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 $S\subseteq \u2102$ such that it is countable and it is dense in $\u2102$ (that is, for every $z\in \u2102$ and $\u03f5\in \mathbb{R},\u03f5>0$, there exists a $w\in \u2102$ such that $|z-w|<\u03f5$ ).

Note that $\mathbb{Q}$ is countable. Then $S=\{z\in \u2102\phantom{\rule{0.2778em}{0ex}}|\phantom{\rule{0.2778em}{0ex}}\forall x,y\in \mathbb{Q}.\phantom{\rule{0.2778em}{0ex}}z=x+iy\}$ is countable too.

It remains to show that $S$ is dense. Note how every $z\in \u2102$ is either composed of only rationals ($z=a+ib$ where $a,b\in \mathbb{Q}$) or at least one irrational (where at least one of $a$ or $b$ is in the set $\mathbb{R}\setminus \mathbb{Q}$).

If $z$ is composed of only rationals, let $w=z$. Thus $|z-w|=0<\u03f5$ as desired.

If $z$ is composed of at least one irrational, define a sequence $({v}_{n}{)}_{n\in \mathbb{N}}$ where each ${v}_{k}\in \u2102$ 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, ${\mathrm{lim}}_{n\to \infty}({v}_{n})=z$, 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 $\u03f5\in \mathbb{R},\u03f5>0$, there exists a $N\in \mathbb{N}$ such that for all $n>N$, $|{x}_{n}-l|<\u03f5$.

Since the above definition of convergence holds for all choices of $\u03f5$, this proves that $w\in S$ exists. Simply let $w={v}_{k}\in ({v}_{n})$ such that $|z-w|<\u03f5$. Because $({v}_{n})$ converges, there will always exist such a $w$. $\square $

- "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. ↑