Proving that $P:\mathbb{C}\rightarrow\mathbb{R}$, defined by $P(z)=Re(z)$ is open but is not closed.

Question

looking in a geometric way i can see that $P$ is open, $P$ works like a projection in the line, but i'm having some dificulties to show this with calculations.

for the second part, i want to consider $F=\{z \in \mathbb{C}|Re(z)=(Im(z))^{-1}, Re(z)\neq 0\}$

the imagem by $P$ of $F$ is open, but i cant prove either way is $F$ closed, i'm triying to use sequences but i get nowhere.

Anyone could help with both parts?

Answer 1

The set $F$ is closed, because$$F=\left\{\frac1x+ix\,\middle|\,x\neq0\right\}$$and if $(x_n)_{n\in\mathbb N}$ is such that $\lim_{n\in\mathbb N}\frac1{x_n}+ix_n=a+bi$, then $\lim_{n\in\mathbb N}\frac1{x_n}=a$ and $\lim_{n\in\mathbb N}x_n=b$. Then$$b=\lim_{n\in\mathbb N}x_n=\frac1{\lim_{n\in\mathbb N}\frac1{x_n}}=\frac1a$$and therefore $(a,b)\in F$.

Now, let $A$ be an open set of $\mathbb C$. For each $z\in A$, take $r_z>0$ such that $D(z,r_z)\subset A$. Then\begin{align}P(A)&=P\left(\bigcup_{z\in A}D(z,r_z)\right)\\&=\bigcup_{z\in A}P\bigl(D(z,r_z)\bigr)\\&=\bigcup_{z\in A}(a-r_z,a+r_z),\end{align}which is an open subset of $\mathbb R$. So, $P$ is an open map.