Past exam of the mathematics course of the University of Cambridge / 2018 / ib / Paper 4 / 13E (Metric and Topological Spaces)

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 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.

The cover $X={\bigcup }_{n\in X}{U}_n$ has no finite subcover, because every $U_n$ is finite. Hence $X$ is not compact.

For the stated pasting lemma for closed subspaces, if $C\subset Z$ is closed, thenis a union of two closed subsets of $Y$; therefore $f$ is continuous.

For $m,n\in X$, let $q=mn$. On $[0,\frac{1}{2}]$ define a path equal to $m$ before the endpoint and equal to $q$ at the endpoint; it is continuous because every open set containing $q$ also contains its divisor $m$. On $[\frac{1}{2},1]$ define it to equal $q$ at the first endpoint and $n$ thereafter. This is likewise continuous. The two closed-interval paths agree at $1/2$, so the pasting lemma joins them into a path from $m$ to $n$. Thus $X$ is path-connected.