Let `S` be the set of all sets and let `R={(A,B):A sub B}`,i.e., . `A` is a proper subset of `B`. Show that `R` is (i) Transitive (ii) Not reflexive (iii) not symmetric.
Clearly ,R satisfies the following properties :
Let `A ,B,C in S ` such that `(A,B)in R ( B,C)in R .`
Now `,(A,B) in R and (B,C) in R`
`implies Asub B and B sub C`
`implies A sub C`
` implies (A,C) in R.`
`therefore`R is transitive .
(ii) Nonreflexivity
Let A be any set in S.
then ` A, cancel sub ` A shows that `(A,A) !in R.`
`therefore `R is not reflexive .
(iii) NOnsymmetry
Now(A,B) in R implies A sub B `
`implies B cancel sub A `
`implies (B,A) !in R.`
` therefore ` R is not symmetric .
hence ,R is transitive but neither reflexive nor symmetric .
(i) `( sub B and B sub C )implies ( A sub C).` SO ,R is transitive
(ii) Clearly `A sub C` is not ture .SO ,R is not reflexive .
(iii) if `A sub B ` then `B sub C` is not true .SO .R is not symmetric .