# Proving a set of accumulation points (Complex Analysis)

Let $z_{i}$ be a complex number. Prove that a finite set of points $z_{1}$...... $z_{n}$ cannot have any accumulation points.

What i tried. Proving by contradiction Let $S$ be the set in the complex plane. I suppose that there is at least one accumulation point $z_{o}$. Then by the definition of accumulation point, we know that the neighborhood of $z_{o}$ must have at least a point that lies in $S$ and those points are $z_{1}$...... $z_{n}$ as given in the question. To arrive at a contradiction, I must show that there is indeed an infinite number of such points and thus infinite number of distances between these points and the accumulation point. I tried using the Eplison delta proof to show this but im stuck at the Eplison delta portion. Would anyone be able to explain the eplison delta proof? Thanks

• There is a finite number of distances between any two different of these numbers, and given any finite set of positive real numbers there is a positive real number less than any number in the set, and . . . Commented Aug 16, 2016 at 15:49

For $z_0$ to be an accumulation point, we must have that every neighborhood of $z_0$ has to contain a point from your set different from $z_0$. Using the fact that you have a finite set, try to come up with a neighborhood around $z_0$ that will not contain any of the $z_1,z_2,\dots,z_n$. That is, try to come up with an appropriate $\varepsilon$ radius for your neighborhood around $z_0$.

• Let epsilon be the minimum distance between your accumulation point and each point in your set. Commented Aug 16, 2016 at 16:05
• What i think is that the neighborhood have to be of the form $N= \{\omega\mid |\omega-z|<eplison\}$ and since eplison can be infinitely small it has to be smaller than $d$ which thus shows that none of the points {z_{1}...z_{n} lies within the neighbourhood which then leads to a contradiction. Commented Aug 16, 2016 at 16:39

Suppose $z$ is an accumulation point of $\{z_{1},\ldots,z_{n}\}$. For $1\leq i\leq n$, let $d_{i}=|z-z_{i}|$. $d_{i}$ is the distance from $z$ to $z_{i}$ in the complex plane. Now, since we have only finitely many such distances, we can take the smallest of all of them: $d=\min\{d_{1},\ldots,d_{n}\}$ (something we couldn't do if we had infinitely many values).

Now, consider the neighborhood $N= \{\omega\mid |\omega-z|<d\}$. I.e., the set of all points in $\mathbb{C}$ that are within $d$ of $z$. By our choice of $d$, none of the points $z_{1},\ldots,z_{n}$ lie in $N$. This contradicts that $z$ is as accumulation point of $\{z_{1},\ldots,z_{n}\}$.

• Is there a need to use eplison delta proofs for this question? Commented Aug 16, 2016 at 16:04
• No. You don't have to. Although, that is sort of what's implicitly going on here. We're taking a neighborhood of radius $d=\epsilon$ and showing that no points from the set lie in it, contradicting the definition.
– ervx
Commented Aug 16, 2016 at 16:04
• What i think is that the neighborhood have to be of the form $N= \{\omega\mid |\omega-z|<eplison\}$ and since eplison can be infinitely small it has to be smaller than $d$ which thus shows that none of the points {z_{1}...z_{n} lies within the neighbourhood which then leads to a contradiction. Commented Aug 16, 2016 at 16:39
• That's correct!
– ervx
Commented Aug 16, 2016 at 18:43