# Describing the sup-closure within a poset without the Axiom of Choice

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.

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 $$a$$.