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Let $(X,\preceq)$ be a partially ordered set. Fix $S\subseteq X$. Say $S$ is sup-closed if whenever $A$ is a nonempty subset of $S$ whose supremum exists, then $\sup A$ is an element of $S$.

Apparently one can describe the sup-closure of $S$ as the following set: $\{x\in X:\text{$x=\sup A$ for some nonempty subset $A$ of S}\}$. I managed to find a proof, but I helped myself to the Axiom of Choice in order to prove that the described set is indeed sup-closed. I am wondering if there is an argument which does not use AC? That is to say, is there a description of the sup-closure within a poset without the Axiom of Choice?

EDIT: Of course one can describe the sup-closure as the intersection of all sup-closed sets containing $S$. I guess the main question is if the particular description written above can be verified without AC.

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There is a maxim in non-choice proofs: When you cannot choose, take everything.

My guess is that you used choice in the proof in the following way: take $A$ to be a non-empty subset of the set you described, such that $\sup A$ exists. For each $a\in A$, choose $S_a$ such that $a=\sup S_a$, then $\sup A=\sup\bigcup S_a$, and therefore in the set.


But you can avoid choice by noting that $S_a=\{x\in S\mid x\leq a\}$ is a uniform choice of sets whose supremum is $a$.

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