7
votes
0answers
110 views

Without AC, which implications between the different definitions of amenability still hold?

More precisely, I would like to know which implications between the following definitions of amenability of a discrete countable (or even finitely generated) group can be proved to hold with only ZF ...
4
votes
1answer
149 views

Is it compatible with ZF to assume that every amenable discrete group is finite?

The question is in the title, amenability being understood as the existence of a left-invariant finitely additive probability measure on the group of interest. The case of countable groups is treated ...
7
votes
1answer
374 views

Non continuous Linear form on $E=C([0,1],\mathbb{R})$ without AC

Let's note $E=C([0,1],\mathbb{R})$ the Banach space of real continuous funtions from the [0,1] interval with the uniform norm. Is it possible to show a non-continuous linear form on $E$ exists ...
14
votes
1answer
351 views

Does ZF imply a weak version of Hahn-Banach?

I have encountered this when I was thinking about differentiability in Banach spaces. There, for $x\in X$ we usually need functionals $u\in X^*$ such that $|u|=1$ and $u(x)=|x|$. This is a simple ...
9
votes
1answer
304 views

Inequivalent complete norms and the axiom of choice

Hi, I've been wondering about the following : Is it possible, without the axiom of choice, to have two inequivalent complete norms on a vector space? All the examples of inequivalent complete norms ...
15
votes
1answer
1k views

Does Arzelà-Ascoli require choice?

Inspired by a recent Math.SE question entitled Where do we need the axiom of choice in Riemannian geometry?, I was thinking of the Arzelà--Ascoli theorem. Let's state a very simple version: ...
5
votes
0answers
1k views

Explicit element of $(\ell^{\infty})^* - \ell^1$? [duplicate]

Possible Duplicate: What’s an example of a space that needs the Hahn-Banach Theorem? It is well known that the dual of $\ell^{\infty}$ properly contains $\ell^1$ (over $\mathbb{N}$, ...