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I am unable to understand how Equality of Sets is a POSET? How can it satisfy anti-symmetric property?

in Set Theory & Algebra by (1.2k points) | 37 views
equality means ‘=’ symbol btw two members !

if you collect all those pairs, then it is automatically reflexive, symmetric, transitive and anti-symmetric.

So it is a poset.

ex:- A is a Set of Sets { {1,2}, {3,4} } then a relation R:$\color{red}A$ → $\color{blue}A$, where R is equality

then elements are like $\large\color{magenta}\{{ (\color{red}{\{1,2\}}, \color{green}{\{1,2\}}), (\color{red}{\{3,4\}},\color{green}{\{3,4\}}) }\large\color{magenta}\}$

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Let set A = { 1,2,3} and set C = {1,1,2,2,2,2,3}

See A is related to C as both are equal and C is related to A as both are equal but at the same time u see set A is equal to set C.

Let ARC denotes A is related to C and CRA (reason I used to C ) be C is related to A.

or, ARC{1,2,3} R (1,1,2,2,2,2,3}

or, CRA is {1,1,2,2,2,2,3} R {1,2,3}
But by definition of equality of set u can see that {1,2,3} is same as {1,1,2,2,2,2,3}

And by definition of anti-symmetry, it is if aRc and cRa then a is equal to c.

So, equality of set is antisymmetry.
Initially it looks that from relation point of view, anti-symmetry is getting violated but the set has their own definition of equality.
by (1.6k points)
So u contradict a powerset is not a set ??

Let A = {null, {1}, { 2 } , {1,2}}

Let B = { 1, 2 } is a set.

Let C be power set of B.

Man ...... why cant u see A is containing a collection of elements ( whatever b da content ) so it is a set .

And C is da powerset of B which also contains every element which A is containing.  So ultimately both A and C are sets.

It is same as Virat Kohli is captain who looks after players including himself ( power set ) bt ultimately he is a player ( set ).
i am not saying power set is not a set !

i am saying, you can’t allow to compare power set with a normal set !

i mean to say, let A={ {1,2},{3,4} } and B={1,2} then you can’t use B ⊂ A
I understood ur example. Ur approach is correct bt since we need to consider the entire domain of universe ( say all sets of integers ) for the set, I dnt think its wrong to bring the set and its multiset in the relation set.

Say for division operation on integers we say it is anti-symmetric because if a divides b (a and b are not equal) then b does not divide a.

But in case of sets, say A={1,2,3} and B={1,2,3,1,1,2}

here A=B as well as B=A, so  how can anti-symmetry be preserved?


4 divides 8 but 8 does not divide 4 implies it not symmetric. Here we don't have to check for anti-symmetry property because the left side of implication is false because as per definition of anti-symmetry : 

if aRc and cRa, then a = c. It is represented as : 

aRc  ∧ cRa → a = c. We know that if left side of implication symbol is false, then the implication statement is always true.

So, the statement becomes :

(4 divides 8 and 8 divides 4 ) → ( 4 = 8 ).

Here 4 divides 8 and 8 divides 4 . This statement is false as 8 does not divide 4 over set of integers and so left side of implication is false, and hence the statement is true. So, it is anti-symmetric…...

Now comes for the sets.

The one Shaik explained i.e. {1,2} R { 1,2}, follows reflexive so this element of relation is also anti-symmetric.

Now coming to my example : 



Courtesy : Kenneth Rosen Discrete Maths.


Way before the multiset chapter was introduced in Kenneth Rosen Discrete Maths book, the above lines were said just at the start of the set chapter regarding equality of sets, i.e. the order and number of times an element occur does not matter.

So, as per definition of anti-symmetry : 

if ( {1,3,5} is related to { 1,3,3,3,5,5,5,5} ∧  { 1,3,3,3,5,5,5,5} is related to {1,3,5} )  → ( {1,3,5} is equal to { 1,3,3,3,5,5,5,5} ).

U can see that both sides of the implication are true and as per logic, True → True is always True.

Hence, the condition of anti-symmtry gets satisfied…..

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