(Verification) Greatest element is Maximal Element in Partially Ordered Set.

Claim

Greatest element is Maximal Element in Partially Ordered Set.

Proof

Let $(P, \le)$ be a partially ordered set P.

Let $x$ be the greatest element in $(P, \le)$

$x \in P$ is greatest element if and only if $y\le x$ for each $y \in P$

Then $\forall y\in P$ if $x\le y$, $x=y$ since $y\le x$ and $x\le y$, which implies that

$x$ is maximal element.

• It's fine, but be careful when writing things like "for $\forall y\in P$"... you really wanted "Then if $x\subset y$ for ANY $y\in P$, then..." or "$\forall y\in P$ if $x\subset y$, then..." – luka5z May 14 '17 at 8:38
• @luka5z modified it into "each" is it legit? – Beverlie May 14 '17 at 8:40
• Nope.Actually, it's the same. Order does matter. It should be rather: "$\forall y\in P$ if $x\subset y$, then $x=y$..." – luka5z May 14 '17 at 8:42
• $\exists x \forall y$ is not the same as $\forall y \exists x$ – luka5z May 14 '17 at 8:43
• @luka5z I understand but little confusing about the predicating order of case of OP.. are they conveying different meaning? – Beverlie May 14 '17 at 8:44

The idea is OK, writeup could be stricter: suppose that $x$ is the greatest element of $(P, \le)$. Then suppose $x \le y$ for some $y \in P$, we need to show $y = x$ (this is the definition of being a maximal element). By $x$ being the maximum element, we know already $y \le x$. But in a PO: $x \le y$ and $y \le x$ implies $y=x$ and we are done.
• one more miscellaneous question: would it be different to denote order with notation mark $\le$ from $\subset$? – Beverlie May 14 '17 at 8:56
• No. I find $\subset$ confusing as it often implies a strict order, like $<$ does, I prefer $\le$ for all partial orders. – Henno Brandsma May 14 '17 at 8:59
• @jackerysmith your messing up the quantors. Not $\forall y \in P x \le y$. That would say it's minimal – Henno Brandsma May 14 '17 at 10:41