I want to approximate function $$z\mapsto \lceil|z|\rceil$$ by complex polynomials. I construct regions $$K_n=\bigcup_{i=0}^\infty K_n^i$$ where each $K_n^i$ is an annulus with a section removed and with outer diameter of $i$, as pictured (and $K_n^0$ is a small closed disc). I then define $f_m$ to be such that $f_m|K_m^i=i$. Complent of $K_n$ is connected, so by Runge theorem for each $f_m$ there is a sequence of complex polynomials $P_{m,n}$ such that $P_{m,n}\to f_m$ as $n\to\infty$. $f_m\to \lceil |z|\rceil$ as $m\to\infty$, do by diagonal argument I get a sequence of polynomials, say $\tilde{P}_n$ as desired.
Is my reasoning correct? Do I need to fill any gaps?
I think the idea is more or less right, but there are three main issues:
a. $K_n$ as defined is not compact (you need this if you want to apply Runge) (Fix: don't take an infinite union, instead a finite union)
b. The $K_n$ don't cover the points $i,2i,3i,\cdots$ (Fix: consider small neighborhoods around these points)
c. The diagonal argument needs to be more precise.
What I suggest is the following. Construct $K_n$ as follows: $$K_n=\bigcup_{i=0}^n S_{n,i}$$ where $S_{n,0}=\overline{B(0,\frac{1}{2n})}$ and for $i>0$, we define $S_{n,i}=A_{n,i}\cup B_{n,i}$ where $$A_{n,i}=\{z: i-1+1/n\leq|z|\leq i, |arg(z)|>\frac{1}{n}\}$$ and $$B_{n,i}=\{z: i-1+1/n\leq|z|\leq i, |arg(z)|<\frac{1}{2n}\}$$
So, $A_{n,i}$ are like the annulus with a section removed and the $B_{n,i}$ are like the neighborhoods around the points that are not considered by $A_{n,i}$
We have that $K_n$ are compact and their complement is connected and more importantly, for any $z\in \mathbb{C}$, there is $N=N(z)$ big enough such that $z\in K_n$ for all $K_n$ with $n>N$.
Define $f_n$ by $f_n(z)=i$ for $z\in S_{n,i}$. Hence, by Runge's theorem there is a polynomial $p_n$ such that $\|f_n-p_n\|_\infty<1/n$. Then for any $z\in\mathbb{C}$, we have that $|f_n(z)-p_n(z)|<1/n$ for $n>N$, Note that $f_n(z)=\lceil |z|\rceil$ for big enough $n$. So, we have $|\lceil |z|\rceil-p_n(z)|<1/n$ for big enough $n$. Taking $n\to\infty$, we get $p_n(z)\to f(z)$, i.e. $p_n\to f$ (pointwisely).