Show that any topological space \(X\) equipped with the indiscrete (trivial) topology is separable.
Solution
If \(X=\varnothing\), then \(X\) itself is a countable dense subset. Assume \(X\ne\varnothing\) and choose \(x\in X\).
1. In the indiscrete topology, the only nonempty open set is \(X\).
2. Hence every nonempty subset of \(X\) meets every nonempty open set, so every nonempty subset is dense.
3. In particular, the singleton \(\{x\}\) is finite and dense. Therefore \(X\) is separable.