Showing something is complete is actally harder than showing something is incomplete, because the trick of looking for an counter example sequence that punch a hole in the space is no useful.

consider the element sequence \(x = (1, 1/2,/3, 1/4, 1/5, 0, \cdots)\) all the way to infinity. Then another sequence that is the truncation of the sequence is in the set \(M\), and it converges to \(x\), but it's not in \(M\). This subspace is incomplete.

This is also not hard to invoke the Cauchy criterion.

Just use \(n, n > \tan(N)\) and exploit the fact that limit of \(\arctan\) is \(\pi/2\). The sequence \(x_n = n\) does the trick. This time we use the Cauchy criterion for the convergence.

Check out these pathologies, they look quite fun.