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I've recently (yesterday) started to studying complex analysis on my own. So i came across this question at the end of the first chapter " Let $G$ be an open set in $ \mathbb{C} $ . Is $\{ z : z \in G \lor \bar z \in G \}$ open?

My attempt :

By the definition a set is open if $ A \subseteq \mathbb{C} \land z \in A $ then A is open if there exists $ \mathcal D( z ; r ) \subseteq A $ , where $ r > 0 $ and $ \mathcal D ( z ; r)$ is an open disc.

I think, for the $ z \in G $ case, it's open as we can find some $ \mathcal D ( z ; \delta ) $ where $\delta$ is $ 0 < \delta < |z-a| - r $ .

But can't make any comment on it's conjugate.

4 Answers 4

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If $G\subset {\mathbb C}$ is open then the set $\bar G$ obtained by reflecting $G$ in the real axis is open. Your set is just $G\cup\bar G$, hence open.

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Geometrically, an open set is a union of open disks. This property ('to be a union of open disks') is preserved under the complex conjugation (for it is the symmetry with respect to the real line): whenever $U \subseteq \mathbf C$ is open, $\overline U$ is also open. Now if $G$ is open, then $\overline G$ is open, then $G \cup \overline G$ is open as a finite union of open sets.

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Geometrically it's simple that consider $\Omega=\{ z : z \in G \lor \bar z \in G \}$ is open, but analytically:

Assumption: $G$ is open in $\mathbb{C}$, that is for every $z\in G$, there exise $r>0$ such that $\mathcal{D}(z;r)\subset G$.

Now let $\Omega=\{ z : z \in G \lor \bar z \in G \}=G\cup\bar{G}$. For every $z\in \Omega$, $z\in G$ or $\bar{z}\in G$.

If $z\in G$, then there exise $r>0$ such that $\mathcal{D}(z;r)\subset G\subset \Omega$.

If $\bar{z}\in G$, then there exise $r>0$ such that $\mathcal{D}(\bar{z};r)\subset G\subset \Omega$. \begin{eqnarray} \mathcal{D}(\bar{z};r)&=&\{w\in\mathbb{C}:\,|w-\bar{z}|

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What you then do is that if $\bar z\in G$ that $\mathcal D(\bar z,\delta)\subset G$. Now consider $w \in \mathcal D(z, \delta)$ instead, that is $|w-z|<\delta$ this means that $|\bar w-\bar z|<\delta$ so $\bar w\in\mathcal D(\bar z, \delta)$. So since $\bar w\in G$ we have that $w$ is in the greater set and therefor $\mathcal D(z, \delta)$ is a subset of the greater set.