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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.

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

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  • $\begingroup$ one more miscellaneous question: would it be different to denote order with notation mark $\le$ from $\subset$? $\endgroup$ – Beverlie May 14 '17 at 8:56
  • $\begingroup$ No. I find $\subset$ confusing as it often implies a strict order, like $<$ does, I prefer $\le$ for all partial orders. $\endgroup$ – Henno Brandsma May 14 '17 at 8:59
  • $\begingroup$ understand. has edited the PO. $\endgroup$ – Beverlie May 14 '17 at 10:38
  • $\begingroup$ @jackerysmith your messing up the quantors. Not $\forall y \in P x \le y$. That would say it's minimal $\endgroup$ – Henno Brandsma May 14 '17 at 10:41
  • $\begingroup$ corrected it. Put forall at the front , then if ~ $\endgroup$ – Beverlie May 14 '17 at 11:05

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