# Is the set of points ${i/n},\,n=1,2,3\ldots$ in the complex plane a closed set?

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?

• Your reasoning is perfectly correct and actually forms a rigorous enough proof to me! What makes proofs rigorous is logical precision and not fancy notation or terminology; the latter exist only to simplify communication and understanding). Commented Mar 22, 2021 at 4:29

## 3 Answers

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 $$0$$.

Yes, your reasoning is correct.

Let $$(X,d_{X})$$ be a metric space.

We say that a set $$E\subseteq X$$ is closed iff it contains all of its adherent points (i.e. $$E\supseteq\overline{E})$$.

Having said that, notice that \begin{align*} \lim_{n\to\infty}\frac{i}{n} = 0 \end{align*} so we can conclude (due the sequential characterization of adherent point) that $$0$$ is an adherent point of $$S$$.

On the other hand, $$0\not\in S$$. Consequently, the given set is not closed.

Hopefully this helps!

You do not have to deal with complex numbers as you have a mapping $$\mathbb{C} \mapsto \mathbb{R}$$ given by $$x: x^2$$ or $$S_{\mathbb{N}}=\{n \in \mathbb{N}:-\frac{1}{n^2}\}$$, but $$S_{\mathbb{N}}$$ is not a closed set so it cannot map to a closed set.