Given a set $S$, the closure of $S$ is sometimes defined as the intersection of all the closed sets that contain it. This type of argument is pervasive in mathematics when one want to construct the smallest object that still verifies some property.
Does this require the axiom of choice? What makes it possible to construct such intersection?
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Not everything uncountable, or infinite, requires the axiom of choice. The axiom of choice comes into play when we have to actually make choices.
In the case of a topological closure we don't have to make choices. We don't have to choose some closed sets which include $S$ in order to generate the closure. We take all of them. Taking all of them doesn't require any choice, because we can define what it means for a set to include $S$ and be closed, so the entire collection is definable and we can prove it's not empty, because the space itself is always a closed set including $S$.
No. Let $S\subseteq X$ be a set, and $\tau$ be the topology on $X$. Then $$\overline{S} = \left\{x\in S: \forall (T\subseteq S : X\setminus T\in \tau)[x\in T]\right\}$$ All we are using here is the axiom schema of specification.