I need to prove that if $\tau_1$ and $\tau_2$ are topologies of a set $X$, then $\tau_1 \cup \tau_2$ is not necessarily a topology on $X$.
I'm looking for counterexamples. I have one: consider $X=\{a,b,c\}$, and $\tau_1=\{\emptyset, X, \{a,c\}\}$ and $\tau_2=\{\emptyset, X, \{a,b\}\}$ then $\tau=\{\emptyset, X, \{a,c\}, \{a,b\}\}$, but note that: $\{a,b\} \cap \{a,c\} =\{a\}\notin \tau$, therefore $\tau$ is not a topology for $X$.
Is this correct?
Some other more interesting counterexample?
Thanks for your help.
Best Answer
Here is an example you might find "more interesting" (though I always like minimalist examples, like yours, very much) :
Take two topological spaces $X$ and $Y$, and on the product, the two following topologies : $$ \tau_X=\{U\times Y | U \mbox{ is an open of } X\} \\ \tau_Y=\{X\times V | V \mbox{ is an open of } Y\} $$ These both clearly are topologies, and their union is not (because $(U\times Y)\cap(X\times V)\notin\tau_X\cup\tau_Y$.