Is $A= \{a + bi \mid a> 4\}$ closed in metric space $(\mathbb{C},d)$ where $d(z,w) = |z| + |w| \:\text{if } z \ne w,\,0$ if $z=w$? I characterized the convergent sequences in this space as those such that $z_n \to z$ if and only if $|z_n| + |z|\to 0$.
Edit: also if $z_n$ is an eventually constant sequence
Hence only the sequences which converge to $0$. Since $0$ is in $A'$, then $A'$ is closed and thus $A$ must be open.
I am not sure if my reasoning is correct.
 A: A set $A\subseteq\mathbb C$ in the metric space $(\mathbb C, d)$ with $d(z,w)=\left\{
\begin{array}{ll}
0 & z=w \\
|z|+|w| & \, \textrm{otherwise} \\
\end{array}
\right.$ is closed if and only if the set of accumulation points of $A$ is a subset of $A$, i.e. $\operatorname{acc}(A)\subseteq A.$
To find out what the accumulation points are, you have to look at sequences $(z_n)$ which converge to $z\in\mathbb C$ in this metric space, and where $z_n\neq z$ for infinitely many $n\in\mathbb N$. Now, $z$ is an accumulation point if and only if such a sequence $(z_n)$ exists.
Now, we are looking at the special case of $A=\{z\in\mathbb C:\operatorname{Re}(z)>4\}$.
A series in this metric space converges not only if $d(z_n,z)=|z_n|+|z|\rightarrow 0$ for $z_n\neq z$, but also if $z_n=z$ for $n\geq N$ for some $N\in\mathbb{N}$. Now, to test if the set is closed, you have to look at a sequence where the first case occurs (we do not want constant sequences for accumulation points). You are right in saying that such a sequence converges if and only if $|z_n|+|z|\rightarrow 0$ for some $z\in A$. Since for all $z\in A$ it is true that $|z|>4>0$, this is not possible for any sequence. So, there are no accumulation points at all. This means that the set is closed, because the definition of a closed set is that $\operatorname{acc}(A)\subseteq A$, which is true for $\operatorname{acc}(A)=\emptyset$.
"Most" subsets of $\mathbb C$ are actually closed in this metric space. The only case where this is not true is a subset which does not include $0$, but which has $0$ as an accumulation point. (For example, $B=\{z\in\mathbb C:z=1/n\text{ for }n\in\mathbb N\}$.) However, your set $A$ does not fall under this category.
Generally, all sets in this metric space can only have the accumulation point $0$, or no accumulation point at all. So, you definitely know that a subset is closed if $0$ is an element of the subset or if there exists a $\delta>0$ so that $|z|>\delta$ for all $z$ in the subset.
