Suppose $(s_n)$ is a sequence with $\displaystyle \lim_{n\to\infty} s_n = L$. Let $S = \{s_n: n \in \mathbb N\} \subseteq \mathbb R$. Prove that if $L \notin S$, then $L \in S'$ ($L$ is an accumulation point).
Proof: Let $\varepsilon > 0$. Since $\displaystyle \lim_{n\to\infty} s_n = L$, there exists an $N \in \mathbb R$ such that for $n \geq N$, we have
$$ \left|s_n - L\right| < \varepsilon$$ $$-\varepsilon < s_n - L < \varepsilon$$ $$L-\varepsilon < s_n < L + \varepsilon$$
By the above algebra, $s_n \in (L-\varepsilon, L + \varepsilon)$, so $s_n \in N(L;\varepsilon)$.
Let $S = \{s_n: n \in \mathbb N\} \subseteq \mathbb R$. Let $L \notin S$.
For some $n \in \mathbb N$, $s_n \neq L$, so we can say that $s_n \in N^*(L,\varepsilon)$ (notation for deleted neighborhood).
From this, we see that $ N^*(L,\varepsilon) \cap S \neq \varnothing$. Therefore, $L \in S'$.