I am proving that $\forall \; P$ closed, $\forall\; V$ open with $P\subset V$, exists $W$ an open set, $F\subset W\subset \overline{W}\subset V$ $\Longleftrightarrow$ $\forall\; P_1,P_2$ closed, $P_1\cap P_2=\emptyset$, then exists $U,V$ open with $P_1\subset U$, $P_2\subset V$ and $\overline{V}\cap\overline{U}$.

Both of the conditions are equivalent to $X$ normal.

The implication $\Longleftarrow$ has no problem.


Let $A,B$ be disjoint closed set. Then $A\subset X\setminus B$ and exists $V$ open such that $A\subset V \subset \overline{V}\subset X\setminus B$. This implies $A\subset V$ and $B\subset X\setminus \overline{V}$. The only thing left to prove is $\overline{V}\cap\overline{X\setminus\overline{V}}=\emptyset$.

However I get $\overline{V}\cap\overline{X\setminus\overline{V}}=\partial(\overline{V})$ \

EDIT: This question appears on Dugundji's "Topology"


Normally, for a topological space to be normal we only demand that for all $A,B \subseteq X$ that are closed and disjoint, we have $U_1,U_2$ open such that $A \subseteq U_1, B \subseteq U_2$ and $U_1 \cap U_2 = \emptyset$, instead of $\overline{U_1} \cap \overline{U_2} = \emptyset$, as you seem to demand.

If this is indeed what is meant, then you are done with your $V$ and $X \setminus \overline{V}$, as these are open and disjoint. Indeed will their closures not be disjoint in general.

However, if disjoint closures is what you want, use $A \subseteq V$ and $B \subseteq (X \setminus \overline{V})$ as the starting point, and apply the left hand side twice again, to find the required $U_1$ and $U_2$ with disjoint closures.

  • $\begingroup$ But how do I prove that $A\subset U_1$ and $B\subset U_2$? $\endgroup$ – user203327 Dec 29 '14 at 21:59
  • $\begingroup$ You apply the left hand side to $A$ and $V$ to get $U_1$, so it's automatical. Likewise for $B$ and $X \setminus \overline{V}$. $\endgroup$ – Henno Brandsma Dec 30 '14 at 6:22

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