# Limit points, accumulation points via open sets

In "Principle of Mathematical Analysis", Chapter 2, Exercise 8, Rudin stated the real-field version of the problem.

I was wondering how to prove the analogous version for the finite complex plane. In "Functions of one complex variable", Conway stated the following two standard definitions.

• For a metric space $$(X,d)$$ a set $$G \subset X$$ is open if for each $$x \in G$$, there exists $$\epsilon > 0$$ such that $$B(x; \epsilon) \subset G$$.
• If $$A \subset X$$, then a point $$x \in X$$ is a limit point of $$A$$ if there is a sequence $$\{ x_n \}$$ of distinct points in $$A$$ such that $$x=\lim_{n \rightarrow \infty} x_n$$.

Also, how about for accumulation points though some authors use the terms accumulation and limit points interchangeably. According to Brown and Churchill: A point $$z_0$$ is said to be an "accumulation point" of a set $$S$$ if each deleted neighborhood of $$z_0$$ contains at least one point of $$S$$.

• It should be worth noting that, as metric spaces, $\mathbb{R}^2$ and $\mathbb{C}$ (both with the euclidean metric) are the same thing. May 18, 2019 at 16:50

Let's use the following definitions on a metric space $$(X,d)$$:

• a point $$x\in X$$ is a limit point of $$A\subseteq X$$ if there exists a sequence $$(x_n)$$ of distinct points in $$A$$ that converges to $$x$$;

• a point $$x\in X$$ is an accumulation point of $$A\subseteq X$$ if every deleted neighborhood of $$x$$ contains points of $$A$$.

Clearly being a limit point implies being an accumulation point: given $$\varepsilon>0$$, the open ball $$B(x,\varepsilon)$$ contains infinitely many terms of the sequence.

Suppose $$x$$ is an accumulation point of $$A$$. Let $$x_1\in B^*(x,1)$$, the deleted open ball of radius $$1$$. Now, suppose we have chosen $$x_1,\dots,x_n$$ such that $$d(x_1,x)>d(x_2,x)>\dots>d(x_n,x)\qquad\text{and}\qquad d(x_k,x)<\frac{1}{k}\quad (k=1,\dots,n)$$ Choose $$x_{n+1}\in B^*\left(x,\min\left\{d(x_n,x),\frac{1}{n+1}\right\}\right)\cap A$$ which is possible because $$x$$ is an accumulation point of $$A$$. Then $$d(x_n,x)>d(x_{n+1},x)$$ and $$d(x_{n+1},x)<1/(n+1)$$.

Thus we get a sequence $$(x_n)$$ of distinct points in $$A$$ that converges to $$x$$. Hence $$x$$ is a limit point of $$A$$.

In order to prove that every point $$x$$ of an open set $$G\subseteq\mathbb{C}$$ is an accumulation point in $$\mathbb{C}$$, take $$x_n=x+1/(n+m)$$, where $$B(x,1/m)\subseteq G$$.

If $$G$$ is open in the complex plane and $$z \in G$$ then $$z+\frac 1 n \to z$$ and $$z+\frac 1 n \in G$$ for $$n$$ sufficiently large.

• How about for accumulation points ? I've edited my question.
– user672937
May 18, 2019 at 12:58
• Every point of $G$ is a limit point ad well as an accumulation point. May 18, 2019 at 13:27