A union of topologies is a topology

Question

I'm looking at a proof that the statement 'A union of topologies is a topology' is false. It goes as follows:

False. For instance, if $T = T_1 \cup T_2$ and $U_1 \in T_1$, but not $T_2$, while $U_1 \in T_2$, but not $T_1$, then $U_1, U_2 \in T$ , but it does not follow that $U_1 ∩ U_2 \in T$.

Surely if $U_1 \in T_1$, but not $T_2$, while $U_1 \in T_2$, but not $T_1$, then their intersection is empty, and the empty set is in the topology?

Answer 1

For $X=\{a,b,c\}$, let $\tau_{1}=\{X,\emptyset,\{a,b\}\}$, $\tau_{2}=\{X,\emptyset,\{b,c\}\}$ but $\{a,b\}\cap\{b,c\}=\{b\}\notin\tau_{1}\cup\tau_{2}$.