Mixing
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
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$begingroup$
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
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add a comment |
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$begingroup$
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 |
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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