In the finite complement topology on \(\mathbb{R}\), to what point or points does the sequence \(x_n = 1/n\) converge?
Solution
1. The sequence \(x_n=1/n\) converges to \(\boxed{\text{every point of }\RR}\).
Let \(x\in\RR\) and let \(U\) be an open neighborhood of \(x\). In the finite complement topology, \(U=\RR-F\) for a finite set \(F\). The terms \(1/n\) are distinct, so at most \(\abs{F}\) indices \(n\) satisfy \(1/n\in F\), and \(x_n\in U\) for all other \(n\). Hence every neighborhood of \(x\) contains \(x_n\) for all but finitely many \(n\), which is convergence \(x_n\to x\).
2. Q.E.D.
Step 1 answers the question.