Existence of a bijective choice function $f:\tau\rightarrow \mathbb{R}$

Question

If $\tau$ is the set of open sets on the real line, it is known that $|\tau| = |\mathbb{R}|$ (see this question).

Thus, it is feasible that there might exist a bijective choice function $f:\tau\rightarrow \mathbb{R}$. A choice function $f$ on $\tau$ is a function such that $\forall U\in \tau, f(U)\in U$.

Does such a bijection exist?

Answer 1

Here is a "back-and-forth" construction that uses AC. I assume there is a less wacky approach out there, but this is the first thing that comes to mind.

Enumerate all reals as $\{r_\alpha: \alpha < \mathfrak c\}$ and enumerate all nonempty open sets as $\{U_\alpha: \alpha < \mathfrak c\}.$ Construct a bijection by iterating the following two steps $\mathfrak c$-many times:

1. For the least-indexed $r_\alpha$ not matched up yet, there are $\mathfrak c$-many open sets containing $r_\alpha$. Fewer than $\mathfrak c$-many open sets have been matched up so far, so we can match up $r_\alpha$ with some yet-unmatched open set containing it.
2. The least-indexed $U_\alpha$ not matched up yet contains $\mathfrak c$-many reals. Fewer than $\mathfrak c$-many reals have been matched up so far, so we can match up $U_\alpha$ with some yet-unmatched real contained in it.

The fact that we went back-and-forth between the reals and open sets, matching up the least indexed element each time, guarantees that we have matched up every real and every open set once we've iterated through $\mathfrak c$ .