A question on Proof of Divergent sequence

Question

Statement: A sequence diverging to $\infty$ is unbounded above but bounded below.

I have no doubt the proof of unbounded above, but there is a question on the the proof of bounded below:
Since the sequence $\{x_n\}_n$ is unbounded above, given $G>0$, however large, there is a natural number $k$ such that
$$x_n>G, \;\forall n\geq k$$
Let $b=\min\{x_1,x_2,\dots,x_{k-1},G\}$. Then $x_n\geq b, \; \forall n\in \mathbb{N}$.

The question is : if we do not include $G$ in calculation of $b$, i.e., if $$b=\min\{x_1,x_2,\dots,x_{k-1}\}$$ what will be the problem?

Answer 1

G might smaller than all the preceding numbers. Then the statement $x_n \ge b$ would be false, since $b > G.$ There's nothing that says that $G$ is a good bound, after all.

Answer 2

For $x_{1}=1,x_{2}=1,x_{3}=1/2,x_{4}=4,x_{5}=5,...$ and if we take $G=1/3$, surely we have $x_{n}>1/3$ for all $n\geq 3$. If we set $b=\min\{x_{1},x_{2}\}$ only, which is just $b=1$, then it fails to have $x_{n}\geq b=1$ for all $n$ because $x_{3}=1/2<1$.