In: Advanced Math
( X, τ ) is normal if and only if for each closed
subset C of X and each open
set U such that C ⊆ U, there exists an open set V satisfies C ⊆ V
⊆clu( V) ⊆ U
Let be a normal topological space. Using the same notation as that in question note that and (the complement of in ) are disjoint closed subsets of (since ) and hence there exist open sets and of so that and
Now, we claim that
For, let if possible Since is open and it follows that which is a contradiction to above and hence and this implies that
Conversely, let be disjoint closed subsets of . Now, as by the hypothesis there exists open set such that and if we take then is open and and and this shows that is a normal topological space.