For a Poset $(P,\sqsubseteq)$ and given is $A \subseteq P$ need to prove is $s=m$

Question

There is a Poset $(P,\sqsubseteq)$ and given is $A \subseteq P$ It has both supremum $s$ and maximum $m$

Need to prove is $s=m$

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My Work:-

Let $x \in A$ then $x \sqsubseteq m$ $\qquad$ { m is max., so it is an upperbound of $A$}

Also $x \sqsubseteq s $ $\qquad$$\qquad$ $\qquad$$\qquad$ {sup $s$ is always one of the upperbounds}

But we also know that Supremum is always the lowest of all upperbounds,

So $s \sqsubseteq m$

But I am not able to prove that $s=m$

Does anyone have any clue for solving this.

Answer 1

You know that $x\sqsubseteq s$ for each $x\in A$, so in particular $m\sqsubseteq s$, since $m\in A$. You showed that $s\sqsubseteq m$, and the relation $\sqsubseteq$ is antisymmetric, so ... ?