Discussion
Let \(X\) denote \(\mathbb{R}^{2}\) endowed with the usual topology. Let \(Y\) denote \(\mathbb{R}\) endowed with the co-finite topology. If \(Z\) is the product topological space \(Y \times Y\), then
Answer: C
The topology of \(Z\) is strictly coarser (weaker) than that of \(X\).
A basic open set of \(Z\) is \(U \times V\) with \(U\) and \(V\) co-finite in \(\mathbb{R}\). The complement of a finite set is open in the usual topology, so \(U \times V\) is open in \(\mathbb{R}^{2}\) with the usual topology. Hence every open set of \(Z\) is open in \(X\).
The inclusion is strict. Any two non-empty open sets of \(Z\) intersect, because co-finite sets in \(\mathbb{R}\) always meet, so \(Z\) is not Hausdorff, while \(X\) is. For instance the open unit disc is open in \(X\) but contains no basic set \(U \times V\), each of which is unbounded.
So the two topologies are comparable and not equal, and it is \(Z\), not \(X\), that is coarser.
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