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

My attempt:
Suppose $L_u$ is the set containing all lower bounds of $U$.
Suppose $x$ is g.l.b. of $U$.
Take arbitrary $b \in B$
Take $u \in U$. We know that $bRu$. Since $u$ was arbitrary, we conclude that $b$ is lower bound for $U$, which means $b \in L_u$
Since $x$ is g.l.b. of $U$, we have $x \in L_u$ and $bRx$.
Since $b$ was arbitrary element of $B$, it follows that $x$ is upper bound of $B$, thus $x \in U$.
Since $x \in L_u$, it also follows that for all $u \in U$, we have $xRu$, hence $x$ is the smallest element of $U$, which implies that $x$ is l.u.b. of $B$. $\Box$
Is it correct?

My attempt:
Consider L to be the set defined as L = {a $\in$ A | $\forall$ u $\in$ U (aRu)}. Or, L is the set of all lower bounds on U. If x is the greatest lower bound of U, x $\in$ L and thus $\forall$ u $\in$ U (xRu). Suppose u $\in$ U. Then, in particular, xRu. Since u was arbitrary, $\forall$ u $\in$ U (xRu). Or, x is the least upper bound of B.
