# Tagged Questions

**7**

votes

**0**answers

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

**1**answer

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

**1**answer

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

**1**answer

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

**1**answer

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

**1**answer

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

**0**answers

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 $\mathbb{N}$, ...