Let $U \subset \mathbb{C}$ be open. I want to show that we can write $U$ as a countable union of open connected sets.

If $U$ is empty there is nothing to prove. Else it contains some ball, and hence a point in $\mathbb{Q} + i\mathbb{Q}$.

Denote $U_{\mathbb{Q}} = U \cap \mathbb{Q} + i\mathbb{Q}$.

For each $q \in U_{\mathbb{Q}}$ let $U_q = \{ z \in \mathbb{C}$ | $\exists \gamma_z:[0,1] \to U$ a path connecting $z$ and $q \}$.

We'll show that $U_{q}$ is open and connected:

First if $z_1, z_2 \in U_q$ then $Im(\gamma_{z_1}), Im(\gamma_{z_2}) \subset U_q$, simply by shortening the paths.

It follows that joining these two paths results in a path in $U_q$ and so $U_q$ is path connected, and thus connected.

Now, if $z \in U_q$ by definition $z \in U$, being in the image of some path. Then choosing a ball around $z$ contained in $U$, and knowing the ball is path connected we can connect any point within to $q$. Thus $U$ is open.

Finally, $\cup_{i \in \mathbb{N}} U_{q_i} \subset U$, $\cup_{i \in \mathbb{N}}\{q_i\} = U_{\mathbb{Q}}$.

Letting $z \in U$ we choose a ball around it contained in $U$, and containing some 'rational' point $q$. Since the ball is path connected there is a path within the ball connecting $z$ and $q$, and so $z \in U_q$.

Is this alright?

  • $\begingroup$ The connected components of $U$ will do this for you. $\endgroup$ – zhw. Dec 12 '17 at 23:37
  • $\begingroup$ @zhw. we're not assuming knowledge except basic analysis/calculus. $\endgroup$ – Mariah Dec 12 '17 at 23:39
  • $\begingroup$ What do you mean with “by shortening the paths”? $\endgroup$ – José Carlos Santos Dec 12 '17 at 23:40
  • $\begingroup$ @JoséCarlosSantos if a point $z$ is on the image of the path $\gamma_{z_1}$ then selecting $t$ s.t $\gamma_{z_1}(t) = z$; $\gamma_{z_1}|_{[t,1]}$ defines the correct path. $\endgroup$ – Mariah Dec 12 '17 at 23:42
  • $\begingroup$ The proof is allright. I don't know how to write this as an answer tho. $\endgroup$ – pancho Dec 12 '17 at 23:57

$\mathbb{C}$ has a countable basis. It is the collection of open balls of rational radii centered at $z=a+bi$, where $a$ and $b$ are rational. Then any open $U \subset \mathbb{C}$ can be written as the countable union of these basis sets.

Another way of thinking about $\mathbb{C}$ is that it's isomorphic to $\mathbb{R}^2$. In fact $\mathbb{R}^n$ has a countable basis.

  • $\begingroup$ thanks, I was looking for a verification though; also added the tag now. $\endgroup$ – Mariah Dec 12 '17 at 23:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.