Proof that $c_0$ is separable (with respect to the $l_infty$ norm)
$begingroup$
I'm looking at the proof that $c_0$ is separable, but I don't understand the proof.In the proof below, it first shows that $S$ is separable, where $S$ is:
Next, it shows that $S$ is dense in $c_0$. However, $S$ is not a countable set, so how does this show that $c_0$ has a countable dense subset? I would greatly appreciate any explanation.
functional-analysis
asked Feb 25 '16 at 21:49
takecaretakecare
2,33221437
$endgroup$
|
show 1 more comment
$begingroup$
I'm looking at the proof that $c_0$ is separable, but I don't understand the proof.In the proof below, it first shows that $S$ is separable, where $S$ is:
Next, it shows that $S$ is dense in $c_0$. However, $S$ is not a countable set, so how does this show that $c_0$ has a countable dense subset? I would greatly appreciate any explanation.
functional-analysis
asked Feb 25 '16 at 21:49
takecaretakecare
2,33221437
$endgroup$
1
$begingroup$
$mathcal{S}$ is not countable, but the set of sequences in $mathcal{S}$ with rational entries is. That should be good enough.
$endgroup$
– carmichael561
Feb 25 '16 at 21:52
$begingroup$
How is that good enough?
$endgroup$
– takecare
Feb 25 '16 at 21:53
$begingroup$
Short version: If $A$ is separable, so is $overline{A}$ (a countable dense subset of $A$ is dense in $overline{A}$). A countable union of separable sets is separable. Finite-dimensional spaces are separable.
$endgroup$
– Daniel Fischer♦
Feb 25 '16 at 21:54
$begingroup$
It's good enough because any element of $c_0$ is eventually small, and the finitely many entries at the beginning can be approximated by rational numbers.
$endgroup$
– carmichael561
Feb 25 '16 at 21:56
$begingroup$
So you mean the set of rational terms of $S$ is also dense in $c_0$?
$endgroup$
– takecare
Feb 25 '16 at 21:56
|
show 1 more comment
$begingroup$
I'm looking at the proof that $c_0$ is separable, but I don't understand the proof.In the proof below, it first shows that $S$ is separable, where $S$ is:
Next, it shows that $S$ is dense in $c_0$. However, $S$ is not a countable set, so how does this show that $c_0$ has a countable dense subset? I would greatly appreciate any explanation.
functional-analysis
asked Feb 25 '16 at 21:49
takecaretakecare
2,33221437
$endgroup$
I'm looking at the proof that $c_0$ is separable, but I don't understand the proof.In the proof below, it first shows that $S$ is separable, where $S$ is:
Next, it shows that $S$ is dense in $c_0$. However, $S$ is not a countable set, so how does this show that $c_0$ has a countable dense subset? I would greatly appreciate any explanation.
functional-analysis
functional-analysis
asked Feb 25 '16 at 21:49
takecaretakecare
2,33221437
asked Feb 25 '16 at 21:49
takecaretakecare
2,33221437
asked Feb 25 '16 at 21:49
takecaretakecare
2,33221437
asked Feb 25 '16 at 21:49
takecaretakecare
2,33221437
asked Feb 25 '16 at 21:49
takecaretakecare
2,33221437
2,33221437
1
$begingroup$
$mathcal{S}$ is not countable, but the set of sequences in $mathcal{S}$ with rational entries is. That should be good enough.
$endgroup$
– carmichael561
Feb 25 '16 at 21:52
$begingroup$
How is that good enough?
$endgroup$
– takecare
Feb 25 '16 at 21:53
$begingroup$
Short version: If $A$ is separable, so is $overline{A}$ (a countable dense subset of $A$ is dense in $overline{A}$). A countable union of separable sets is separable. Finite-dimensional spaces are separable.
$endgroup$
– Daniel Fischer♦
Feb 25 '16 at 21:54
$begingroup$
It's good enough because any element of $c_0$ is eventually small, and the finitely many entries at the beginning can be approximated by rational numbers.
$endgroup$
– carmichael561
Feb 25 '16 at 21:56
$begingroup$
So you mean the set of rational terms of $S$ is also dense in $c_0$?
$endgroup$
– takecare
Feb 25 '16 at 21:56
|
show 1 more comment
1
$begingroup$
$mathcal{S}$ is not countable, but the set of sequences in $mathcal{S}$ with rational entries is. That should be good enough.
$endgroup$
– carmichael561
Feb 25 '16 at 21:52
$begingroup$
How is that good enough?
$endgroup$
– takecare
Feb 25 '16 at 21:53
$begingroup$
Short version: If $A$ is separable, so is $overline{A}$ (a countable dense subset of $A$ is dense in $overline{A}$). A countable union of separable sets is separable. Finite-dimensional spaces are separable.
$endgroup$
– Daniel Fischer♦
Feb 25 '16 at 21:54
$begingroup$
It's good enough because any element of $c_0$ is eventually small, and the finitely many entries at the beginning can be approximated by rational numbers.
$endgroup$
– carmichael561
Feb 25 '16 at 21:56
$begingroup$
So you mean the set of rational terms of $S$ is also dense in $c_0$?
$endgroup$
– takecare
Feb 25 '16 at 21:56
1
1
$begingroup$
$mathcal{S}$ is not countable, but the set of sequences in $mathcal{S}$ with rational entries is. That should be good enough.
$endgroup$
– carmichael561
Feb 25 '16 at 21:52
$begingroup$
$mathcal{S}$ is not countable, but the set of sequences in $mathcal{S}$ with rational entries is. That should be good enough.
$endgroup$
– carmichael561
Feb 25 '16 at 21:52
$begingroup$
How is that good enough?
$endgroup$
– takecare
Feb 25 '16 at 21:53
$begingroup$
How is that good enough?
$endgroup$
– takecare
Feb 25 '16 at 21:53
$begingroup$
Short version: If $A$ is separable, so is $overline{A}$ (a countable dense subset of $A$ is dense in $overline{A}$). A countable union of separable sets is separable. Finite-dimensional spaces are separable.
$endgroup$
– Daniel Fischer♦
Feb 25 '16 at 21:54
$begingroup$
Short version: If $A$ is separable, so is $overline{A}$ (a countable dense subset of $A$ is dense in $overline{A}$). A countable union of separable sets is separable. Finite-dimensional spaces are separable.
$endgroup$
– Daniel Fischer♦
Feb 25 '16 at 21:54
$begingroup$
It's good enough because any element of $c_0$ is eventually small, and the finitely many entries at the beginning can be approximated by rational numbers.
$endgroup$
– carmichael561
Feb 25 '16 at 21:56
$begingroup$
It's good enough because any element of $c_0$ is eventually small, and the finitely many entries at the beginning can be approximated by rational numbers.
$endgroup$
– carmichael561
Feb 25 '16 at 21:56
$begingroup$
So you mean the set of rational terms of $S$ is also dense in $c_0$?
$endgroup$
– takecare
Feb 25 '16 at 21:56
$begingroup$
So you mean the set of rational terms of $S$ is also dense in $c_0$?
$endgroup$
– takecare
Feb 25 '16 at 21:56
|
show 1 more comment
1 Answer
1
active
oldest
votes
$begingroup$
Define $S_k = {(x_1, x_2, ..., x_k, 0, 0, 0,...): x_iin mathcal{Q}, forall 1leq ileq k }$, and $S= bigcup_{kgeq 1} S_k$. For each $k$, $S_k$ is countable and so is their countable union $S$. Now you want to show that $S$ is dense in $c_0$.
Take any element $yin c_0$, $y=(y_1, y_2, y_3, ...)$, $forall epsilon >0, exists N$, s.t. $|y_n|<epsilon$ for $ngeq N$. Since we know $mathcal{Q}$ is dense in $mathcal{R}$, we can take $x=(x_1, ..., x_N, 0, 0, ...)in S$ with $|x_i - y_i|<epsilon, forall 1leq ileq N$. In this way, we can verify that $||x-y||_{infty} <epsilon$. Therefore $S$ is dense in $c_0$, and $c_0$ admits a countable dense subset, $c_0$ is separable by definition.
answered Nov 12 '18 at 2:27
Linfeng LiLinfeng Li
825
$endgroup$
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%2f1672344%2fproof-that-c-0-is-separable-with-respect-to-the-l-infty-norm%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
$begingroup$
Define $S_k = {(x_1, x_2, ..., x_k, 0, 0, 0,...): x_iin mathcal{Q}, forall 1leq ileq k }$, and $S= bigcup_{kgeq 1} S_k$. For each $k$, $S_k$ is countable and so is their countable union $S$. Now you want to show that $S$ is dense in $c_0$.
Take any element $yin c_0$, $y=(y_1, y_2, y_3, ...)$, $forall epsilon >0, exists N$, s.t. $|y_n|<epsilon$ for $ngeq N$. Since we know $mathcal{Q}$ is dense in $mathcal{R}$, we can take $x=(x_1, ..., x_N, 0, 0, ...)in S$ with $|x_i - y_i|<epsilon, forall 1leq ileq N$. In this way, we can verify that $||x-y||_{infty} <epsilon$. Therefore $S$ is dense in $c_0$, and $c_0$ admits a countable dense subset, $c_0$ is separable by definition.
answered Nov 12 '18 at 2:27
Linfeng LiLinfeng Li
825
$endgroup$
add a comment |
$begingroup$
Define $S_k = {(x_1, x_2, ..., x_k, 0, 0, 0,...): x_iin mathcal{Q}, forall 1leq ileq k }$, and $S= bigcup_{kgeq 1} S_k$. For each $k$, $S_k$ is countable and so is their countable union $S$. Now you want to show that $S$ is dense in $c_0$.
Take any element $yin c_0$, $y=(y_1, y_2, y_3, ...)$, $forall epsilon >0, exists N$, s.t. $|y_n|<epsilon$ for $ngeq N$. Since we know $mathcal{Q}$ is dense in $mathcal{R}$, we can take $x=(x_1, ..., x_N, 0, 0, ...)in S$ with $|x_i - y_i|<epsilon, forall 1leq ileq N$. In this way, we can verify that $||x-y||_{infty} <epsilon$. Therefore $S$ is dense in $c_0$, and $c_0$ admits a countable dense subset, $c_0$ is separable by definition.
answered Nov 12 '18 at 2:27
Linfeng LiLinfeng Li
825
$endgroup$
add a comment |
$begingroup$
Define $S_k = {(x_1, x_2, ..., x_k, 0, 0, 0,...): x_iin mathcal{Q}, forall 1leq ileq k }$, and $S= bigcup_{kgeq 1} S_k$. For each $k$, $S_k$ is countable and so is their countable union $S$. Now you want to show that $S$ is dense in $c_0$.
Take any element $yin c_0$, $y=(y_1, y_2, y_3, ...)$, $forall epsilon >0, exists N$, s.t. $|y_n|<epsilon$ for $ngeq N$. Since we know $mathcal{Q}$ is dense in $mathcal{R}$, we can take $x=(x_1, ..., x_N, 0, 0, ...)in S$ with $|x_i - y_i|<epsilon, forall 1leq ileq N$. In this way, we can verify that $||x-y||_{infty} <epsilon$. Therefore $S$ is dense in $c_0$, and $c_0$ admits a countable dense subset, $c_0$ is separable by definition.
answered Nov 12 '18 at 2:27
Linfeng LiLinfeng Li
825
$endgroup$
Define $S_k = {(x_1, x_2, ..., x_k, 0, 0, 0,...): x_iin mathcal{Q}, forall 1leq ileq k }$, and $S= bigcup_{kgeq 1} S_k$. For each $k$, $S_k$ is countable and so is their countable union $S$. Now you want to show that $S$ is dense in $c_0$.
Take any element $yin c_0$, $y=(y_1, y_2, y_3, ...)$, $forall epsilon >0, exists N$, s.t. $|y_n|<epsilon$ for $ngeq N$. Since we know $mathcal{Q}$ is dense in $mathcal{R}$, we can take $x=(x_1, ..., x_N, 0, 0, ...)in S$ with $|x_i - y_i|<epsilon, forall 1leq ileq N$. In this way, we can verify that $||x-y||_{infty} <epsilon$. Therefore $S$ is dense in $c_0$, and $c_0$ admits a countable dense subset, $c_0$ is separable by definition.
answered Nov 12 '18 at 2:27
Linfeng LiLinfeng Li
825
answered Nov 12 '18 at 2:27
Linfeng LiLinfeng Li
825
answered Nov 12 '18 at 2:27
Linfeng LiLinfeng Li
825
answered Nov 12 '18 at 2:27
Linfeng LiLinfeng Li
825
825
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%2f1672344%2fproof-that-c-0-is-separable-with-respect-to-the-l-infty-norm%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
1
$begingroup$
$mathcal{S}$ is not countable, but the set of sequences in $mathcal{S}$ with rational entries is. That should be good enough.
$endgroup$
– carmichael561
Feb 25 '16 at 21:52
$begingroup$
How is that good enough?
$endgroup$
– takecare
Feb 25 '16 at 21:53
$begingroup$
Short version: If $A$ is separable, so is $overline{A}$ (a countable dense subset of $A$ is dense in $overline{A}$). A countable union of separable sets is separable. Finite-dimensional spaces are separable.
$endgroup$
– Daniel Fischer♦
Feb 25 '16 at 21:54
$begingroup$
It's good enough because any element of $c_0$ is eventually small, and the finitely many entries at the beginning can be approximated by rational numbers.
$endgroup$
– carmichael561
Feb 25 '16 at 21:56
$begingroup$
So you mean the set of rational terms of $S$ is also dense in $c_0$?
$endgroup$
– takecare
Feb 25 '16 at 21:56