Complex conjugation continuous function

Question

How can I show that complex conjugation is a continuous function? I tried looking at open sets $U$ and then the preimage. Can assume preimage is not empty so that if $z$ is in preimage then $f(z) \in U$. How to go from here?

Answer 1

HINTS:

1) Complex conjugation as a function $c:\Bbb C\rightarrow\Bbb C$ is an involution, i.e. $c^2=\text{id}_{\Bbb C}$. Thus, it coincides with its inverse. Thus the pre-image of an open set $U\subset\Bbb C$ is the same as its image. Now: what happens when you take the complex conjugates of a ball centered at $z_0$ of radius $r$?

2) Under the Argand-Gauss identification $\Bbb C=\Bbb R^2$ the standard topology in $\Bbb C$ is the product of the standard topologies in $\Bbb R$. Now, if you have a function $$ f:S\longrightarrow X\times Y $$ between topological spaces, you know that $f$ is continuous if and only if $\text{pr}_X\circ f$ and $\text{pr}_Y\circ f$ are continous, where $\text{pr}_X$ and $\text{pr}_Y$ denote the projection onto $X$ and $Y$ respectively. What you have if you compose complex conjugation with the two projections on the real and imaginary axes?

Answer 2

Let $z_0$ be a complex number. To make $|\overline{z}-\overline{z_0}|<\epsilon$, it is enough to make $|z-z_0|<\epsilon$, since $|\overline{z}-\overline{z_0}|=|z-z_0|$. So in the definition of continuity (for metric spaces) we are choosing $\delta=\epsilon$.