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

Question

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 ?

Answer 1

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 ... ?

Answer 2

I have been struggling myself with this proof for some hours now. And as many of you, I started by saying that $b$ was an arbitrary element of $B$. However, I would like us to rethink this, because this is false! Note that the goal is to prove that $x$ is the l.u.b. of B, so this means that $x$ is the smallest element of $U$ (not $B$), the set of all upper bounds for $B$. Thus, our proof should start by assuming some $u\in U$. The arbitrary $b$ appears later. This is the proof I wrote:

Suppose that $x$ is g.l.b. of $U$. Suppose $u_2\in U$, and so we have to prove that $xR_2$ and $x\in U$. Suppose $b\in B$, so our new goals are $xRu_2$, $bRx$, and $x\in A$. Let $L_U$ be the set of l.b. of $U$. Since x is g.l.b. of $U$, then it is the largest element of $L_U$. Thus, $\forall m_1\in U (xRm_1)$ and $x\in A$. Given that $\forall m_1\in U (xRm_1)$ and $u_2\in U$, then $xRu_2$. Also, since $u_2\in U$, it follows that $\forall b\in B (bRu_2)$ and $u_2\in A$. Since $b\in B$ and $\forall b\in B (bRu_2)$, then $bRu_2$. Given that $x,u_2\in A$, then $u_2Rx$. Since $R$ is a partial order, then it is transitive on $A$. Given that $bRu_2$ and $u_2Rx$, then $bRx$. Since $u_2$ was an arbitrary element of $U$, then $x\in U$. Since $u_2$ was an arbitrary element of $U$, then $x$ is a l.u.b. of $B$.