Open set in the complex plane, then the set of conjugates is also open Take a set $A \subset \mathbb{C}$ such that $A$ is open. Show that the set $\{z \in \mathbb{C} : \bar{z} \in A\}$ is also open. 
I have been using the definition of the open ball, but seem to be confusing my self. Any hints on how to show this?
 A: Lord Shark's comment is the most direct way to solve the problem, but if you are having trouble with that, try the following hint (draw yourself a picture to understand).
Hint: Denote by $f$ the map which associates a point to its conjugate.
Let $\bar{w}$ be a point in $$f(A) = \{ z \in \mathbb{C} \colon \bar{z} \in A \}.$$
Then, $w$ is a point in $A$.
Let $B \subset A$ be an open ball around $w$ (such a ball exists since $A$ is open).
What can you say about $f(B)$? Is it open? Is it contained in $f(A)$?
A: Lemma:  For $c,d \in \mathbb C$ if $d(c,d) = m$ then $d(\overline c, \overline d) = m$.
Pf:  If $c = c_1 + c_2 i$ and $d = d_1 + d_2 i$ then $d(c,d) = \sqrt{(d_1 - c_1)^2 + (d_2-c_2)^2} = \sqrt{(d_1-c_1)^2 + (-d_2 - (-c_2))^2} = d(\overline c, \overline d)$.
So it is true that around ever point $x \in A$ there is an $\epsilon > 0$ so that $N_{\epsilon}(x) = \{y \in \mathbb C|d(x,y) < \epsilon\} \subset A=\{w|w\in A\}$.  
But for every $z \in \{z \in \mathbb{C} : \bar{z} \in A\}$ there is a $x \in A$ so that $w = \overline x$ and there is an $\epsilon$ so that $N_{\epsilon}(x) = \{y \in \mathbb C|d(x,y) < \epsilon\} \subset A=\{w|w\in A\}$.  But that means that there is a set $N_{\epsilon}(\overline x) =\{\overline y \in \mathbb C| d(x,y)=d(\overline x, \overline y) < \epsilon\} \subset \{z \in \mathbb{C} : \bar{z} \in A\} $ .
So $ \{z \in \mathbb{C} : \bar{z} \in A\} $ is open.
