Let $M$ be a metric space and and $x$ a point in $M$. I have two statements: (1) $x$ not isolated and (2) Every neighborhood of $x$ contains an infinite number of points of M. I want to prove that stament 1 and 2 are equvalent:
1 $\implies$ 2
We assume $x$ is not isolated. This means that for any $\epsilon > 0$, there exists an open ball $S_\epsilon(x)$ that contains at least one point of $M$ different from $x$. Let's call this point $y$, where $y \neq x$ and $y \in S_\epsilon(x)$.
Now, consider the distance $\epsilon_1 = \frac{d(x,y)}{2}$. Since $x$ is not isolated, there must exist a point $y_2 \in S_{\epsilon_1}(x)$ such that $y_2 \neq x$. The choice of $\epsilon_1$ ensures that $y_2 \neq y$. We can continue this process, defining $\epsilon_i = \frac{d(x, y_{i-1})}{2}$, and finding points $y_i \in S_{\epsilon_i}(x)$ such that $y_i \neq x$ for all $i$.
This iterative process creates ever smaller radiuses, ensuring that $S_{\epsilon}(x)$ contains infinitely many points of $M$.
2 $ \implies $ 1
Using the contrapositive statement: $\neg 1 \implies \neg 2$.
Suppose that $x$ is isolated. This means that for any $\epsilon > 0$, there exists an open ball $S_\epsilon(x)$ such that $S_\epsilon(x) = \{x\}$. This implies that every neighborhood of $x$ contains only the point $x$ itself and no other points. The neighborhood of $x$ is therefore finite.
Does this suffice as a proof? I am mainly concerned about the second implication. It somehow feels unfinished, but I don't know what I am missing or if I used the contrapositive statement incorrectly.
