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I am trying to construct a choice function on the open subset of $\mathbb{R}^2$ (with the topology induced by the euclidiean distance)

I cannot use maximum and minimum because if the set is open I might not have these. My first try is to find a choice function on the open sets of $\mathbb{R}$ to take the product of this choice function to have one on $\mathbb{R}^2$.

I know that the open sets of $\mathbb{R}$ are the union of open interval so if $I=]a,b[$ then $i=\frac{sup(I)+inf(I)}{2} \in I$ so I have a choice function on any non-empty interval but I do not know how to generalize to an explicit choice function on the non-connected open set. Is it possible to construct one ? (I know that it is not possible to construct explicitly a choice function on the borelian).

Camolistia
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    Not clear. Do you want a choice function whose domain is the set of open components ($\subseteq$-maximal connected open subsets) of a non-empty open subset of $\Bbb R^n$? If so, let ${x_m:m\in\Bbb N}$ be an enumeration of $\Bbb Q^n$ and for a component $C$, let $g(C)=\min {m: x_m\in C}$ and let $f(C)=x_{g(C)}$. – DanielWainfleet Feb 14 '23 at 10:10
  • I want the domain of my choice function to be the set of open subset of $\mathbb{R}^2$, not only the open components – Camolistia Feb 14 '23 at 10:38
  • The understanding of your question is impeded by some grammatical mistakes. I think both in your comment above and in the first sentence of the question you meant "open subsets" in the plural? – joriki Feb 14 '23 at 11:23
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    As in my previous comment, then define $g(C)$ and $f(C)$ for any non-empty open $C\subset \Bbb R^n$. – DanielWainfleet Feb 14 '23 at 11:26
  • yes ! i mean open subsets in plural (I'm not a native speaker and struggle with grammar) – Camolistia Feb 14 '23 at 12:11

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