I am trying to work through a proof on a modification of the ascoli theorem that is supposed to hold in ZF (even without assuming countable choice).
My problem is within the part (1) $\Rightarrow$ (2b):
Let $F$ be a set of continous functions from $\mathbb{R} \to \mathbb{R}$ and let each sequence in $F$ have a subsequence converging continously to $g$. We want to show each countable subset $G$ of $F$ is equicontinous.
Suppose not, then there is a countable subset $G \subseteq F$ with elements $f_n$, $n \in \mathbb{N}$ which isn't equicontinous. By definition there exists $x \in \mathbb{r}$ and $\varepsilon > 0$ such that for all $\delta > 0$ there is $y \in \mathbb{R}$ such that $$|x - y| < \delta \text{ and } |f_n(x) - f_n(y)| \geq \varepsilon$$
In the next step we define $$\nu(n) := \min\left\{m \in \mathbb{N} \mid \text{there exists } y \in [x - 2^{-n}, x + 2^{-n}] \text{ with } |f_m(x) - f_m(y)| \geq \varepsilon \right\}$$ Because of the well ordering of the naturals $\nu(n)$ is well defined.
Now set $g_n := f_{\nu(n)}$ and define $$x_n := \min \left\{ y \in [x - 2^{-n}, x + 2^{-n}] \mid |g_n(x) - g_n(y)| \geq \varepsilon \right\}.$$
Why does this minimum exists? Is it maybe just an infimum? If so, does the property every bounded subset of $\mathbb{R}$ has an infimum hold in ZF?
The completeness of $\Bbb R$ holds in $\sf ZF$, the axiom of choice has absolutely nothing to do with it.
And it seems to me that $x_n$ is the infimum of a closed set, which is indeed the minimum.