# Is 0 the greatest lower bound for the set of all positive real numbers?

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 $$z \lt y$$.

• Assume that there is a greater bound. Jul 30, 2019 at 21:55
• That's what I was asking about. However, I'm not sure what to do next. I've edited some of the question.
– Tim
Jul 30, 2019 at 21:56
• Try letting $z=y/2$. Jul 30, 2019 at 21:58

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.

You have already shown that $$0$$ is a lower bound for the set of positive numbers.

In order to show that $$0$$ is the greatest lower bound you need to prove that if $$y>0$$ then $$y$$ is not a lower bound for the set of positive numbers.

Let $$y>0$$ be a real number and consider $$z=y/2$$

Note that $$z>0$$ and $$z so the set of positive numbers have an element element which is less than $$y$$ so $$y$$ is not a lower bound of the set of positive numbers.

Yes you are on the right path

We know that $$0$$ is a lower bound. Assume by way of contradiction that there is a non zero GLB $$y \in \mathbb{R}$$.

Note that $$y <0$$, for otherwise if $$y>0$$ then $$y/2 \in P$$ with $$y/2 . Namely, $$y$$ is not even a lower bound for $$P$$

So $$y <0$$. But this contradicts the fact that $$y$$ is the greatest lower bound (since $$0$$ is lower bound of $$P$$ greater than $$y$$)

• I don't think you need to say anything about $y/2$ when $y\lt0$, since the OP has already established that $0$ is a lower bound. Jul 30, 2019 at 22:00
• Thank you. I see it now.
– Tim
Jul 31, 2019 at 4:31