Heine borel theorem on the complex plane I'm trying to understand this proof of the Heine-Borel theorem on the complex plane. I'm reading Lang's Complex Analysis (page 22):
 
I didn't understand the converse. Why there is a convergent subsequence $\{z_{n_1}\}$? Why there is a convergent sub-subsequence $\{z_{n_2}\}$? Why $a+ ib\in S$?
Thanks
 A: $\{x_n\}$ is a bounded real sequence, so by the real Heine-Borel theorem it has a convergent subsequence $\{x_{n_1}\}$. Then look just at the corresponding complex sequence $\{z_{n_1}\}$. It isn't convergent, but its real part $\{x_{n_1}\}$ is convergent and its imaginary part $\{y_{n_1}\}$ is still bounded. So use real Heine-Borel again and get a subsequence $\{z_{n_2}\}$ of the subsequence $\{z_{n_1}\}$, and now we see that the real and imaginary parts of $\{z_{n_2}\}$ both converge.
A: Here is how to prove that $c:=a+ib\in S$. Suppose $c\not\in S$. Since $S$ is closed, $\mathbb C\backslash S$ is open. Thus, there exists an $\varepsilon>0$ such that $B(c,\varepsilon)\subseteq \mathbb C\backslash S$. On the other hand, since $|z_{n_2}-c|\to0$, we have $|z_{n_2}-c|<\varepsilon$ for all sufficiently large $n_2\in\mathbb Z_{\ge1}$. This contradiction shows that $c\in S$.

Disclaimer: this proof is modelled after the one found at Lokenath Debnath, Piotr Mikusinski - Introduction to Hilbert spaces with applications, pp. 16.
