> [!definition] > > Let $(X, \topo)$ be a [[Topological Space|topological space]]. $X$ is **normal** if it is $T_1$ and that disjoint [[Closed Set|closed sets]] can be [[Separation Axioms|separated]] by [[Open Set|open]] [[Neighbourhood|neighbourhoods]]: > $ > \forall A, B \text{ closed}, A \cap B = \emptyset, \exists U \in \cn(A)^o, V \in \cn(B)^o: U \cap V = \emptyset > $ > [!theorem] Alternate Characterisation > > ![[normal_alternate.png|300]] > > A $T_1$ space $X$ is normal ($T_4$) if and only if for all $A \subset X$ closed and $V \in \cn(A)^o$, there exists $U \in \cn(A)^o$ such that $A \subset U \subset \ol{U} \subset V$. > > *Proof*. Suppose that $X$ is normal. Let $A \subset X$ be closed and $V \in \cn(A)^o$. Since $V^c$ is closed and $A \cap V^c = \emptyset$, there exists $U \in \cn(A)^o$ and $V' \in \cn(V^c)^o$ such that $U \cap V' = \emptyset$. As $(V')^c \supset U$ is closed, $(V')^c \supset \ol{U}$ and we have found $U \in \cn(A)^o$ such that > $ > A \subset U \subset \ol{U} \subset (V')^c \subset V > $ > > Suppose that the above conditions hold. Let $A, B \subset X$ be disjoint closed sets, then $B^c \in \cn(A)^o$. There exists $U \in \cn(A)^o$ such that > $ > A \subset U \subset \ol{U} \subset B^c > $ > and we have found $U \in \cn(A)^o$ and $V = \ol{U}^c \in \cn(B)^o$ such that $U \cap V = \emptyset$.