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Given the well known spaces of sequences: $$ l_\infty =\{(x_n), n\in \mathbb{N}, x_n \in \mathbb{R} : \sup_n |x_n|<\infty\} $$ $$ l_1= \{(x_n), n\in \mathbb{N}, x_n \in \mathbb{R} : \sum_n |x_n|<\infty\} $$ we can prove, by Hahn-Banach theorem, that $l_1\subsetneqq( l_\infty)^*$, where $( l_\infty)^*$ is the dual space of $ l_\infty$. But it seems that it's impossible to show explicitly an element of $( l_\infty)^*/l_1$.

Here I found that "it is impossible for an explicit example to be constructed", but there is not a proof of this statement. I found the same statement also here, but again without proof. I tried to find a proof but it's too hard for me. Someone can give me a sketch of the proof, or indicate an accessible source where I can find such proof? I'm interested to this question because it's connected with that Do we really need reals?, being an example of a non constructive existence theorem with an explicit proof of the fact that a constructive approach is impossible. Some one know other similar results?

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Emilio Novati
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1 Answers1

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The answer is given in both your sources, in the second one more explicitly. Shelah showed that, assuming ZF is consistent, ZF cannot prove that $(\ell^\infty)^\ast \neq \ell^1$. In particular, some non-constructive methods are needed to separate $(\ell^\infty)^\ast$ from $\ell^1$. Shelah's result is even stronger: even ZF+DC cannot prove that $(\ell^\infty)^\ast \neq \ell^1$ (Dependent Choice is a weak form of AC).

Here ZF is Zermelo–Fraenkel, the usual constructive set of axioms used for set theory. The non-constructive Axiom of Choice is also usually assumed, and it implies in particular the Hahn–Banach theorem, which in turn shows that $(\ell^\infty)^\ast \neq \ell^1$. Without AC, however, the Hahn–Banach theorem cannot be proved, as Shelah's result shows. Shelah proved his result by constructing a model of ZF+DC in which $(\ell^\infty)^\ast = \ell^1$. The exact paper of Shelah is probably referenced in Schechter's book Handbook of Analysis and its Foundations, but it might prove to be too technical for you (or me).

Yuval Filmus
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  • If I understand: because AC it's essential to prove $(l^\infty)^* \ne l^1$ and AC is not constructive then the elements that make $(l^\infty)^*$ a proper superset of $l^1$ cannot be defined in a constructive way. But isn't this reasoning a bit circular? Maybe i've to think about a lot more .... And waht about other exemples? – Emilio Novati Dec 28 '14 at 22:29
  • Another question (maybe stupid) is Hahn- Banach Th. equivalent to AC ? – Emilio Novati Dec 28 '14 at 22:35
  • Sorry! I have just now find that question: http://math.stackexchange.com/questions/285940/how-to-construct-an-explicit-element-of-ell-infty-mathbb-n-setminus?rq=1, so mine is a duplicate. – Emilio Novati Dec 28 '14 at 22:38
  • The Mathoverflow question you link to has an answer stating that the Hahn–Banach theorem is weaker than AC, though no reference is given. Must be common knowledge. – Yuval Filmus Dec 28 '14 at 22:42
  • If you want to prove that something cannot be done constructively, you have to put this statement in formal terms. Otherwise it cannot be proved. In this case, we identify "non-constructive" with "cannot be proved in ZF". If you have other definitions for "non-constructive", you should check whether they are implied by this one. – Yuval Filmus Dec 28 '14 at 22:46