= Solution
Since $U_m\cap U_n=U_{\gcd(m,n)}$ when the greatest common divisor is at least $2$, and is empty otherwise, finite intersections of the $U_n$ are again among the allowed unions. Arbitrary unions are built into the definition; the empty union is empty, and $X=\bigcup_{n\in X}U_n$. Thus these sets form a <topology>.
The space is \b[not Hausdorff]: every neighborhood of $4$ contains its divisor $2$, and every neighborhood of $2$ contains $2$, so $2$ and $4$ cannot have disjoint neighborhoods.
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