In my current obsession to learn equivalences of the Ultrafilter Lemma, I found the Handbook of Analysis and its Foundations, by Eric Schechter - by the way, a marvelous book in my opinion.
In this book (section 28.29), the proof that Banach-Alaoglu Theorem implies (in ZF) the Ultrafilter Lemma goes like this:
- we fix a set $\Omega$ and a proper filter $\mathcal{F}$ on $\Omega$;
- we set $X=\{f\colon \Omega\to\mathbb{R}|f$ is bounded$\}$ equipped with the sup norm;
- by Banach-Alaoglu, the closed unit ball on $X^*$ is $w^*$-compact (compact in the weak$^*$ topology), call it $V$;
- the map $\varphi\colon \Omega\to V$ defined as $\varphi_\omega(x)=x(\omega)$ for $\omega\in \Omega$ and $x\in X$ is an injection, so we may suppose that $\Omega\subset V$;
- the set $\mathcal{K}=\{w^*\text{-cl}(F): F\in\mathcal{F}\}$ is a family of $w^*$-closed subsets of $V$ with the finite intersection property, hence there exists $v\in \bigcap\mathcal{K}$;
- we define $\mu\colon \wp(\Omega)\to\mathbb{R}$ by setting $\mu(S)=v(1_S)$, where $1_S\colon \Omega\to \{0,1\}$ is the characteristic function of $S$;
in order to see that $\mathfrak{u}=\{G\subset\Omega:\mu(G)=1\}$ is an ultrafilter that contains $\mathcal{F}$, the author uses the following fact
for any fixed $F\in\mathcal{F}$, since $v\in w^*\text{-cl}(F)$, there is some net $(\omega(\alpha):\alpha\in A)$ in $F$ such that $\varphi_{\omega(\alpha)}\stackrel{w^*}{\longrightarrow}v$.
My problem concerns this last step. I know that in any topological space $Y$, $y\in Y$ and $A\subset Y$,
$(\dagger)$ $\quad$ $y\in\overline{A}$ if and only if there exists a net of elements in $A$ converging to $y$.
However, the only proof I know about this fact uses the Axiom of Choice.
Finally, my questions: There is a way to prove $(\dagger)$ without the Axiom of Choice? If not, what am I missing in order to prove the 8th step?
Thank you.
I am aware of this post, but my question concerns specifically about the proof in Schechter's book.
