The Complex Numbers are Separable, a Different Proof

Written by Charlotte Ausel, 2024-02-22.

About the Proof

I wrote this proof for my course Honours Complex Variables a few weeks ago, which is a course (aka module) on complex analysis. This proof was quite fun because I went for quite a unique method, by doing a quick dip into the theory of basis representation of numbers. I liked it so much I decided to share it with you!

Claim

The complex numbers are separable. In other words, there exists a subset of $S \subseteq ℂ$ such that it is countable and it is dense in $ℂ$ (that is, for every $z \in ℂ$ and $ϵ \in ℝ, ϵ > 0$ , there exists a $w \in ℂ$ such that $| z - w | < ϵ$ ).

Proof

Note that $ℚ$ is countable. Then $S = {z \in ℂ | \forall x, y \in ℚ . z = x + i y}$ is countable too.

It remains to show that $S$ is dense. Note how every $z \in ℂ$ is either composed of only rationals ( $z = a + i b$ where $a, b \in ℚ$ ) or at least one irrational (where at least one of $a$ or $b$ is in the set $ℝ ∖ ℚ$ ).

If $z$ is composed of only rationals, let $w = z$ . Thus $| z - w | = 0 < ϵ$ as desired.

If $z$ is composed of at least one irrational, define a sequence $(v_{n})_{n \in ℕ}$ where each $v_{k} \in ℂ$ is defined such that $v_{k} = p_{k} + i q_{k}$ is the basis representation of $z$ where $p_{k}$ and $q_{k}$ are truncated after $k$ basis places (if you're not used to talking in arbitrary bases, just replace each mention of 'basis' with 'decimal').

The following are claimed without proof, but a proof is available on ProofWiki. First, that every real number has at least one basis representation[1]. For irrational numbers, their decimal expansion never terminates, or in other words, it has 'infinite' digits.[2]

Thus, by the monotone convergence theorem, $\lim_{n \to \infty} (v_{n}) = z$ , because $(v_{n})$ is bounded above by $z$ and is monotone (since with each successive digit, the magnitude of a real number increases or stays the same, and thus, $| v_{k} |$ will increase as $k$ increases).

Recall the definition of convergence: the sequence $(x_{n})$ converges to $L$ iff for all $ϵ \in ℝ, ϵ > 0$ , there exists a $N \in ℕ$ such that for all $n > N$ , $| x_{n} - l | < ϵ$ .

Since the above definition of convergence holds for all choices of $ϵ$ , this proves that $w \in S$ exists. Simply let $w = v_{k} \in (v_{n})$ such that $| z - w | < ϵ$ . Because $(v_{n})$ converges, there will always exist such a $w$ . $□$

References

"Basis Representation Theorem." ProofWiki, 2021-07-27. Accessed 2024-02-22. [Online] Available here. ↑
"Basis Expansion of Irrational Number." ProofWiki, 2023-12-03. Accessed 2024-02-22. [Online] Available here. ↑