Prove that if $A=B$ then $\max A=\max B$

Question

Prove if $A=B$ then $\max A=\max B$.

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(It follows from Question regarding $\gcd(a,b)=\gcd(b,r)$).

What I have done:

I will assume that $A$ and $B$ are equipped with a partial order $\leq$.

Let $\max A=a=\{x\in A\mid a\leq x\to x=a\}$ be the maximum element of $A$. Since $x\in A$, by hipothesis, $x\in B$, and so $a\leq x$ for all $x\in B$. Calling $a=b$ we have that $b\leq x$, so $x=b$. We have shown $\max A\subseteq\max B$, the same idea for the converse.

Is it correct?

Answer 1

Let $A = B$ be posets with the same order. Then pick some $a \in \operatorname{maximals}(A)$. Since $A = B$, we have $a \in B$. Now let $x \in B$. Since $A = B$ we have $x \in A$ and so by the fact that $a \in \operatorname{maximals}(A)$ we have $x \le a$. Now since $x \in B$ was arbitrary, we conclude $a \in \operatorname{maximals}(B)$. We have proven $\operatorname{maximals}(A) \subseteq \operatorname{maximals}(B)$. The converse is similar.