# Proving a complex set is open

Prove that the set $\mathcal{U}=\left\{z\in\mathbb{C}\,\colon\,\Re{(z)}>0\right\}$ is open. Let $a\in \mathcal{U}$, we must show that there exists an $r>0$ such that the disk $$D(a, \,r) = \left\{z\in\mathbb{C}\,\colon\,\lvert z-a\rvert<r\right\}$$ is a subset of $\mathcal{U}$. Now take $0<r<\Re{(a)}$ and $z_0\in D(a, \,r)$ so that $\lvert z_0-a\rvert<r<\Re{(a)}$. We must show $\Re{(z_0)}>0$. Now recall

\begin{align} \Re{(a)}>\lvert z_0-a\rvert&\ge\Re{(z_0-a)}\\ &= \Re{(z_0)}-\Re{(a)} \end{align}

$$\implies \Re{(z_0)}< 2\Re{(a)}$$

Now I'm not really sure how to continue.

• Let the real parts of $a$ and $z_0$ be $\alpha$ and $x$ respectively. First show that $|z_0 - a| < r \implies (x - \alpha)^2 < r^2 \iff -r < x - \alpha < r$. Oct 14, 2015 at 10:53
• But you haven't shown that $x>0$... Oct 14, 2015 at 12:49
• Just note that $|z_0-a|\geqslant\Re{(a-z_0)}=\Re{(a)}-\Re{(z_0)}.$
– CIJ
Oct 14, 2015 at 13:05

What you should do is consider the components of $w=z_0-a$. Since $|w|<r$ you would inevitably have that $|\Re w|<r$. Consequently $\Re z_0 = \Re w + \Re a$, and $|\Re w| < r = \Re a$ so $\Re z_0 > 0$.
The the last step is because $|\Re w|<r$ means that $-r< \Re w < +r$ (where the last inequality is of no interrest here) consequently we have by adding $r=\Re a$ $$0 = r-r = \Re a - r < \Re w + \Re a = \Re z$$
• You have not shown $\Re{(z_0)}>0$. You have shown that $\Re{(z_0)} < 2\Re{(a)}$ exactly as I have. If $\Re{(z_0)} < 2\Re{(a)}$ then even though $r$ is positive, we have a less than sign and $\Re{(z_0)}$ could still be negative because a negative number is less than a positive number. Oct 14, 2015 at 12:47
• @MichaelHowlard I think I have, note the absulute sign around $\Re w$. It means that $-r < \Re w < r$ and therefore $\Re a-r < \Re z_0 = \Re a + \Re w < \Re a + r$. Oct 14, 2015 at 12:51