Question: Determine the boundedness of $X = \bigcup_{n=1}^\infty \left[n^2, n^2+1\right]$, and, if they exist, its infimum and supremum (justifying your answer).
I know that $X$ is not bounded above, but is bounded below such that $\inf S=[1,2]$.
How can I prove that it's not bounded above? I feel like I could use the Archimedean property but am not sure how to apply it.
Nor do I know how to prove the infimum. I know $[1,2]\in X$ as for $n=1$, $[1^2, 1^2 +1] = [1,2]$. Is it sufficient to then show that $n \ge 1 \implies n^2\ge 1^2 = 1 \implies n^2+1\ge1+1=2$ therefore $\inf S=[1,2]$?
The set $X$ is not bounded above because if $M\in\mathbb R$, there is (by the Archimedian property) a number $n\in\mathbb N$ such that $n>M$. But then $n^2>M$ and $n^2\in X$.
On the other hand, $1\in X$ and every element of $X$ is greater than or equal to $1$. Therefore $1=\min X$. In particular, $1=\inf X$.