Problem 1.10 Give an example of a set X and a collection of subsets of X s.t $\mathcal T$ which is not a topology on X but satisfies all the conditions for a topology except for arbitrary union over index sets; however union over countable index sets must hold.
let $ X=[0,1]$ then let $\mathcal T $ be all the countable subsets of X with X and $\emptyset $ clearly a countable union of countable sets is countable. and finite intersection of countable sets is countable. but if we take the the union of every singleton in the interval $[0,\frac{1}{2}]$ clearly the arbitrary union is an uncountably infinite set.
I don't like this example can anyone find a better one?
One example that is seen 'in the wild' is $\mathcal B(\mathbb R),$ the Borel $\sigma$-algebra on $\mathbb R,$ which the closure of all the open sets in $\mathbb R$ under countable unions, intersections and complements. It's a bit tricky to show it's not closed under arbitrary unions. You can observe that any singleton is in $\mathcal B(\mathbb R)$, so if it were closed under arbitrary unions, every subset of $\mathbb R$ would be in $\mathcal B(\mathbb R).$ But this can't be cause the cardinality of $\mathcal B(\mathbb R)$ is only $2^{\aleph_0},$ which can be shown by a transfinite induction argument. (Alternatively, it is probably easier to show that the Vitali set is not in $\mathcal B (\mathbb R).$ )