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The question I am working on is, "Find the duals of these posets.

a) $(\{0,1,2\},≤)$

b) $(\Bbb Z,≥)$

c) $(P(\Bbb Z),⊇)$

d) $(\Bbb Z^+,|)$

In my textbook, they say to find the dual of a poset, you simply find $R^{-1}$. To find the inverse of a relation, would that involve me finding the inverse of, say, $\le$ by switching the inequality to $\ge$? If so, why procedure find the inverse of a relation? I am having a little difficulty comprehending that. In time prior, when I had to find the inverse of a relation, I simply flipped the numbers around in an ordered-pair.

Edit:

I think the ones I am having most trouble with are parts c and d. Does part c mean that the relation $R$ is a subset of $P(\Bbb Z) \times P(\Bbb Z)$, where $A$ and $B$ elements of $P(\Bbb Z)$, and the relation is $R=\{(A,B)|A \subseteq B\}$? And the inverse of that relation would be $R^{-1}=\{(A,B)|A \supseteq B\}$?

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FWIW: I think your take on part (c) is right on. –  amWhy Nov 13 '12 at 23:13
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I agree. You understanding seems spot on. –  Baby Dragon Nov 13 '12 at 23:39
    
Wow! Thank you so much! –  Mack Nov 14 '12 at 0:10
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+1 EMACK {{{}}} –  amWhy Nov 16 '12 at 1:18
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2 Answers

up vote 1 down vote accepted

Let $R\in X\times X$ be a relation that is a poset. In other notation, one may write this as $R=\{(x,y):x\leq_R y\}$. Note this is a tautology and with a change in notation. The change in notation is that $\leq_R$ is simply a way to write $R$ that emphasized that you have a partial order. In note that $$x\leq_R y$$ if and only if $$(x,y)\in R$$ if and only if $$(y,x)\in R^{-1}$$ if and only if $$y\leq_{R^{-1}} x$$. This last expression is simply the dual poset.

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Oh, I see: $x \ge y$ is the same thing as $y \le x$. Also, I made an edit to my post, could you perhaps check it out for me? Thank you –  Mack Nov 13 '12 at 22:42
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$(d)\;(\Bbb Z^+,|)$

Here, our set is the set of all positive integers $\mathbb{Z}^+$. We have the relation $R$ given by $\;x\sim y\;$ if and only if $\;x$ divides $y$. More concisely, $$\text{for}\; x, y \in \mathbb{Z}^+,\quad R = \{(x, y): x\mid y\}.$$

You know that $$(x, y) \in R \iff x\mid y,$$

if and only if $$(y, x) \in R^{-1}\;\iff y\mid x.$$

This gives us the relation $R^{-1}$ defined on $\mathbb{Z}^+$ such that $$R^{-1} = \{(y, x): y|x\};\; \text{ in ohter words}\;\; R^{-1} = \{(y,x): \exists n \in \mathbb{Z}^+ \text{ such that}\; x = ny\}.$$

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Shouldn't it be that $x=ny$? Because $x|y$ is equivalent to $y=nx$, then taking the inverse of that would be $x=ny$, which is equivalent to$y|x$ –  Mack Nov 15 '12 at 20:46
    
@Emack Yes, typos...on my part. (I've been making a lot lately!) See edited text. –  amWhy Nov 16 '12 at 1:06
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