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$\boldsymbol{ba(\mathcal{P}(\mathbb{N}))}$ consisting of all bounded, finitely additive set functions defined on the $\sigma$-algebra $\mathcal{P}(\mathbb{N})$ of all subsets of the natural numbers. Thus, if $\mu: \mathcal{P}(\mathbb{N})\to \mathbb{K}$, then $\mu\in ba(\mathcal{P}(\mathbb{N}))$ if and only if $\mu(\varnothing)=0$;

$$ \mu\left(\bigcup\limits^{n}_{i=1}A_{i}\right)=\sum\limits^{n}_{i=1}\mu(A_{i}), $$

where $A_{1},A_{2},\ldots ,A_{n}$ are pairwise disjoint members of $\mathcal{P}(\mathbb{N})$; and $|\mu|(\mathbb{N})<\infty$, where for $E\in \mathcal{P}(\mathbb{N}):$

$$ |\mu|(E):=\sup\left\{ \sum\limits^{n}_{i=1}|\mu(A_{i})|: \ \bigcup\limits^{n}_{i=1}A_{i}=E \text{ and } A_{i}\cap A_{j}=\varnothing \text{ if } i\neq j \right\}. $$

If $\mu,\lambda\in ba(\mathcal{P}(\mathbb{N}))$ and $\alpha\in\mathbb{K}$ we define: $ (\mu+\lambda)(E):=\mu(E)+\lambda(E) $; $(\alpha\mu)(E):=\alpha\mu(E)$ and $\|\mu\|:=|\mu|(\mathbb{N})$. With this, $(ba(\mathcal{P}(\mathbb{N})), \|\cdot\|)$ is a Banach space.

Prove directly that $ba(\mathcal{P}(\mathbb{N}))$ is not separable.

Norbert
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2 Answers2

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This is just the dual space of $\ell^{\infty}$ (See The Duals of $l^\infty$ and $L^{\infty}$). If this were separable, then $\ell^{\infty}$ would be separable, which it is not.

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Prahlad Vaidyanathan
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  • Does this count as "direct"? I thought the idea was to find a big discrete set explicitly enough(ie zorn allowed) – Pipicito Sep 17 '13 at 06:03
  • Not sure how that would work. I tried various things, but since $\ell^1$ sits $w-\ast$-densely inside this, it was hard to find such a "big" discrete set. So, I gave up and posted this. – Prahlad Vaidyanathan Sep 17 '13 at 06:07
  • Why is the first variant of your answer is not correct? – Norbert Sep 17 '13 at 06:53
  • Because the number of finite subsets of $\mathbb{N}$ is countable. I thought it would give an uncountable family initially, until I thought through it more carefully. – Prahlad Vaidyanathan Sep 17 '13 at 08:22
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It may be worthwhile to have a look at these links in MO:

Explicit element of $(\ell^\infty)^*-\ell^1$?

What's an example of a space that needs the Hahn-Banach Theorem?

There is argued that to exhibit (explicitly, in a constructive sense) an element of $(\ell^\infty)^*-\ell^1$ you need some form of choice. So, there is not really a way to come up with an explicit element in $(\ell^\infty)^*-\ell^1$. Eric Schechter's book "Handbook of Analysis and its Foundations" is suggested as a possible reference for a proof of this.

I think that this is relevant to understand what would be a direct proof of the non-separability of the space here discussed.

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Pipicito
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