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I have the set $A=\{1,2,3,4\}$ and my relation is on $\mathcal{P}(A)$ where $X\subseteq Y \wedge XRY$
I wrote $\mathcal{P}(A)$ and get
$$\mathcal{P}(A) =\{(1,1),(2,2),(3,3),(4,4),(1,2),(1,3),(1,4),(2,1),(2,3),(2,4),(3,1),(3,2),(3,4),(4,1),(4,2),(4,3)\}$$ my question is if there is minimal or maximal terms in this power set? becuase the definition is that (let say we want the minimal ) for minimal is for all $a\in A \rightarrow xRa$ and there is few minimals terms.
Thanks.

Ofir Attia
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    Note that $\mathcal{P}(A) \neq A \times A$. Recall that $|\mathcal{P}(A)| = 2^{|A|}$. – Adriano Jul 31 '13 at 19:58
  • why its not equal? – Ofir Attia Jul 31 '13 at 20:10
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    Their cardinalities happen to be the same ($2^4=4^2$), but that's just a coincidence. The actual elements in each set are different (for example, $(2,3)\in A \times A$ yet $(2,3) \notin \mathcal{P}(A)$, and ${2,3,4} \in \mathcal{P}(A)$ yet ${2,3,4} \notin A \times A$. – Adriano Jul 31 '13 at 22:35
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    I don't really understand your ordering relation $R$. Given any two elements of $\mathcal{P}(A)$, how can we determine which is "better" than the other? For example, one possible relation is: $$ \forall X,Y\in \mathcal{P}(A),~~ XRY \iff X \subseteq Y $$ – Adriano Jul 31 '13 at 22:37
  • Adriano,this is equal! |p(a)|=16 – Evyatar Elmaliah Jul 31 '13 at 20:49
  • This is completely irrelevant. $\wp(A)\ne A\times A$, which is what the OP wrote. The fact that they have the same cardinality is quite beside the point. (And what you wrote here should have been a comment.) – Brian M. Scott Aug 01 '13 at 01:38
  • @BrianM.Scott what is $\mathcal P(A)$? – Ofir Attia Aug 02 '13 at 09:52
  • @Ofir: It’s the set of all subsets of $A$. If $A$ were ${a,b,c}$, for instance, it would be $$\Big{\varnothing,{a},{b},{c},{a,b},{a,c},{b,c},{a,b,c}\Big};.$$ – Brian M. Scott Aug 02 '13 at 09:54
  • ${ \phi, {1},{2},{3},{4},{1,2},{1,3},{1,4},{2,3},{2,4},{3,4},{1,2,3,4}}$ – Ofir Attia Aug 02 '13 at 10:01
  • I need to add ${1,2,3},{1,2,4},{1,3,4},{2,3,4}$? – Ofir Attia Aug 02 '13 at 10:05
  • @Ofir: Yes, you do; that brings you to $16=2^4$ subsets, which is correct. – Brian M. Scott Aug 02 '13 at 10:16

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