Set-theoretic equality of splitting fields within a fixed algebraic closure Let $F$ be a field and let $f(x)\in F[x]$ be a polynomial. Recall the following two facts:
(1) algebraic closures are unique up to isomorphism
(2) splitting fields are unique up to isomorphism
Fix an algebraic closure $\overline F$ of $F$. Is it true that any two (necessarily isomorphic) splitting fields $E,E'\subseteq \overline F$ for $f$ are equal $as$ $sets$ in addition to being isomorphic? 
My intuition tells me that this should be the case: fixing an algebraic closure allows us to fix roots of $f$ within this particular ambient field, and since any splitting field is the smallest field containing these roots, any two such fields must actually be equal as sets, in addition to being isomorphic. Does this sound about right? Is this totally trivial? 

I don't think they are equal as sets. Remember that in our way to get splitting field for $\;f(x)\in  F[x]\;$ , we first get (assuming $\;f\;$ is irreducible, otherwise we take one of its irreducible factors) the quotient ring (field) $\;K:=F[x]/\langle f(x)\rangle\;$ . 
Here, we already have no more $\;F\;$ but an isomorphic copy of $\;F\;$ within $\;K\;$ , and there could be several ways to obtain that copy...
