". I don't know if that helps. Recall the definition of the diameter.
$endgroup$
– Shaun
Nov 29 '18 at 0:40
add a comment |
$begingroup$
I would place the emphasis on what comes beforehand (i.e., "for all $E$ with $text{diam}(E)<delta$, if $x, y in E$, then $d(x, y)≤text{diam}(E)<delta$"), especially this bit: "if $x, y in E$, then". I don't know if that helps. Recall the definition of the diameter.
$endgroup$
– Shaun
Nov 29 '18 at 0:40
$begingroup$
I would place the emphasis on what comes beforehand (i.e., "for all $E$ with $text{diam}(E)<delta$, if $x, y in E$, then $d(x, y)≤text{diam}(E)<delta$"), especially this bit: "if $x, y in E$, then". I don't know if that helps. Recall the definition of the diameter.
$endgroup$
– Shaun
Nov 29 '18 at 0:40
$begingroup$
I would place the emphasis on what comes beforehand (i.e., "for all $E$ with $text{diam}(E)<delta$, if $x, y in E$, then $d(x, y)≤text{diam}(E)<delta$"), especially this bit: "if $x, y in E$, then". I don't know if that helps. Recall the definition of the diameter.
$endgroup$
– Shaun
Nov 29 '18 at 0:40
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
By definition
$$
text{diam} (E)=sup_{x,yin E}d(x,y).
$$
For the proof, $epsilon/2$ is an upper bound of the set ${d(f(x), f(y))mid x,yin E}$. But the supremum is the least upper bound whence
$$
text{diam}(f(E))=sup_{x,y in E}d(f(x),f(y))leq epsilon/2<epsilon
$$
answered Nov 29 '18 at 0:38
Foobaz JohnFoobaz John
21.6k41352
$endgroup$
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%2f3017961%2frudin-chapter-4-problem-9-rephrasing-the-definition-of-uniform-continuity-un%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
$begingroup$
By definition
$$
text{diam} (E)=sup_{x,yin E}d(x,y).
$$
For the proof, $epsilon/2$ is an upper bound of the set ${d(f(x), f(y))mid x,yin E}$. But the supremum is the least upper bound whence
$$
text{diam}(f(E))=sup_{x,y in E}d(f(x),f(y))leq epsilon/2<epsilon
$$
answered Nov 29 '18 at 0:38
Foobaz JohnFoobaz John
21.6k41352
$endgroup$
add a comment |
$begingroup$
By definition
$$
text{diam} (E)=sup_{x,yin E}d(x,y).
$$
For the proof, $epsilon/2$ is an upper bound of the set ${d(f(x), f(y))mid x,yin E}$. But the supremum is the least upper bound whence
$$
text{diam}(f(E))=sup_{x,y in E}d(f(x),f(y))leq epsilon/2<epsilon
$$
answered Nov 29 '18 at 0:38
Foobaz JohnFoobaz John
21.6k41352
$endgroup$
add a comment |
$begingroup$
By definition
$$
text{diam} (E)=sup_{x,yin E}d(x,y).
$$
For the proof, $epsilon/2$ is an upper bound of the set ${d(f(x), f(y))mid x,yin E}$. But the supremum is the least upper bound whence
$$
text{diam}(f(E))=sup_{x,y in E}d(f(x),f(y))leq epsilon/2<epsilon
$$
answered Nov 29 '18 at 0:38
Foobaz JohnFoobaz John
21.6k41352
$endgroup$
By definition
$$
text{diam} (E)=sup_{x,yin E}d(x,y).
$$
For the proof, $epsilon/2$ is an upper bound of the set ${d(f(x), f(y))mid x,yin E}$. But the supremum is the least upper bound whence
$$
text{diam}(f(E))=sup_{x,y in E}d(f(x),f(y))leq epsilon/2<epsilon
$$
answered Nov 29 '18 at 0:38
Foobaz JohnFoobaz John
21.6k41352
answered Nov 29 '18 at 0:38
Foobaz JohnFoobaz John
21.6k41352
answered Nov 29 '18 at 0:38
Foobaz JohnFoobaz John
21.6k41352
answered Nov 29 '18 at 0:38
Foobaz JohnFoobaz John
21.6k41352
21.6k41352
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%2f3017961%2frudin-chapter-4-problem-9-rephrasing-the-definition-of-uniform-continuity-un%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
$begingroup$
I would place the emphasis on what comes beforehand (i.e., "for all $E$ with $text{diam}(E)<delta$, if $x, y in E$, then $d(x, y)≤text{diam}(E)<delta$"), especially this bit: "if $x, y in E$, then". I don't know if that helps. Recall the definition of the diameter.
$endgroup$
– Shaun
Nov 29 '18 at 0:40