# DM-Sets,Relations and Functions-Q6

Let $f$ be one to one function.

S1: $f(S\cup T) = f(S)\cup f(T)$

S2: $f(S\cap T) \subseteq f(S)\cap f(T)$

Which of the following is TRUE?

(A). Only S1

(B). Only S2

(C), Both S1 and S2

(D). Neither S1 nor S2

asked Jul 5, 2019
reshown Jul 6, 2019

+1 vote

Both S1 and S2 are true. Please note that this statements are true for any function. Function need not be one-one.

answered Jul 8, 2019 by (34,710 points)
selected Jul 9, 2019
sir please give me one example in S2 in which LHS would be a proper subset of RHS and a function need to one to one
We can not get such example for one-one function as  for one to one function f(S1 intersection S2) = f(S1) intersection f(S2).
so the equation becomes wrong na at every time it has to be equal but in ques the equation says it can be proper subset of RHS
Equation can not be wrong. Try to understand the definition of subset OR equal.
For A subset or equal to B to be true A =B is sufficient.
A should be the answer S2 should be true when it is not a one to one mapping but there is a one to one mapping so S2 equations would like LHS=RHS
answered Jul 6, 2019 by (850 points)
provide a counter example for S2
for which case when it onto or one to one
when function is one to one then
suppose 1->a
2-> b
3-> c
then two subsets 12 and 23 then  f(12) give a,b and for f(23) give b,c intersection of both gives b only now when f(12 intersection 23) it gives f(2) it will gives b only it is true .
now when function is not one to one
a-> 1
b->2
c-> 2
now consider two subsets ab and c
then do ab intersection c it is phie f(phie)=phie
and if we do f(ab)intersection f(c) then 12 intersection 2 it would come 2 now equation be like f(A interection B)<= f(A)intersection f(B)
correct me if i am wrong
so phi ⊆ {2}
this satisfies
but the function which i had taken that was not onto i was saying if it was not a one to one function then S2 would be true
S and T mean what here... nothing is given about them. And i think it is similar to a gate question in which S2 also had equal to operator not the subset operator
I think C is correct.
answered Jul 6, 2019 by (34,740 points)

$S2 : f(S \cap T) \subseteq f(S) \cap f(T)$

Let's assume $\exists y \in f(S \cap T )$ but doesn't exist in $f(S) \cap f(T)$

and also f(x) = y

So if y  $\in f(S \cap T )$ then x $\in (S \cap T )$

but we are also stating that $y \notin f(S) \cap f(T)$

this is only possible when x $\notin (S \cap T )$

So this is a contradiction

Hence we can say that S2 is true.

Note: Notice that y can only exist because of x because f is one to one

answered Jul 7, 2019 by (1,710 points)
now see my answer
–1 vote

see this i have provided a counter example clearly functions is not one to one

answered Jul 7, 2019 by (850 points)
This is not a counter example

Provide an example where function is one to one and S2 condition is not satisfied
Then only we can say say S2 is false
only i wanted to say in S2 LHS=RHS
can you provide me any counter example in which function is one to one and
LHS is a proper subset of RHS as it is written as LHS  can bv proper as well improper suset in the question.