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!


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


Note that is countable. Then S = { z | x , y . z = x + i y } is countable too.

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

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 where each v k 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 ( 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 , ϵ > 0 , there exists a 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 S exists. Simply let w = v k ( v n ) such that | z w | < ϵ . Because ( v n ) converges, there will always exist such a w.


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