Past exam of the mathematics course of the University of Cambridge / 2020 / ia / Paper 1 / 10F (Analysis I)

Choose a closed interval containing the bounded sequence. Bisect it and retain a half containing infinitely many terms; repeat. The resulting intervals are nested and their lengths tend to zero. By the nested interval theorem they have one common point $x$, and choosing successively indexed terms from these intervals produces a subsequence converging to $x$. This proves the Bolzano-Weierstrass theorem.

Boundedness of $(z_n)$ makes both real sequences $(x_n)$ and $(y_n)$ bounded. The Bolzano-Weierstrass theorem first gives a subsequence $(z_{n_j})$ for which $(x_{n_j})$ converges. Applying it again to $(y_{n_j})$ gives a subsubsequence for which the imaginary parts also converge. Since convergence of complex numbers is equivalent to convergence of their real parts and imaginary parts, this subsubsequence converges in $\mathbb{C}$.

Construct the nested index sequences inductively. Start with $N^{(0)}=(1,2,\dots )$. Having chosen $N^{(i-1)}$, apply the Bolzano-Weierstrass theorem to the bounded sequence of $i$th coordinates along those indices, and let $N^{(i)}$ be an increasing subsequence along which that coordinate converges. Because each new sequence is nested inside all earlier ones, coordinates $1,\dots ,i$ all converge along $N^{(i)}$.

Now take the diagonal subsequence argumentNestedness impliesso $(m_j)$ is strictly increasing. For every fixed $i$, the tail $(m_j{)}_{j\ge i}$ is a subsequence of $N^{(i)}$; hence $(x_{m_j}^{(i)})$ converges. This single diagonal subsequence works for every coordinate.