If $b$ is the largest element of $B$ then it is also maximal element of $B$

Question

Suppose $R$ is a partial order on $A$, $B \subseteq A $ and $b\in B$. Prove that if b is largest element of B, then it is a maximal element of $B$.

I have to prove now that $$\lnot \exists x \in B (bRx \land b \neq x).$$ This can be written as

$$\forall x \in B ((b,x) \notin R\,\, or\,\, b=x) .$$

Let $b$ be largest element of $B$. So $ \forall x \in B (x,b) \in R$. Now assume $x\neq b $. Also assume that $(b,x) \in R,$ $\forall x \in B$. So $b=x$ which is a contradiction as $b\neq x$. Hence $(b,x) \neq R$.

Is this right? Thank you.

Answer 1

It is too complex.
It is the order dual of your next question.
As such, it is the same question with the order reversed.
The proof is the same with the order reversed.

Clearly b is an upper bound of B.
Let x be an upper bound of B. Since b in B, bRx.
Thus b is the least upper bound.