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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:

  1. $s_n$ converges implies that that the infimum and supremum exist since there are finitely many points outside the convergence interval $L-\epsilon < s_n < L +\epsilon$.
  2. 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 $0$

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Do you know that $( s_n )$ converges? –  Arthur Fischer Oct 5 '12 at 14:58
    
The assumptions are not clearly stated. It looks as if one of them is that $(s_n)$ converges. And you want to pick a fixed $\epsilon$, probably you intend $|s|/2$, not the unreasonable $s_n/2$. –  André Nicolas Oct 5 '12 at 15:04
    
@CodeKingPlusPlus: Please check my solution for your question. –  blindman Oct 5 '12 at 16:20
    
@ArthurFischer Yes, $s_n$ does converge. I just fixed it in the question. –  CodeKingPlusPlus Oct 7 '12 at 20:46
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1 Answer

In the following we assume that $s_n\ne 0$ for all $n\in \mathbb{N}$. Let $\displaystyle s=\lim_{n\rightarrow\infty}s_n\ne 0$. Then $$ \lim_{n \rightarrow \infty}|s_n|=|s|>0. $$ Let $\displaystyle\varepsilon =\frac{|s|}{2}$. Then there exists $N\in\mathbb{N}$ such that $$ ||s_n|-|s||<\frac{|s|}{2} \quad \forall n\geq N. $$ Hence $$ |s_n|>\frac{|s|}{2} \quad \forall n\geq N. $$ Let $$\bar{s}=\min\left\{|s_1, |s_2|, \ldots, \frac{|s|}{2}\right\}>0,$$ then $|s_n|\geq \bar{s}$ for all $n\in \mathbb{N}$. So $$ \inf\{|s_n|:\in\mathbb{N}\}\geq \bar{s}>0. $$

Note. If there exists $n_0\in \mathbb{N}$ such that $s_{n_0}=0$ then $$ 0\leq \inf\{|s_n|: n\in \mathbb{N}\}\leq |s_{n_0}|= 0, $$ which implies that $$ \inf\{|s_n|: n\in \mathbb{N}\}=0. $$

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