The union of a sequence of funtions and its convergent point
$begingroup$
If ${h_n}_{n∈N} ⊂ C ([a, b])$ is $| cdot |_{infty}$ convergent to $h$, then $A ={h_n}∪{h}$ is $| cdot |_∞$-compact, $|cdot |_∞$-closed, $| cdot |_∞$-bounded and uniformly equicontinuous.
Is it proved in the same way that I prove $A$ is compact because of the Arzela-Ascoli theorem?
arzela-ascoli
edited Dec 8 '18 at 5:32
RRL
50k42573
asked Dec 8 '18 at 4:43
sofia de la morasofia de la mora
31
$endgroup$
add a comment |
$begingroup$
If ${h_n}_{n∈N} ⊂ C ([a, b])$ is $| cdot |_{infty}$ convergent to $h$, then $A ={h_n}∪{h}$ is $| cdot |_∞$-compact, $|cdot |_∞$-closed, $| cdot |_∞$-bounded and uniformly equicontinuous.
Is it proved in the same way that I prove $A$ is compact because of the Arzela-Ascoli theorem?
arzela-ascoli
edited Dec 8 '18 at 5:32
RRL
50k42573
asked Dec 8 '18 at 4:43
sofia de la morasofia de la mora
31
$endgroup$
add a comment |
$begingroup$
If ${h_n}_{n∈N} ⊂ C ([a, b])$ is $| cdot |_{infty}$ convergent to $h$, then $A ={h_n}∪{h}$ is $| cdot |_∞$-compact, $|cdot |_∞$-closed, $| cdot |_∞$-bounded and uniformly equicontinuous.
Is it proved in the same way that I prove $A$ is compact because of the Arzela-Ascoli theorem?
arzela-ascoli
edited Dec 8 '18 at 5:32
RRL
50k42573
asked Dec 8 '18 at 4:43
sofia de la morasofia de la mora
31
$endgroup$
If ${h_n}_{n∈N} ⊂ C ([a, b])$ is $| cdot |_{infty}$ convergent to $h$, then $A ={h_n}∪{h}$ is $| cdot |_∞$-compact, $|cdot |_∞$-closed, $| cdot |_∞$-bounded and uniformly equicontinuous.
Is it proved in the same way that I prove $A$ is compact because of the Arzela-Ascoli theorem?
arzela-ascoli
arzela-ascoli
edited Dec 8 '18 at 5:32
RRL
50k42573
asked Dec 8 '18 at 4:43
sofia de la morasofia de la mora
31
edited Dec 8 '18 at 5:32
RRL
50k42573
asked Dec 8 '18 at 4:43
sofia de la morasofia de la mora
31
edited Dec 8 '18 at 5:32
RRL
50k42573
edited Dec 8 '18 at 5:32
RRL
50k42573
edited Dec 8 '18 at 5:32
RRL
50k42573
50k42573
asked Dec 8 '18 at 4:43
sofia de la morasofia de la mora
31
asked Dec 8 '18 at 4:43
sofia de la morasofia de la mora
31
asked Dec 8 '18 at 4:43
sofia de la morasofia de la mora
31
31
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
No need for Arzela-Ascoli. In any metric space $x_n to x$ implies ${x,x_1,x_2,cdots}$ is compact and this follows easily from definition of compactness (using open covers). Just note that $x$ belongs to one of the sets in the open cover and this set also contains $x_n$ whenever $x$ is sufficently large. It follows that there is a finite subcover.
answered Dec 8 '18 at 5:06
Kavi Rama MurthyKavi Rama Murthy
55.2k42056
$endgroup$
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%2f3030702%2fthe-union-of-a-sequence-of-funtions-and-its-convergent-point%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
$begingroup$
No need for Arzela-Ascoli. In any metric space $x_n to x$ implies ${x,x_1,x_2,cdots}$ is compact and this follows easily from definition of compactness (using open covers). Just note that $x$ belongs to one of the sets in the open cover and this set also contains $x_n$ whenever $x$ is sufficently large. It follows that there is a finite subcover.
answered Dec 8 '18 at 5:06
Kavi Rama MurthyKavi Rama Murthy
55.2k42056
$endgroup$
add a comment |
$begingroup$
No need for Arzela-Ascoli. In any metric space $x_n to x$ implies ${x,x_1,x_2,cdots}$ is compact and this follows easily from definition of compactness (using open covers). Just note that $x$ belongs to one of the sets in the open cover and this set also contains $x_n$ whenever $x$ is sufficently large. It follows that there is a finite subcover.
answered Dec 8 '18 at 5:06
Kavi Rama MurthyKavi Rama Murthy
55.2k42056
$endgroup$
add a comment |
$begingroup$
No need for Arzela-Ascoli. In any metric space $x_n to x$ implies ${x,x_1,x_2,cdots}$ is compact and this follows easily from definition of compactness (using open covers). Just note that $x$ belongs to one of the sets in the open cover and this set also contains $x_n$ whenever $x$ is sufficently large. It follows that there is a finite subcover.
answered Dec 8 '18 at 5:06
Kavi Rama MurthyKavi Rama Murthy
55.2k42056
$endgroup$
No need for Arzela-Ascoli. In any metric space $x_n to x$ implies ${x,x_1,x_2,cdots}$ is compact and this follows easily from definition of compactness (using open covers). Just note that $x$ belongs to one of the sets in the open cover and this set also contains $x_n$ whenever $x$ is sufficently large. It follows that there is a finite subcover.
answered Dec 8 '18 at 5:06
Kavi Rama MurthyKavi Rama Murthy
55.2k42056
answered Dec 8 '18 at 5:06
Kavi Rama MurthyKavi Rama Murthy
55.2k42056
answered Dec 8 '18 at 5:06
Kavi Rama MurthyKavi Rama Murthy
55.2k42056
answered Dec 8 '18 at 5:06
Kavi Rama MurthyKavi Rama Murthy
55.2k42056
55.2k42056
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%2f3030702%2fthe-union-of-a-sequence-of-funtions-and-its-convergent-point%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
