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I have to prove that $\ell^{\infty}(\mathbb N)$ is not separable.

My attempt

Consider a SUBSET $V$ of $\ell^{\infty}(\mathbb N)$ consisting of bounded sequences that have only $0$, $1$ entries, e.g. $(0,1,1,0,0,0,0,1,0,0,\dots)$

Now assume that this SUBSET is uncountable.

If we take a radius $r=1/4$ then balls with origins that are elements of $V$ would be disjoint and because any base of $\ell^{\infty}(\mathbb N)$ must contain a subset of each element of the set of these balls, base of $\ell^{\infty}(\mathbb N)$ can't be countable so it doesnt satisfy the second axiom fo countability and thus is not separable (since $\ell^{\infty}(\mathbb N)$ is a metric space).

Could someone check this? And I still need to prove that $V$ is uncountable...

Martin Sleziak
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luka5z
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    What you have is right. A bit problematic is using the word "subspace" for $V$, since in the context of (topological) vector spaces, "subspace" usually denotes linear subspaces. Use "subset" to avoid that. Regarding the uncountability of $V$, can you draw a connection between $V$ and $\mathfrak{P}(\mathbb{N})$? – Daniel Fischer Mar 06 '14 at 20:53
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    Sorry, by separable, do you mean that you'd like to prove that $\ell^\infty$ has no countable dense subset? If so, it seems you're going in the wrong direction... – Ben Grossmann Mar 06 '14 at 20:56
  • Exactly, no countable dense subset. Could you tell me what is wrong? thanks – luka5z Mar 06 '14 at 20:58
  • Daniel, you mean connection between the power set of $N$ and $V$? Your remark about subspace vs subset is 100% correct. – luka5z Mar 06 '14 at 20:59
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    Thre is no neeed to assume that your set $V$ is uncountable, for it is uncountable. You can do a standard diagonal argument to show this; notice that there is an obvious bijeection between your $V$ and the set of subsets of $\mathbb N$. – Mariano Suárez-Álvarez Mar 06 '14 at 21:02
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    My approach would be as follows: given a countable subset ${x_n}$ of $\ell^\infty$, find a $y$ such that $|y-x_n| \geq 1$ for each $n$. – Ben Grossmann Mar 06 '14 at 21:02
  • @luka5z :Don't you need finitely-many 1's in your sequence of 0's and 1s,, so that it is in $l^{\infty}$? – Gary. Jun 27 '15 at 04:03
  • No. From definition $\ell ^{\infty} (\mathbb{N})$ is space of all sequences of natural numbers, such that each element of that sequence is finite. – luka5z Jun 27 '15 at 15:37

1 Answers1

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As the OP guessed, the right subset to consider is the set $A\subset\ell^\infty$ consisting of all the sequences with zeros and ones, i.e., $\{a_n\}\in A$ if and only if $a_n\in\{0,1\}$, for all $n$.

Clearly, if $\{a_n\},\{b_n\}\in A$ and $\{a_n\}\ne\{b_n\}$, then $$ \|\{a_n\}-\{b_n\}\|_\infty=1, $$ since there exists at least one $n$, for which $a_n\ne b_n$, and hence $|a_n-b_n|=1$.

Also $|A|=\big\lvert 2^{\mathbb N}\big\rvert= 2^{\aleph_0}$. Hence $A$ is uncountable, and in fact equinumerous to $\mathbb R$.

The open sets $$ \big\{B\big(\{a_n\},\tfrac{1}{3}\big): \{a_n\}\in A\big\}, $$ are uncountably many mutually disjoint open sets. The fact that such a collection of open sets exists means that $\ell^\infty$ is not separable.

Yiorgos S. Smyrlis
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