Since the definition of a normal space is:
I know that:
$1-$A space $X$ is said to be normal if all singleton sets are closed and if for any two disjoint closed subsets $A, B\subseteq X$ there are open sets $U,V \subseteq X$ such that $$A \subseteq U, B \subseteq V, \textbf{ and } U \cap V = \emptyset.$$
$2-$ A space $X$ is a Hausdorff if for every pair of distinct points $a,b \in X,$ there are open sets $U,V \subseteq X$ such that $$a \in U, b \in V, \textbf{ and } U \cap V = \emptyset.$$
My intuition is YES any normal space Hausdorff space. Am I correct? If so, Is there a good way to write a justification for this ?
Best Answer
It's straightforward from the given definition (which should rather use 'and' instead of 'or' connecting the two conditions):
Just take $A:=\{a\}$ and $B:=\{b\}$ for given distinct points $a,b$.