Is this definition of the closure of a set circular? Let $(X,\mathcal{T})$ be a set and collection of open sets forming a topology on $X$.  For $A\subset X$ we define the closure of $A$, as 
$$
\overline{A} = \bigcap_{F\ \mathrm{is\ closed};\, F\supset A} F.
$$
This is a definition I have seen more than once.  My puzzlement is that if we assume this definition is proper, then one of the $F$'s turns out to be the closure of $A$.  So in a sense $\overline{A}$ appears on both sides of the equal sign in a definition.  I am sure I haven't found some paradox in topology so can someone correct my misunderstanding? 
 A: This is not circular at all. The power set of $X$ exists by the Axiom of Powers, you are taking the intersection of the family $\{F \in \mathscr{P}(X): F \text{ is closed }, F \supset A\}$ which is again a set by the Axiom of Specification. This set is not empty since $X$ is closed. Note that we never mentioned closures here, you can decide if a set is closed or not simply by inspecting $\tau$.
If you are not comfortable enough with this amount of set theory I would recommend you read some set theory book before attempting to study Topology. I recommend Naive Set Theory from Halmos (the first 30 or so pages are more relevant in this case).
A: No it is not. We know what closed sets are (complements of open sets), and that they are closed under arbitrary intersection. So the right hand side is just a special closed set $C$, and is well-defined. Just from the axioms. 
The way we made it, ensures that every closed set $C'$ that contains $A$ (there is at least one, namely $X$) is in the family we take an intersection of, so $C \subseteq C'$ for all such $C'$.
So it is the minimal closed set that contains $A$ (this is what the previous paragraph means). So this is a well-defined set, we can just define from $A$ and the topology alone. So we can call it $f(A)$ or, as is usual, $\overline{A}$ (sometimes $\operatorname{Cl}(A)$ as well). 
We do not assume the closure is well-defined. We show it is well-defined from the axioms and then introduce a notation for it, as it turns out to be a useful set to consider in relation to $A$.
A: The above definition for $\bar{A}$ is essentially saying that $\bar{A}$ is the $\textbf{smallest}$ closed set which contains A. But for sets which don't have any sort of measurable structure on them (and may have infinite or uncountably infinite numbers of members) how do we understand the notion of $\textbf{smallest}$?
The above definition describes how to construct this smallest closed set and so this is defining what we mean by smallest in this context. You are absolutely right that $\bar{A}$ will be one of these closed sets, but we won't know which one until we have taken the intersection of $\textbf{all}$ of the possible closed sets, what we are left with is then the closure of A,  $\bar{A}$. 
Without going through the process of taking all intersection we can't possibly know which closed set is indeed the closure we are looking for.
