# Proof verification for proving greatest lower bound for a set with a lower bound.

Given that every set with an upper bound has least upper bound, I have to show similar result can be applied to sets with lower bounds. Here's how I did it:

Let $-S = \{-s|s \in S\}$. Suppose $-S$ has an upper bound, then $-S$ must have a least upper bound, namely, $L \in \mathbb{R}$. Hence, $\forall -s \in -S$, $-s < L$ such that there's no $L'$ so that $-s<L'<L$. This implies $s\geq-L$ for all $s \in S$. Hence, we have $-L$ as a lower bound for $S$. To show $-L$ is the greatest lower bound, suppose there is $-L'$ such that $s>-L'>-L$. But this implies $-s<L'<L$, which contradicts that $L$ being the least upper bound of the set $-S$. Hence, $-L$ is the greatest lower bound of the set $S$.

It looks good except for one detail.

I would first say that if $L$ has a lower bound then that number multiplied by $-1$ is an upper bound for $-L$.

That way the proof is complete, namely that if $L$ has a lower bound then $L$ has a greatest lower bound.

Let $S$ be a set with upper bound. Then $S$ has the least upper bound, defined by $L$. Namely,

(1) For for $\forall x\in S$, $x\le L$;

(2) For $\forall \varepsilon>0$, there is $x\in S$ such that $L-\varepsilon<x\le L$.

Thus

(1) For $\forall x\in -S$, then $-x\in S$ and hence $-x\le L$ or $x\ge -L$;

(2) For $\forall \varepsilon>0$, there is $t\in S$ such that $L-\varepsilon<t\le L$. Let $t=-x$ and then $-L\le -t<-L+\varepsilon$.

Thus $-L$ is the largest lower bound of $-S$.