A set is said to have the countable-closed topology if the closed sets are the countable sets together with the empty set. If \(X\) is a topological space with an uncountably infinite number of points is the diagonal \(\Delta:=\{(x,x)\mid x\in X\}\) closed in the finite complement topology? Justify your answer carefully.
Solution
The statement defines the countable-closed topology and then asks about the finite-complement topology. The answer is no for each of three readings: the cofinite topology on \(X\times X\), the product of cofinite topologies on \(X\), and the product of countable-closed topologies on \(X\).
1. If “the finite complement topology” means the cofinite topology directly on the set \(X\times X\), then \(\Delta\) is not closed.
In the cofinite topology on an infinite set, the proper closed subsets are precisely the finite subsets. Since \(X\) is uncountable, the diagonal \[ \Delta=\{(x,x):x\in X\} \] is in bijection with \(X\) and is therefore uncountable, hence infinite. It is also a proper subset of \(X\times X\) because \(X\) has at least two distinct points, so an off-diagonal pair belongs to its complement. Thus \(\Delta\) is neither finite nor all of \(X\times X\), and hence is not closed in the cofinite topology on \(X\times X\).
2. If \(X\) carries the cofinite topology and \(X\times X\) carries the product topology, then \(\Delta\) is not closed.
Choose distinct points \(a,b\in X\). Suppose there were a basic product neighborhood \[ (a,b)\in U\times V \] with \[ (U\times V)\cap\Delta=\varnothing, \] where \(U,V\) are nonempty cofinite open subsets of \(X\). The intersection \(U\cap V\) is cofinite, hence nonempty because \(X\) is infinite. Choose \(z\in U\cap V\). Then \[ (z,z)\in U\times V \] and also \((z,z)\in\Delta\), a contradiction. Therefore every basic neighborhood of the off-diagonal point \((a,b)\) meets \(\Delta\). Hence \[ (a,b)\in\overline\Delta\setminus\Delta, \] so \(\Delta\) is not closed.
3. If the intended topology was instead the countable-closed topology defined in the first sentence, with the product topology on \(X\times X\), then \(\Delta\) is again not closed.
The open nonempty subsets of the countable-closed topology are the complements of countable subsets of \(X\). If \(U,V\) are two such nonempty open sets, then \[ X\setminus(U\cap V) =(X\setminus U)\cup(X\setminus V) \] is countable. Since \(X\) is uncountable, \(U\cap V\) is nonempty.
Now repeat the argument of 2: for any off-diagonal point \((a,b)\) and any basic product neighborhood \(U\times V\) of it, choose \(z\in U\cap V\). Then \((z,z)\in(U\times V)\cap\Delta\). Thus every off-diagonal point lies in \(\overline\Delta\), so in particular \(\Delta\) is not closed.
4. Thus the source ambiguity does not affect the conclusion: the diagonal is not closed.
3. All three yield the same negative answer.