About cluster points in a proof

Question

I am reading in a proof, but there's only one part that I do not understand

Theorem: Let $\Omega$ be an open subset of $\mathbb{C}$, let $z_1,z_2,\dots$ be a sequence of different points from $\Omega$ with no cluster point in $\Omega$, [...]

Proof: Since the sequence $\{z_k\}$ does not cluster in $\Omega$, there exist $a\in \Omega$ and $r>0$ such that $B(a,r)$ does not intersect the set $\{z_k\}$. [...]

The definition of a cluster point is as follows

Let $A\subseteq \mathbb{C}$, a point $a\in\mathbb{C}$ is called an cluster point of $A$ if $(B(a,r)\cap A)\setminus \{a\}\neq \emptyset$ for all $r>0$.

So, saying that $A$ has no cluster points, it must mean, by negating this definition that

$$\forall a\in \mathbb{C}\exists r>0:(B(a,r)\cap A)\setminus \{a\}= \emptyset \tag{1}$$

I do not understand why the proof is true, and/or what is the connection between this and $(1)$?

Answer 1

Since $\Omega$ is open in $\mathbb{C}$, it's uncountable, and hence we can choose a point $z$ not in our sequence $z_1,\cdots$. Now since $z$ isn't a cluster point of the sequence, there is some neighborhood $B(z,r)$ such that $B(z,r)\setminus \{z\}$ doesn't intersect our sequence. But $z$ doesn't either, so $B(z,r)$ doesn't intersect the sequence, as desired.

Answer 2

Just choose any $a\in \Omega$ that is not in your sequence, i.e. $a\neq z_i$ for all $i$. Since $a$ is not a cluster point, we have that there is $r>0$ with $B(a,r)$ not intersecting any of the $z_i$.

Answer 3

Let's favor prose over notation.

Definition: A point $c$ is a cluster point of a sequence $\{z_{k}\}_{k}$ if every neighborhood of $c$ contains a point of the sequence.

This definition implies: If $c$ is a cluster point of $\{z_{k}\}_{k}$, then every neighborhood of $c$ contains infinitely many points of the sequence.

Now suppose $\{z_{k}\}_{k}$ has no cluster points in a set $\Omega$. Pick any point $\omega$ in $\Omega$ and any neighborhood of $\omega$ in $\Omega$. This neighborhood, then, contains at most finitely points of the sequence. Therefore, we can find a ball $B(a, r)$ in that neighborhood that contains no points of the sequence.