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 $S$.

  • 1
    $\begingroup$ It should be worth noting that, as metric spaces, $\mathbb{R}^2$ and $\mathbb{C}$ (both with the euclidean metric) are the same thing. $\endgroup$
    – Thorgott
    May 18, 2019 at 16:50

2 Answers 2


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.

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

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