Let $P$ be defined by $P:=\{x \in \mathbb R | 0 \lt x\}$. $P$ in this case represents the set of all positive real numbers. I want to show that $0$ is the greatest lower bound of P. I'm pretty sure I know how to show it is a lower bound. Indeed, if $x \in P$, then $0 \lt x$, so by definition, $0$ is a lower bound for $P$. However, I'm not entirely sure how I should be proving that it is the greatest.
Should I use proof by contradiction in this case and assume that there is a lower bound $y \in \mathbb R$ of $P$ such that $0 \lt y$? If I do, I'm not sure where to go from there. All I end up getting is something like $y \in P$ and that's it. I'm not sure how to prove rigorously that there exists some number $z \in P$ such that $z \lt y$.
Thanks in advance.
Best Answer
There are two steps to this proof, showing 0 is a lower bound, and then showing no number greater than 0 can be a lower bound. The first step is self-explanatory. For the second step, we assume that 0 is not the greatest lower bound. Then there is a positive real number $\epsilon$ such that all $x$ satisfy $x > \epsilon$. But $\epsilon/2$ is also a positive real number, and it is less than $\epsilon$. This contradiction means 0 must be the greatest lower bound.