Show that any open set in $\mathbb{C}$ is a unioun of countably many open balls

Question

I know that I can identify one point in the form of z=a+ib where a,b $\in$ $\mathbb{Q}$ from each ball and count them. Yet, I am not clear about how to identify the balls.

Any help is appreciated.

Answer 1

You can consider the sets $B_{1/n}(a+ib) \cap K$ for $n \in \mathbb{N}$ and $a,b \in \mathbb{Q}$.

Answer 2

A fellow student of Dr. Li, I see!

You have the correct idea (with some minor error), let $K=\{x+iy:x,y\in\mathbb{Q}\}$. This set is $\textit{dense}$ in $\mathbb{C}$; that is, every ball about some $z\in\mathbb{C}$ contains some point $k\in K$. Notice also, $K$ is $\textit{countable}$.

We want to construct our union of balls centered at these points so that we may start properly accounting for them.

I'll sketch the proof for you:

Let $X$ be an open set of $\mathbb{C}$. Then, for some $r>0$, $B_r(x)\subset X$, for all $x\in X$. We $\mathbf{need}$ this $r$ to be rational, else we'd have an uncountable number of balls (i.e. not the result we're looking for). [Show we can choose have some rational $r$]. Then, let $k\in K$. Next, you need to find a way to construct balls around the $k$ values so that they completely "cover" $B_r(x)$. [Find an appropriate set to place $k$ in and radius, $s$, to do the "covering"]. Then, you can take $X=\cup_{k\in X}B_s(k)$.

Once you fill in the blanks, you're result should be plain to see (and you'll have finished Q5 on the homework :D).

Happy hunting!

Answer 3

Call the balls with rational radius and center in ${\mathbb Q}+i{\mathbb Q}$ special. The special balls form a countable family $\bigl(B_k\bigr)_{k\in{\mathbb N}}$. We need the following

Lemma. If $B(z,r)\subset{\mathbb C}$ is an arbitrary open ball then there is a special ball $B_k$ such that $$z\in B_k\subset B(z,r)\ .\tag{1}$$

Proof. Choose a point $w\in{\mathbb Q}+i{\mathbb Q}$ with $|w-z|<{r\over4}$ and a rational number $\rho$ with ${1\over2}r<\rho<{3\over4}r$. The ball $B':=B(w,\rho)$ then is special, and satisfies $(1)$.

Now let an open set $\Omega\subset{\mathbb C}$ be given. Then for each $z\in\Omega$ there is an open ball $B(z,r_z)$ such that $B(z,r_z)\subset\Omega$. Choose for each $B(z,r_z)$ a special ball $B_{k_z}$ using the Lemma. Let $$A:=\bigl\{k_z\in{\mathbb N}\bigm| z\in\Omega\bigr\}\ \subset{\mathbb N}\ .$$ Then one has $$\Omega=\bigcup_{k\in A} B_k\ .$$