A set is said to be closed if it contains all of its boundary points, where a boundary point of a set in the complex plane is any point such that every neighborhood of the boundary point contains at least one point in the set and at least one point not in the set. At first glance, I assumed that the set $S$ defined by $S=\{ i/n\,\vert \,n\in\mathbb{Z}^+\}$ is closed, since each point $i/n$ is a boundary point of the set and is an element of the set. However, the more I think about it, the more I feel that the complex number $z=0$ is also a boundary point of the set, since any neighborhood of $0$ must contain a complex number that is arbitrarily close to $0$ on the positive imaginary axis (and therefore in $S$) and $0$ itself (which is not in $S$). Since this boundary point $z=0$ is not contained in the set, $S$ does not contain all of its boundary points, and is therefore not closed. I know that this is not a rigorous proof of my claim, but is this reasoning correct?
Is the set of points ${i/n},\,n=1,2,3\ldots$ in the complex plane a closed set
complex numberscomplex-analysiselementary-set-theorygeneral-topology
Best Answer
Yes, your reasoning and your conclusion are correct.
Rigorous? Basically. I guess you could add in the detail of showing that any open disk around $0 \in \mathbb{C}$ contains points in your sequence. Such a disk is defined by the inequality $|z| < \varepsilon$ for some $\varepsilon > 0$. Now for any $n > \tfrac{1}{\varepsilon}$, $$ \biggl\lvert \frac{i}{n} \biggr\rvert = \frac{1}{n} < \varepsilon, $$ so the point $\tfrac{i}{n}$ in your set is in the neighborhood of $0$.