# Prove that $\ell_{\infty}$ and $\mathcal{L}^{\infty}$ are non-separable normed spaces [closed]

Prove that $\ell_{\infty}$ and $\mathcal{L}^{\infty}$ are non-separable normed spaces.

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What's the motivation? What have you tried? Is this homework? You should avoid posting questions as a command. – Tyler Apr 14 '11 at 21:08
Proposition 1.26, 1.27 (p.15) in Functional analysis and infinite-dimensional geometry By Marián J. Fabian et al. books.google.co.uk/… – Martin Sleziak Apr 14 '11 at 21:11
Dear Ig_7, I'm closing this question as too localized: if this is homework, please see the FAQ for how you should ask it. – Akhil Mathew Apr 14 '11 at 23:44
I am for not deleting this thread because of the answer below. – Jonas Meyer Aug 1 '12 at 6:25

## closed as too localized by t.b., Jonas Meyer, Akhil MathewApr 14 '11 at 23:42

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I'll just show it for $l_\infty$. I'm pretty sure that's enough. Also, showing that they are normed spaces is pretty straight forward.
Anyway, a space is separable if there is a countably dense subset. We can show $l_\infty$ is not separable by showing the existence of an uncountable subset such that any two elements of the subset cannot be closer than some fixed number. The reason for this is, put a ball of that fixed radius around each element of the uncountable set. Then, by density of the countable subset, we can put an element of the countable subset inside each one of those balls. But then the countable subset must be uncountable, a contradiction.
So, for $l_\infty(\mathbb{C})$, let $\{I_\alpha\}$ be the uncountable set of all subsets of $\mathbb{N}$. And let $\{a_n^\alpha\}$ be the sequence such that $a_n^\alpha=1$ if $n\in I_\alpha$ and $a_n^\alpha=0$ if not. Then the set of all such sequences (indicator sequences for subsets of $\mathbb{N}$ by the way), is uncountable. But, the distance between any two of them, which is equal to $sup_n\{a_n^\alpha-a_n^\beta\}$ must be equal to 1, since if it isn't, $I_\alpha=I_\beta$. So, we have our contradiction.
The case of $L^{\infty}$ is even easier. Just consider $\{\chi_{[-t,t]}\}_{t \in \mathbb{R}}$. – t.b. Apr 14 '11 at 21:38