Bounded nonempty countable sets necessarily contain its supremum?

Question

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

SBF · Accepted Answer · 2013-01-11T22:18:27.720

3

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.

edited Jan 11 '13 at 22:18

answered Jan 11 '13 at 21:51

SBF

36,041

now i get it. thank you so much – Yunhui Shi Jan 11 '13 at 22:29
@YunhuiShi: you're welcome! – SBF Jan 11 '13 at 22:31

score 3 · Answer 2 · answered Jan 11 '13 at 22:10

3

For example: $$\mathbb Q\cap(0,1)$$

answered Jan 11 '13 at 22:10

Asaf Karagila

393,674

Bounded nonempty countable sets necessarily contain its supremum?

2 Answers2