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In a given matrix below, say rows and columns are formed of {a,b,c,d}

$\begin{bmatrix}\ 1\ 0\ 0\ 0\ \\1\ 1\ 0\ 1\\1\ 1\ 1\ 1\\0\ 0\ 0\ 1\end{bmatrix}$

Then the pairs in the relation are (a,a) (b,a) (b,b) (b,d,) (c,a) (c,b) (c,c) (c,d) (d,d) And therefore the relation is an order

Now, how do I determine the least and minimal and greatest and maximal elements? This is not so easy to me as it when a poset is in single numbers. I have the answer, but I do not know how to arrive to the answer. I should say solution.

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2 Answers

up vote 3 down vote accepted

HINT: Since there are only four elements, a brute force approach is perfectly reasonable: work out the Hasse diagram. I’ll write $\le$ for the relation. From the first row of the matrix (or from your list of ordered pairs) it’s clear that $c\le x$ for every $x\in\{a,b,c,d\}$, so we can put $c$ at the bottom of the diagram. From the second row we see that $b\le a$ and $b\le d$, so $a$ and $d$ cannot be immediately above $c$ in the diagram. This suggests the following diagram, and it’s easy to check that it’s actually right:

             a   d  
              \ /  
               b  
               |  
               c

Once you have this, it’s pretty easy to spot the maximal and minimal elements and the least and greatest elements (if they exist): just pay attention to the definitions.

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Thank you I will try to understand this –  Aaron Nov 14 '12 at 23:31
    
Got it. Thank you. –  Aaron Nov 14 '12 at 23:32
    
@Aaron: You’re welcome. –  Brian M. Scott Nov 14 '12 at 23:37
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Hint: From your reading of the pairs, $c$ is less than everything and $a$ is not less than anything else. $d$ is also not less than anything else.

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What do you mean by c is less than "everything" ? –  Aaron Nov 14 '12 at 23:22
    
@Aaron: all the elements in your set: $a,b,c,d$ –  Ross Millikan Nov 14 '12 at 23:28
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