Showing that a subset of the complex plane is open.

Question

Let $B=\{z\in\mathbb{C}: -\operatorname{Im}(z)< \operatorname{Re}(z)\}$. I need to show that $B$ is open by finding an $\epsilon$-ball around each $z\in B$ which is contained in $B$. For any $z\in B$ let $\epsilon$ be the shortest distance from $z$ to the line $-\operatorname{Im}(z)= \operatorname{Re}(z)$. Then take the ball to be $B(z, \epsilon)$. For an arbitrary $w\in B(z,\epsilon)$ we have $$ |z-w|<\epsilon. $$ Assume that $-\operatorname{Im}(w)\geq \operatorname{Re}(w)$. I have tried arguing towards a contradiction to get $|z-w|\geq \epsilon$ but I could not achieve a contradiction.

Answer 1

If $|z-w| < \delta$, then $|Re(z) - Re(w)| < \delta$, from which it follows that $Re(w) > Re(z) -\delta$. And the same for $Im$. For clarity, rewrite the condition (known for $z$ and desired for $w$) as $Re(z) + Im(z) > 0$. You have $Re(z) + Im(z) = \epsilon > 0$; use the inequalities above to see if you can choose $\delta$ such that $w \in B(z,\delta)$ will force $Re(w)+Im(w) > 0$.