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

(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?

Andrés E. Caicedo
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manooooh
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  • you can just switch A and B in the definition of max, and get equality immediately. – zwim Feb 22 '20 at 17:14
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    Note that not every partial order has a maximum element. For example consider $A = B = \mathbb{R}$ with the usual ordering. Also your definition of $\operatorname{max}(A)$ is incorrect and self-referential. The correct definition should be: $\operatorname{max}(A) = a$ iff $\forall x \in A(x \le a)$. – Ldddd Feb 22 '20 at 17:15
  • @ShervinSorouri in my book it says that an element $m\in A$ is maximal iff for all $x\in A$, if $m\leq x$ then $x=m$. Didn't I write the same definition on my question? – manooooh Feb 22 '20 at 17:16
  • Why the negative votes? I have shown my work, why? – manooooh Feb 22 '20 at 17:18
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    @manooooh, oh, I thought you meant maximum. But then, still there are posets with no maximal elements, the same example in my previous quote works. Also in your proof you are setting an element of $\max A$ equal to $\max A$, which is self-referential and in contradiction with axioms of set theory. – Ldddd Feb 22 '20 at 17:20
  • @ShervinSorouri oh, thank you. What would be a correct proof so? – manooooh Feb 22 '20 at 17:21
  • @zwim, As the author is trying to prove that $\max$ is a function on the class of posets, what you suggest does not work. – Ldddd Feb 22 '20 at 17:22
  • Technically, what I am trying to prove is "higher". I suppose that is equal to "maximum", i.e. $\max$. – manooooh Feb 22 '20 at 17:23

1 Answers1

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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.

Ldddd
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  • Thank you! Question: If $A=B$ are posets then $(A,\leq_1)$ and $(B,\leq_2)$ i.e. can it be ${\leq_1}\neq{\leq_2}$? Or it must be ${\leq_1}={\leq_2}={\leq}$? – manooooh Feb 22 '20 at 17:37
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    No problem. No it can be that they have different orderings. But in the context of this question, I didn't mention it. I have added that in. – Ldddd Feb 22 '20 at 17:39
  • How do you go from $a \in \operatorname{maximals}(A)$ to $a\in B$? – manooooh Feb 22 '20 at 17:41
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    That's not part of the argument. I used $A=B$ to obtain $a \in B$. – Ldddd Feb 22 '20 at 17:43