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I was reading a proof of this result, which says that there exists a continuous function separating 2 disjoint closed sets in a normal space, call them E and F.

The proof would construct an increasing sequence of closed sets indexed by the dyadic rationals. All of which contain one of the sets, say E, and are contained in the complement of the other set F.

In my mind's eye, it seems as if the increasing sets "work their way" to F from E. Does this mean that points "arbitrarily close" to F (in every open set containing F?) have a nonempty intersection with the closed sets in the dyadic rationals, or can the closed sets in the proof be separated from F by some open set that never intersects with them?

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It depends on a particular construction. Let $E,F\ne\varnothing$ and $f$ be the constructed continuous function such that $f(E)=\{0\}$ and $f(F)=\{1\}$. Then $f^{-1}(1)$ contains no neighborhood of $F$ iff each neighborhood of $F$ intersects with a closed sets in the dyadic rationals. In particular, if $F$ is open, then $F$ intersects with no closed set in the dyadic rationals.

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