Closure of anti symmetric relations.

Question

Can someone please help me prove that anti symmetric relations are closed under set difference ?

I tried to represent set difference as

A - B = A intersection (B' ).

But we know that anti symmetric is closed under intersection and not under complement. With this property I am not able to prove closure of anti symmetric property under set difference.

Answer 1

$A\setminus B$ is a subset of $A$, and the class of antisymmetric relations is closed under taking subsets. (It doesn't matter whether $B$ is antisymmetric or not).

Answer 2

$a\in A\setminus B$ it's $a\in A$ and $a\not\in B$, which is $a\in A\cap B'$ and we are done!

Answer 3

Anti symmetric relations are closed under set difference, Set Intersection and every subset of Anti symmetric is Anti symmetric.

Every time you try to add something in an Anti Symmetric set... There is a chance to make it not Anti Symmetric. Operations like supersets, Union.

Let's say 2 antisymmetric relations R1 = { (1,2) } R2 = { (2,1) }

But the moment we do intersection

R1 ∩ R2 = { (1,2) , (2,1) } which makes a not anti symmetric.

Likewise while performing complement ... We might end up having Symmetric Relations....