ettolrach
(she/her)ettolrach
(she/her)Written by Charlotte Ausel, 2024-02-22.
The complex numbers are separable. In other words, there exists a subset of such that it is countable and it is dense in (that is, for every and , there exists a such that ).
Note that is countable. Then is countable too.
It remains to show that is dense. Note how every is either composed of only rationals ( where ) or at least one irrational (where at least one of or is in the set ).
If is composed of only rationals, let . Thus as desired.
If is composed of at least one irrational, define a sequence where each is defined such that is the basis representation of where and are truncated after 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, , because is bounded above by and is monotone (since with each successive digit, the magnitude of a real number increases or stays the same, and thus, will increase as increases).
Recall the definition of convergence: the sequence converges to iff for all , there exists a such that for all , .
Since the above definition of convergence holds for all choices of , this proves that exists. Simply let such that . Because converges, there will always exist such a .