1
$\begingroup$

I'm trying to show that the set $\{a+bi\mid a,b\in\mathbb{Q}\}$ is dense in $\mathbb{C}$ with the standard metric.

I understand that to prove this I want to show that every element in $\mathbb{C}$ is a limit point of my set.

Can I say that since both $a,b\in\mathbb{Q}$ which is dense in $\mathbb{R}$ and $\mathbb{R}$ $\subset$ $\mathbb{C}$ $\implies$ there must exist some $\epsilon > 0$ such that $d(a,b) < \epsilon$ thus my set is dense in $\mathbb{C}$?

  • 1
    Yes, you can say that $|z_1-z_2| \le |x_1-x_2|+|y_1-y_2|$ (triangle inequality) – 2017-02-03

2 Answers 2

3

You just need to show that, for each $x+yi$ (with $x,y\in\mathbb{R}$) and each $r>0$, there is a complex number $p+qi$ with rational coordinates such that $d(x+yi,p+qi)

Find $p$ and $q$ rational such that $|x-p|

The triangle inequality $|z_1+z_2|\le|z_1|+|z_2|$ is valid in $\mathbb{C}$: here I used $z_1=x-p$ and $z_2=(y-q)i$.

The proof is now complete, because every open set contains a complex number with rational coefficients. If you need to find a sequence, choose $p_n,q_n$ so that $|x-p_n|<1/(2n)$ and $|y-q_n|<1/(2n)$. Then $p_n+q_ni\to x+yi$.

  • 0
    Thank you for the clarity, I understand now! – 2017-02-03
1

For reales $a,b,c,d$:

$$|(a+ib)-(c+id)|=\sqrt{(a-c)^2 +(d-b)^2}\le |a-c|+|d-b|$$

  • 0
    This shoule be: $|(a+ib)-(c+id)|=\sqrt{(a-c)^2 +(b-d)^2}\le |a-c|+|b-d|$ – 2017-02-03
  • 0
    This should state that $a, b, c, d$ are reals. – 2017-02-03
  • 0
    @SomeStrangeUser Yes, I missed typed. – 2017-02-03