Prove b is the g.l.b of B Let R be a partial order on A, $B \subseteq A$ and $b \in B$   
Prove that if b is the smallest element of B, then it is also g.l.b of B  
ATTEMPT
Let b be smallest element of B. So $\forall x \in B,  bRx$. This means b is lower bound. Now let $c$ be another lower bound of B. so cRb. Also we have bRx. So cRx. But this would mean that c is smallest element which is a contradiction.
Is this correct ?
Thanks
 A: Clearly b is a lower bound of B.
Let x be a lower bound of B. Since b in B, xRb.
Thus b is the greatest lower bound.
A: Let $R$ be a partial order on $A$; let $B \subseteq A$; let $b$ be the smallest element in $B$.  Then $b$ is a lower bound for $B$.  Suppose $l \in A$ is a lower bound for $B$.  Then $lRb'~\forall b' \in B$.  In particular, $lRb$ since $b \in B$. If $l \neq b$, then $l$ is smaller than $b$, so it is not a greatest lower bound for $B$ since $b$ is a lower bound for $B$.  Hence, $b$ is the greatest lower bound for $B$.    
A: Suppose that $b$ is the smallest element of $B$. Let $L$ be the set of all lower bounds of $B$. Clearly $b \in L$, so $L \neq \emptyset$. Now let $y \in L$ be arbitrary. Then $\forall x \in B (y R x)$, since $y$ is a lower bound of $B$. Letting $x = b$ in particular, we have $ y R b$. Since $y \in L$ was arbitrary, $\forall y \in L (y R b)$. Thus, $b$ is the largest element of L. Since $b$ is the largest element of the set of lower bounds $L$, $b$ is the greatest lower bound of $B$.
