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?
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.