Complex analysis: limit point at infinity Let $S\subset\overline{\mathbb C}$.
Are the following two statements equivalent?

$\infty$ is a limit point of $S$.

and

For every $M>0$, there exists $l\in S$ such that $|l|>M$.

Intuitively this is correct, since we have points arbitrarily far away from the origin, so we have points arbitrarily close to $\infty$.
However, I don’t fully trust my intuition. Moreover, even if my intuition is correct, I cannot rigorously prove that the statements are equivalent.
Is my intuition correct? If yes, how can I rigorously prove the equivalence of these two statements?
Any help will be appreciated.

I assume that by $\infty$ being a limit point of $S$ you mean that every neighbourhood of $\infty$ contains a point in $S$ that is not $\infty$. Recall that the topology on the Riemann sphere is defined so that every open neighbourhood of $\infty$ is of the form $\{\infty\} \cup (\mathbb{C} \backslash C)$ for some closed and compact set $C \subseteq \mathbb{C}$. 
Now, assume that $\infty$ is a limit point of $S$. Let $M > 0$ and consider the neighbourhood $\{\infty\} \cup (\mathbb{C}\backslash \overline{B_M(0)})$ where $\overline{B_M(0)}$ is the closed ball of radius $M$ around $0$. By assumption, there is some $l \in S$ with $l \in \mathbb{C}\backslash \overline{B_M(0)}$ and so $|l|>M$. 
For the other direction, assume that $\infty$ is not a limit point of $S$. Then there is some open neighbourhood $\{\infty\} \cup (\mathbb{C} \backslash C)$ that contains no elements of $S$. This implies $S \subseteq C$. Since $C$ is compact, it is bounded by the Heine-Borel theorem and we are done. 
