# Prove the infimum and supremum of the positive rational numbers

I am having this set:

$$X= \mathbb{Q^+} = \{x \in \mathbb{R} \ \ |x \in \mathbb{Q} \ \text{and} \ x>0 \}$$

How can I prove that $\inf X= 0$ and there is no supremum ?
(I think there is no maximum and no minimum in X?)

My attempt: $$X=]0, \infty[$$ $$\implies \inf \ B = 0$$

• For all $y \in B: y\ge0$
• For all $x >0 \ \exists y \in \ B:x>y$

My Questions: Is this a correct way to prove the infimum of B? How can I prove that there is no supremum?

• You need to show that $0$ is a lower bound and for an $\epsilon>0$, find some $x \in X$ such that $x<0+\epsilon$. It is fairly straightforward to show that $X$ is unbounded above. Nov 17, 2014 at 16:19
• @copper.hat thanks for your comment. Can you write it down correctly, so that I can understand your hint? Nov 17, 2014 at 16:22
• You can use the completeness axiom. You know that $\mathbb{Q^+}$ is bounded below, so you know that inf must exist. Similarly, it is not bounded above, and hence the least upper bound (ie supremum) cannot exist. Nov 17, 2014 at 16:24
• @DrkVenom Thanks for help. I used this know to proove that there is a infimum and NO Sup. How can I proove that this infimum is 0? Nov 17, 2014 at 16:28
• Show that $0$ is a lower bound. Then show that no number greater than $0$ can be a lower bound. Nov 17, 2014 at 16:44

Zero is a lower bound, because $x >0$ for all $x \in X$. For any $\epsilon>0$, we can find some $n \in \mathbb{N}\subset X$ such that ${ 1\over n} < 0+\epsilon$. Suppose $L$ is another lower bound for $X$. Then if $L>0$, the previous sentence gives a contradiction (take $\epsilon = {L \over 2}$), hence we must have $L \le 0$. Consequently zero is the greatest lower bound, that is, $\inf X = 0$.
Suppose $U \in \mathbb{R}$ is an upper bound. We have $\mathbb{N}\subset X$, and we can find some $n \in \mathbb{N}$ such that $U<n$, which contradicts $U$ being an upper bound.