Is this true or false? I know how to prove that finite sets contain its supremum.
Let $A=\{a_1,…,a_n\}$ be a finite subset of $\mathbb R$. Since $A$ is finite, $\max(A)$ exists (can be proved by induction). And $\max(A)=\sup(A)$. So finite sets contain its supremum.
But countable sets includes finite sets and countable how to prove a bounded countable infinite set contains its supreme. I don't think the maximum method works any more for countable infinite set. Does this has something to do with closure? need some hints please
Best Answer
False: $\left\{1-\frac1n\right\}_{n\in\Bbb N}$.
Does it has something to do with closure? Yes, a closed bounded set necessarily contains its supremum, even if it is uncountable.