Understanding Lang's proof that every closed and bounded set is compact

Question

In his book Serge Lang provides the following proof that every bounded set that is closed is also compact.

" Let S be a closed and bounded set. This implies that there exists $B$ s.t. $|z|\leq B$ where $z\in s$. For all $z$ write $z=x+iy$, then $|x|\leq B$ and $|y| \leq B$.

Let ${z_n}$ be a sequence in $S$, and write $z_n=x_n+iy_n$.

There is a subsequence {$z_{n_1}$} s.t. {${x_{n_1}}$} converges to a real number $a$ and similarly there is a subsequence {${z_{n_2}}$} s.t. {${y_{n_2}}$} converges to a real number $b$.

Then {$z_{n_2}= x_{n_2}+i{y_2}$} converges to $a+ib$ which implies that $S$ is compact."

What I don't understand is two things.

1. How does proving {$x_{n_1}$} converges to $a$ prove that {${x_{n_2}}$} converges to $a$ aswell?
2. How is it implied that there exist such subsequences such that {${x_{n_1}}$} and {${y_{n_2}}$} converge to these values? Afterall isn't this what we are suppose to prove?

If anybody could shine some light, I will greatly appreciate it.

Answer 1

This is indeed a mistake. You should take a convergent subsequence of $y_{n_1}$ which converges, not just a subsequence of $y_n$. In that case $x_{n_2}$ is a subsequence of $x_{n_1}$ and hence it converges to $a$ as well.

As for your second question, I believe the author assumes you already know that a bounded sequence of real numbers has a convergent subsequence (the Bolzano-Weierstrass theorem), and now he uses it to prove this property holds for complex sequences as well.

Answer 2

1. The sequence $\{x_{n_2}\}$ is a subsequence of the sequence $\{x_{n_1}\}$. Since $\{x_{n_1}\}$ converges to $a$, any of its subsequences also converges to $a$.
2. By the Bolzano-Weierstrass theorem, every bounded sequence of real numbers has a convergent subseqeunce.

Answer 3

As $\mathbb{R}^2=\mathbb{C}$, we may proceed by noting that:

Every closed and bounded set in $\mathbb{R}^2$ is compact.

Here is a proof:

Let $F\subseteq \mathbb{R}^2$ be both, closed and bounded. Let $(x_n)_n\subseteq F$ be a sequence. As $(x_n)_n$ is bounded, it follows from the Bolzano Weirstrass theorem that $(x_n)_n$ has a convergent subsequence, which converges to some $x\in \mathbb{R}^2$. As $F$ is closed, $x\in F$. $\square$