Complex Analysis: How to show a set is open in the complex plane.

Question

I need to show that the set $S = \{(x,y) \in \mathbb{C} : x>0\; \text{and}\; y> \frac{1}{x^2} \} $ is open.

So I've figured out that if I let $z_0 = (a,b) \in S$ and let $\epsilon =min\{ a, \frac{2}{a^2}\}$, then $(x,y) \in D_{\epsilon}(a,b)$ which means S is open.

I think can prove the x component by showing that $|x-a|<\epsilon$ where $\epsilon = a$ leads to $ \frac{x}{2} < \epsilon$ so $ x=\frac{a}{2} \in d_{\epsilon}(a,b)$.

Is this even a valid method? If so how would I show the y-component?

Answer 1

Here is another approach using continuous functions:

At first, your problem has not much to do with the complex plane, so lets just consider everything in $\mathbb R^2$, or in this case more convienient, in $\mathbb R^2_+:=\{(x,y)\in\mathbb R^2\mid x,y>0\}$. Consider the following continuous function

$$ f: \mathbb R^2_+\rightarrow \mathbb R, (x,y)\mapsto x^2-\frac 1y. $$

Note that $f$ is positive, exactly if $y >\frac 1{x^2}$. So $f^{-1}[(0,\infty)]=S$, and we can conclude, that $S$ as a preimage of an open set under a continuous function must be open itself. So $S$ is open in $\mathbb R^2_+$ and $\mathbb R^2_+$ is open in $\mathbb R^2$, so we finally can conclude that $S$ is open in $\mathbb R^2$.

If it was not your intial objective to show that the function $f$ or something likewise was continuous, then this would be my way to go.