# Prove if $x$ is greatest lower bound of $U$ then $x$ is the least upper bound of $B$

This is one of the problem I have been solving from Velleman's How to prove book;

Suppose $R$ is a partial order on $A$ and $B \subseteq A$. Let $U$ be the set of all upper bounds for B. Prove that if $x$ is the greatest lower bound of $U$, then $x$ is the least upper bound of $B$.

Even though I have worked out the proof with an concrete example, I'm not able to prove it. This is one of my attempt:

Suppose $x$ is the greatest lower bound of $U$. Let $a$ be an arbitrary element in $A$. $a$ will be in $U$ if $\forall b \in B(bRa)$. $a$ will be in the lower bound of $U$ if $\forall u \in U (aRu)$. Since $x$ is the greatest lower bound of $U$ it follows that $aRx$. Suppose $b$ be an arbitrary element in $B$. From $\forall b \in B(bRa)$, it follows that $bRa$. From $bRa$ and $aRx$, it follows that $bRx$.

I'm stuck in this position. Can somebody point me out what I'm doing wrong ?

You need to show two things:

• that $x$ is an upper bound for $B$, i.e., that $b\mathrel{R}x$ for each $b\in B$, and
• that $x$ is the least upper bound for $B$, i.e., that if $y\in A$ and $b\mathrel{R}y$ for all $b\in B$, then $x\mathrel{R}y$.

You’ve tried to do the first, but your argument is a bit confused. For one thing, it starts in the wrong place: you want to show that $x$ is an upper bound for $B$, so you should be starting with an arbitrary $b\in B$ and trying to show that $b\mathrel{R}x$. This isn’t too hard: if $b\in B$, then $b\mathrel{R}u$ for each $u\in U$, so $b$ is a lower bound for $U$, and therefore $b\mathrel{R}x$. Thus, $x$ is an upper bound for $B$, and it only remains to show that $x$ is the least upper bound for $B$.

Suppose that $y$ is any upper bound for $B$. Then $b\mathrel{R}y$ for each $b\in B$, so $y\in U$. But then ... ?

• Why are you concluding $u R x$ ? Shouldn't it be $b R x$ ? – Sibi Apr 12 '15 at 20:38
• @Sibi: It should indeed; that was a typo. – Brian M. Scott Apr 12 '15 at 20:58
• Any ideas for the second part ? All I can come up with is $y R x$ from the assumption that $x$ is the greatest lower bound of $U$. Unfortunately, that is not leading me anywhere. I can apply transitive property, but it just gives me $bRx$ which is proved already. – Sibi Apr 12 '15 at 21:24
• @Sibi: The fact that $x$ is a lower bound for $U$, together with the fact that $y\in U$, tells you that $x\le y$, which is exactly what you want. – Brian M. Scott Apr 12 '15 at 21:30