Is the field of complex numbers complete?

Question

I know that the complex numbers can't be ordered, does this imply that the field $\mathbb{F}$ of $\mathbb{C}$ is not complete? do we need any other proof other than what I stated? Can we use the same logic for quaternions?

Answer 1

In the context of the real numbers, every bounded sequence admits a convergent subsequence. Since every Cauchy sequence is bounded, it admits a convergent subsequence, whence we conclude that it converges (why?). Consequently, the set of real numbers is complete. Based on the completeness of the reals, you can prove the completeness of the complex numbers as follows.

Suppose that $z_{n} = (x_{n},y_{n})\in\mathbb{C}$ is a Cauchy sequence of complex numbers. This means that: \begin{align*} (\forall\varepsilon\in\mathbb{R}_{>0})(\exists n_{\varepsilon}\in\mathbb{N})(\forall m,n\in\mathbb{N})(m\geq n \geq n_{\varepsilon} \Rightarrow |z_{m} - z_{n}| < \varepsilon) \end{align*}

Given that $|z_{m} - z_{n}| \geq |x_{m} - x_{n}|$ and $|z_{m} - z_{n}| \geq |y_{m} - y_{n}|$, we conclude that $x_{n}\in\mathbb{R}$ and $y_{n}\in\mathbb{R}$ are Cauchy sequences of real numbers. Consequently, they converge to $x$ and $y$, let's say. If we denote by $z = (x,y)$, it results that $z_{n}$ converges to $z$ (why?), whence we conclude that $\mathbb{C}$ is complete.

Hopefully this helps!

Answer 2

A simple Google search yielded this:"In mathematics, a complete field is a field equipped with a metric and complete with respect to that metric. Basic examples include the real numbers, the complex numbers, and complete valued fields (such as the p-adic numbers)." If this is what you mean by "complete", then yes the field of complex numbers is complete. Being equipped with a metric and being complete with respect to that metric means the field is equipped with a distance function and that the field is complete with respect to that distance function. I hope this answers your question.