Is a space with no disjoint closed sets normal?

Question

Let $X=\{a,b,c\},T=\{\phi,X,\{b\},\{c\},\{b,c\}\}$

The closed sets are $X,\phi,\{a,c\},\{a,b\},\{a\}$.

Since there are no disjoint closed sets, is it correct to say that the space is normal?

Answer 1

Actually, what happens is that there are no two disjoint non-empty closed subsets of $X$. It follows from this that if $A_1,A_2\subset X$ are open disjoint non-empty sets, there are no closed disjoint sets $F_1$ and $F_2$ such that $A_1\subset F_1$ and $A_2\subset F_2$. Actually, in order for this justification to be complete, on has to add some example of a pair of open disjoint non-empty subsets of $X$. You can take, say, $\{b\}$ and $\{c\}$.

Answer 2

Yes if $X$ normal is defined as: for every pair of disjoint closed sets $A$ and $B$ there are disjoint open supersets, then yes in a space where there are no disjoint non-empty sets we have normality "voidly". The only remaining cases are where $A=\emptyset$ or $B=\emptyset$ and in that case we use the separating neighbourhoods $\emptyset$ and $X$, which are always open.

That's all there is to it. That's the reason that normality is often combined with $T_1$ (singletons are closed), which ensures that the condition actually means something (at least if we have two or more points...). (Often this combo is called $T_4$).