Given $S=\displaystyle \bigcap^{\infty}_{k=1}\left(1-\frac{1}{k}, 1+\frac{1}{k}\right)$, what is $\sup(S)$ and $\max(S)$?
I reasoned that this is empty, since as $k$ goes to infinity, then $\frac{1}{k}$ goes to $0$. So ultimately, the intersection of all the intervals in $S$ is $(1,1)$, which should be empty. But the solution the supremum and maximum does exist and they are $1$. Why is this?
Since $1-\frac{1}{k} < 1 < 1+\frac{1}{k}$ for all $k>0$, $1$ is indeed in the intersection. Now, suppose there is some $1\ne a \in S$ and suppose $a<1$ (the case $1<a$ is similar.) Then, there exists $k\in \Bbb{N}$ such that $\frac{1}{k}<1-a$. Thus $a<1-\frac{1}{k}$ and $a\notin S$.
Therefore, $S=\{1\}$ and $sup(S)=inf(S)=1$.