TITLE: Bounded nonempty countable sets necessarily contain its supremum?
QUESTION [5 upvotes]: 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

REPLY [3 votes]: For example: $$\mathbb Q\cap(0,1)$$