1

Fix a point $z \in \mathbb{C}$. Let $S \subset \mathbb{C}$ be a nonempty closed set. Let $$T = \{ w \mid d(z, w) = d(z, S)\}.$$ Then $S \cap T$ is a nonempty closed set. Let $f\colon [0, 1] \to T$ be defined by $t \mapsto z + d(z, S)e^{2\pi it}$. Then $f^{-1}(S \cap T)$ is closed subset of $[0, 1]$. Thus $f(\inf f^{-1}(S \cap T)) \in S$.

By this way we can make a choice function for the entire family of nonempty closed subsets of $\mathbb{C}$. Is this way independent of $\mathsf{AC}$? And can this method extended to other Euclidean spaces, or even metric spaces and topological spaces?

Ris
  • 1,292

1 Answers1

1

Yep, this works (except I think you mean for $f$ to be $t\mapsto z+d(z,S)e^{2\pi it}$). There are a variety of similar techniques you can use for $\mathbb{R}^n$ for any $n$; the general idea is you can isolate a specific point by taking maxima or minima of certain functions (which must be achieved since the set is closed). For instance, given a nonempty closed set $S\subseteq\mathbb{R}^n$, you can replace $S$ with $S\cap [-N,N]^n$ for the least $n$ such that it is nonempty to assume $S$ is bounded. Then, you can take the smallest first coordinate of any point in $S$, then the smallest second coordinate among points with that first coordinate, and so on, to pick a single element of $S$.

To choose points from nonempty closed sets in arbitrary metric spaces is equivalent to AC, though, since you can just equip any set with the discrete metric so every set is closed.

Eric Wofsey
  • 330,363