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I said if $b$ is an ement of $\displaystyle G \cap \cap_{n \in \mathbb{K}}F(n)$ then $b$ is in both $G$ and $F(n)$ for every $n \in \mathbb{K}$. Say for example that there are $m$ lots of $n$ in $\mathbb{K}$ and denote these by $n_{1}, n_{2}, \cdots n_{m}$. We can pair each of $f(n_1), f(n_2), f(n_3), \cdots, f(n_m)$ with $G$ i.e. $b \in G \cap F(n_{1})$, $b \in G \cap F(n_{2})$, $b \in G \cap F(n_{3})$, $\cdots$, $b \in G \cap F(n_{m}).$ That's $b \in \cap_{n \in \mathbb{K}}(F(n) \cap B).$ Working backwards, implies $b \in G \cap F(n_{1})$, $b \in G \cap F(n_{2})$, $b \in G \cap F(n_{3})$, $\cdots$, $b \in G \cap F(n_{m})$. And we just get rid of the pairing - $b \in G$ and $b \in [f(n_1), f(n_2), f(n_3), \cdots, f(n_m)]$ i.e. $b \in G \cap \cap_{n \in \mathbb{K}}F(n)$. Therefore $\displaystyle G \cap \cap_{n \in \mathbb{K}}F(n) = \cap_{n \in \mathbb{K}}(F(n) \cap B).$

I'm pretty sure the above is pretty crap. Could someone please post up a proof - I'm completely lost on how to write maths proofs. I can prove bit more difficult stuff, but the fact that it appears something so obvious puts me off completely (I'm not sure how much to leave out etc).