A complex function defined on the natural numbers is a…

A complex function defined on the natural numbers is a sequence. Prove that a bounded sequence has at least one accumulation point.

Step 1 · Bound the real and imaginary parts

Suppose $|z_n|\le M$ for all $n$. Writing $z_n=x_n+iy_n$: since $|x_n|,|y_n|\le|z_n|\le M$, both $(x_n)$ and $(y_n)$ are bounded real sequences.

Step 2 · Extract a convergent subsequence of $(x_n)$

By the real Bolzano–Weierstrass theorem, there is a subsequence $(x_{n_k})$ with $x_{n_k}\to x_0$ for some real $x_0$.

Step 3 · Extract a convergent sub-subsequence of $(y_{n_k})$

The sequence $(y_{n_k})$ is also bounded, so it has a further subsequence $(y_{n_{k_j}})$ with $y_{n_{k_j}}\to y_0$ for some real $y_0$.

Step 4 · Recombine

Along the doubly-extracted indices $n_{k_j}$: $\ x_{n_{k_j}}\to x_0$ (a subsequence of a convergent sequence still converges to the same limit) and $y_{n_{k_j}}\to y_0$. Set $z_0=x_0+iy_0$. Then:

$$ |z_{n_{k_j}}-z_0|\le|x_{n_{k_j}}-x_0|+|y_{n_{k_j}}-y_0|\ \xrightarrow[j\to\infty]{}\ 0 $$

Given $\varepsilon>0$, choose $J$ so that both terms are below $\varepsilon/2$ for $j>J$; then $|z_{n_{k_j}}-z_0|<\varepsilon$.

Conclusion

$$ \boxed{z_{n_{k_j}}\to z_0 \text{ as } j\to\infty} $$

so $z_0$ is an accumulation point of $(z_n)$.