Let $S_n$ be A Sequence that Converges

Question

I am confused with these problems

Let $S_n$ be a sequence that converges How do I:

a) Show that if $S_n \geq a$ for all but finitely many $n$, then $lim_{n\to\infty} S_n\geq a$

b) Show that if $S_n \leq b$ for all but finitely many $n$, then $lim_{n\to\infty} S_n\leq b$

c) Conclude that if all but finitely many $S_n$ belong to $[a,b]$, then $lim_{n\to\infty} S_n$ belongs to $[a,b]$

For $(a),(b)$ $(c)$ use contradiction, I mean, suppose $\lim_{n\to\infty}S_n<a$ and $\lim_{n\to\infty}S_n>b$, and $\lim_{n\to\infty}S_n\notin [a,b]$. — DiegoMath, Sep 13 '14 at 18:14
note, I forgot to say that $S_n$ is a sequence that converges — Pasie15, Sep 13 '14 at 18:20
Ok the way I am think is like this, Suppose $lim_{n\to\infty}S_n > a$ then $lim_{n\to\infty}S_n = L$ for $L > a$. Thus as $n\to\infty $S_n$ gets closer to L, and so $a\geq L$ — Pasie15, Sep 13 '14 at 18:32

Jeybe · Accepted Answer · 2020-09-09T01:42:02.650

By contradiction, suppose that $\lim_{n\rightarrow \infty}S_{n}=S<a.$ Since $S_n \geq a$ for all but finitely many $n$, let $N_0$ the largest integer with this property. Since $S_n$ converges, then $\forall \, \epsilon >0, \exists \, N\in \mathbb{N}$ such that $n>N \Rightarrow |S_n - S|<\epsilon.$ Let $M=\max(N_0,N)$.Then $$\forall \, n>M \Rightarrow S_n \geq a \wedge |S_n - S|<\epsilon, $$

and let $\epsilon=a-S>0$ (since $S<a$). So you've $S_n - S<a-S$ then $S_n<a$ for all $n>M$. Which is a contradiction, because we already show that $S_n \geq a$ $\forall n >M.$ Thus $S\geq a.$

Let $S_n$ be A Sequence that Converges

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