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i have $A$=$\left \{1,2,3,4,5,6 \right \}$ $R= \left \{ (a,b)|a\in A, b\in P(A),a\in b \right \}$ and i want to answer about what properties does this relation has, my Dilemma is about transentive and symmetry

Note: in my course antisimmetric property is defined as followed:regular antisemmetric :=$(a,b)\in R \Rightarrow (b,a)\not\in R$ Wide antisemmetric:= $(a,b),(b,a) \in R \Rightarrow a=b $

I need to distinguish between element and a set of element right? how it's effect symmetry property and transitive property

I conculeded : R is not semmetric beacuse $1R\left \{ 1,2 \right \} but \left \{ 1,2 \right \} \not R 1 $ and it is not Reflexive infect it is anti-Reflexive beacuse $\forall a \in A (a,a)\not \in R$ but i awere that this is problematic beacue $\left \{ 1,2 \right \} \not \in A$

Thank you.

cisco_guy
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  • Uh... before we go too much further... how rigorously are you defining your natural numbers? Is the object you label as $1$ being considered as an urelement, i.e. something which is only considered as an element and not as a set itself? Or is it being considered as a set which is rigorously defined as ${\emptyset}$ while $2$ is rigorously defined as ${\emptyset,{\emptyset}}$ and so on? – JMoravitz Apr 23 '20 at 16:11
  • There might be two different final answers depending on the naivety of how you are treating your natural numbers. – JMoravitz Apr 23 '20 at 16:14
  • By the looks of it, you're not actually dealing with a relation on $A$ (i.e. a subset of $A \times A$), but a relation between $A$ and $P(A)$ (i.e. a subset of $A \times P(A)$). It doesn't really make sense to consider relation between two different sets being (anti)reflexive, (anti)symmetric, or (anti)transitive. – user771918 Apr 23 '20 at 16:16
  • Thank you guys, the Natural numbers in my Course is with the $0\in N $ , when i wrote 1 i considered as an element by itself not as a set while $\left { 1 \right }\in P(A)$ – cisco_guy Apr 23 '20 at 16:17
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    $1$ when treated as an urelement is not an element of $P(A)$, though ${1}$ is. In fact, when treating the natural numbers as urelements, none of the elements in the codomain of the relation are in the domain of the relation. When treating it like this... you technically have the relation is transitive, but trivially/vacuously so... since you never have a scenario where a first thing is related to a second thing and that same second thing is related to a third since if a first thing is related to a second thing, that second thing is not in the domain of the relation. – JMoravitz Apr 23 '20 at 16:19
  • Yes, user771918 1 , the relation is between $A$ and $P(A)$, i see there is a probleme with my conclution but i am not sure what will be the right and preciese answer or the explanation of it – cisco_guy Apr 23 '20 at 16:20
  • Thank you JMoravitz, and what about the symmetry part and (anti)symmetry?(as my course define it by 2 terms that i Noted) – cisco_guy Apr 23 '20 at 16:22
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    Things get more exciting when you treat them as sets... because you get statements like how $1\in 3$ since this is otherwise written as ${\emptyset}\in{\emptyset,{\emptyset},{\emptyset,{\emptyset}}}$ which is true as well as $3\in 5$... in fact, when treating them as sets via the Von Neumann construction of the naturals, you should find that it is in fact also true that it is transitive, but it is no longer because it is vacuously true. – JMoravitz Apr 23 '20 at 16:23
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    As far as symmetry... in neither interpretation will you have any symmetric pairs. If $x\in y$ and $y\in x$ simultaneously that violates the Axiom of Regularity, just as how this relation is not reflexive since that would have $x\in x$. As such, in either interpretation it is irreflexive and antisymmetric, either vacuously true because of a domain argument in the case of your natural numbers being urelements, or otherwise. – JMoravitz Apr 23 '20 at 16:26
  • JMoravitz , Interesting fact, i understand what you are implying although i didn't learn about "Von Neumann construction of the naturals" i will check it out. – cisco_guy Apr 23 '20 at 16:26

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