I am trying to prove: $\inf\{|s_n| : n \in \mathbb{N}\} > 0$. Where $s_n \neq 0 \land \lim s_n \neq 0 \land s = \lim s_n$. My understanding of infimum is that it is the greatest lower bound for a set. My intuition is as the following:
- $s_n$ converges implies that that the infimum and supremum exist since there are finitely many points outside the convergence interval $L-\epsilon < s_n < L +\epsilon$.
- Since $s_n \neq 0$ Then the absolute value of the infimum is greater than $0$.
Correct me if I am wrong. I have no idea how to formalize this, other than:
Let $\epsilon = \dfrac{s_n}{2}$ then since $s_n \to s$ we can write:
$|s_n -s| < \dfrac{|s|}{2}$ By the definition of the limit
Thus for $n > N \implies |s_n| > \dfrac{s}{2}$ We are in our convergence interval
Therefore there are finitely many points less than $\dfrac{|s|}{2}$ and $\inf\{s_N\} \neq 0$. So, the infimum of $|s_n|$ must be greater than $0$