# Three closed sets in $T_4$ space which have an empty intersection.

Let $X$ be a $T_4$ space and $F_1 ,F_2, F_3 \subset X$ are closed sets with empty intersection $F_1 \cap F_2 \cap F_3 = \emptyset$. Prove that they have neighborhoods $U_1,U_2,U_3$ with empty intersection $U_1 \cap U_2 \cap U_3 = \emptyset$.

I started by letting $V_1 = F_2 \cap F_3 ,\ V_2 = F_1 \cap F_3 ,\ V_3 = F_2 \cap F_1$, these are closed sets so that for each $i=1,2,3$ we have $F_i \cap V_i = \emptyset$, thus $F_i , V_i$ can be separated by disjoint open sets $Q_i, Q_i '$ so $F_i \subset Q_i ,\ V_i \subset Q_i'$ and $Q_i \cap Q_i' =\emptyset$.

Clearly, I can't say anything about $Q_1 \cap Q_2 \cap Q_3$ yet, but I don't know how to proceed from here.

## 1 Answer

HINT: Use normality to get an open nbhd $U_1$ of $F_1$ such that $(\operatorname{cl}U_1)\cap(F_2\cap F_3)=\varnothing$, and let $C_1=\operatorname{cl}U_1$. Now replace $F_1$ by $C_1$ and do the same thing with $F_2$: let $U_2$ be an open nbhd of $F_2$ such that $(\operatorname{cl}U_2)\cap(C_1\cap F_3)=\varnothing$. Let $C_2=\operatorname{cl}U_2$, replace $F_2$ by $C_2$, and repeat the process once more with $F_3$.