I am well aware of the fact that any two splitting fields of a set of polynomials are isomorphic. However, I am wondering when two splitting fields are actually the same.
Fix an algebraic closure of $E$ of $F$. Then, if $\{ f_i \}$ are polynomials in $F[x]$ and $K_1$ and $K_2$ are splitting fields of $F$ contained in $E$ I want to say that $K_1 = K_2$. In the case of a finite set of polynomials, I can prove this by looking at the product of polynomials in $E[x]$. If $K_1 \neq K_2$, then the product of polynomials would have too many roots over $E[x]$ which would lead to a contradiction.
However, I am not sure in the case that $\{ f_i \}$ is an infinite as I cannot just take the product of polynomials. For reference, this question was asked here: Set-theoretic equality of splitting fields within a fixed algebraic closure in the case of a single polynomial. Does this result extend to the infinite case?
If $\alpha\in K_1$ is a root of $f_i$ for some $i$, then $\alpha\in K_2$, as $f_i$ splits in $K_2$. That is:$$0=f_i(\alpha)=\lambda \prod_{j=1}^{\deg f_i}(\alpha-\beta_j),$$ with $\beta_j\in K_2$ and $\lambda\neq0$, so $\alpha=\beta_j$ for some $j$.
Thus $K_1\subseteq K_2$ and vice versa by the same argument.