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

Question

Not a duplicate of this or this.

This is exercise $4.4.21.c$ from the book How to Prove it by Velleman $($$2^{nd}$ edition$)$:

Suppose $R$ is a partial order on $A$ and $B\subseteq A$. Let $U$ be the set of all upper bounds for $B$.

$(c)$ Prove that if $x$ is the greatest lower bound of $U$, then $x$ is the least upper bound of $B$.

Here is my proof:

Suppose $x$ is the $g.l.b.$ of $U$. We define $L=\{y|\forall z\in U(yRz)\}$ to be the set of all the lower bounds of $U$ and so we have $\forall l\in L(lRx)$. Let $b$ be an arbitrary element of $B$. Since $b$ is a lower bound for $U$, $b\in L$. From $\forall l\in L(lRx)$ and $b\in L$, $bRx$. Since $b$ is arbitrary, $x\in U$. Let $u$ be an arbitrary element of $U$ and so $\forall b\in B(bRu)$. Since $x\in L$, $\forall z\in U(xRz)$. Since $u\in U$, $xRu$. Since $u$ is arbitrary, $x$ is the smallest element of $U$. Since $x\in U$ and $\forall u\in U(xRu)$, $x$ is the $l.u.b.$ of $B$. Therefore if $x$ is the $g.l.b.$ of $U$ then $x$ is the $l.u.b.$ of $B$. $Q.E.D.$

Is my proof valid$?$

Thanks for your attention.

Answer 1

Today I solved this problem again and I noticed that

"Let $u$ be an arbitrary element of $U$ and so $\forall b\in B(bRu)$. Since $x\in L$, $\forall z\in U(xRz)$. Since $u\in U$, $xRu$. Since $u$ is arbitrary, $x$ is the smallest element of $U$."

is a little redundant and could be written as

"Since $x\in L$, from the definition of $L$ we obtain $\forall z\in U(xRz)$ which means that $x$ is the smallest element of $U$."

which is correct and simpler.

Answer 2

I think it would be more appropriate to begin by assuming that there is some $u_\in U$, since $x$ is the l.u.b. of $B$ means that $x$ is the smallest element of $U$. 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 $B$.