Suppose we have a fixed, bounded infinite sequence$\ (b_{n}) \ $and for each$\ n\in\mathbb{N}^{+}, \ $we set$\ w_{n} \ =inf\{b_{n},b_{n+1},b_{n+2},...\} \ $and$\ g_{n} \ =sup\{b_{n},b_{n+1},b_{n+2},...\}. \ $We also let$\ g=inf\{g_{n}:n\in\mathbb{N}^{+}\} \ $and$\ w=sup\{w_{n}:n\in\mathbb{N}^{+}\}. \ $Prove if $\ w = g, \ $then$\ (b_{n}) \ \rightarrow w=g. $
My attempt at proving this:
We know that if $\ a_{n}\le\ k_{n}\le\ c_{n}\ \ $for all $\ n \ $sufficiently large and $$\lim_{n\to \infty}a_{n} \ =\lim_{n\to \infty}c_{n}=a $$
Then, $$\lim_{n\to \infty}k_{n}\ =a $$
It is stated above that: $$\lim_{n\to \infty}w_{n}\ =\lim_{n\to \infty}g_{n} $$
Therefore, if we can show the following is true (for all for all $\ n \ $sufficiently large):
$$\ g_{n}\le\ b_{n}\le\ w_{n}\ \ $$
Then we have successfully shown that: $$\lim_{n\to \infty}b_{n}=w=g\ $$
We know that $\ g_{n} \ $is monotone decreasing (non-increasing) and that $\ w_{n} \ $is monotone increasing (non-decreasing), and therefore, for $\ n \ $sufficiently large, we can assume $\ g_{n}\le\ b_{n}\le\ w_{n}\ \ $ and thus, we can conclude that: $$\lim_{n\to \infty}b_{n}=w=g\ $$
You mixed up some notations. You have by the definitions
$$w_n \le b_n \le g_n$$
for all $n$. You wrote it down the other way around. Then, as you correctly said,$ g_n$ is non-increasing, so having an infimum means it has the limit, same goes to $w_n$, which concludes your proof, since those limits are assumed to be the same.