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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?

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  • $\begingroup$ 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
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$\mathbb{C}$ 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.

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

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