Prove Supremum of all subsets of a Poset is same with respect to some provided conditions.

Question

There is a poset $(P, \sqsubseteq )$, and let $A \subseteq B \subseteq C \subseteq P$ , $\qquad$ Also given is Sup$(A)$ and Sup$(C)$ exist, and Sup$(A)$ = Sup$(C)$.

I have to prove that Sup$(B)$ exists and Sup$(B)$ = Sup$(A)$.

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My Work:-

Let $x \in A$ , So then $x \sqsubseteq$ sup$(A)$

As $x \in A$ then $x \in B$, $x \in C$ because $A \subseteq B \subseteq C$

so it means $x \sqsubseteq$ Sup$(C)$

Now same $x$ is in $B$ and as we have proved that it has Sup$(A)$ and Sup$(C)$, So obviously it has a Supremum in $B$ as well, Hence Sup$(B)$ exists.

And also obviously for the same element Sup$(A)$ = Sup$(C)$ , so Sup$(B)$ would also be same.

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Query:- I am not sure If I have proved it in a correct way. If anybody has a comment or hint to solve this then please share it.

Answer 1

It does not look right to me. Yes it looks obvious but you have to prove it and that is the point of the question.

Suppose $Sup(B)$ does not exist.

Then for all $x\in B$, we also have $x\in C$ since $B \subseteq C$, so $x \sqsubseteq Sup(C)$.

Since $Sup(B)$ does not exist we can find a $y$ that $x\sqsubseteq y\sqsubseteq Sup(C)$ for all $x\in B$.

Now consider $A$, for all $z\in A$, we also have $z\in B$ and therefore $z \sqsubseteq y\sqsubseteq Sup(C)=Sup(A)$ contradiction.

Hence $Sup(B)$ exist. The other part is easy, since $Sup(B)$ exist it must be at least $Sup(A)$ and at most $Sup(C)$ so it must be equal to them.

Answer 2

You did not show that $x\sqsubseteq\sup C$ for each $x\in B$: you started with an $x\in A$, so you showed it only for elements of $A$.

To show that $\sup A$ is an upper bound for $B$ you should begin with an arbitrary $x\in B$; then $x\in C$, so $x\sqsubseteq\sup C=\sup A$, and $\sup A$ is indeed an upper bound for $B$. If $\sup A\ne\sup B$, $B$ must have an upper bound $u$ such that $u\sqsubset\sup A$. Then $u$ is not an upper bound for $A$, so there is some $a\in A$ such that $a\not\sqsubseteq u$. But $a\in B$, since $A\subseteq B$, and $u$ is an upper bound for $B$, so $a\sqsubseteq u$. This contradiction shows that $\sup B=\sup A$.