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Let $X$ and $Y$ be finite sets and $f:X \to Y$ be a function. Which one of the following statements is TRUE?

1. For any subsets $A$ and $B$ of $X, |fA \cup B| = |f(A)| + |f(B)|$
2. For any subsets $A$ and $B$ of $X, f(A \cap B) = f(A) \cap f(B)$
3. For any subsets $A$ and $B$ of $X, |f(A \cap B| = \min \{|f(A)|, |f(B)|\}$
4. For any subsets $S$ and $T$ of $Y, f^{-1}(S \cap T) = f^{-1}(S) \cap f^{-1}(T)$
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D)

3 out of 4 options can be eliminated with the help of a counter example.

Let $X = \{ a , b , c \}$ and $Y = \{ 1 , 2 \}$
A Function $f$ maps each element of $X$ to exactly one element in $Y$.
Let $f(a)=1 , f(b)=1 , f(c) =1$ and
$A = \{ a \}, B = \{ b , c \}$
A)
$|f ( A \cup B )| = |f (\{a,b,c\})| = |\{1\}| = 1$
$|f(A)|+|f(B)| = 1 + 1 = 2$ , LHS != RHS.

B)
$f ( A \cap B ) = f (\{\}) = \{\}$.
$f(A) \cap f(B) = \{ 1\} \cap \{ 1\} = \{ 1\}$
LHS!=RHS
C)
$|f ( A \cap B )| = |f (\{\})| = |\{ \}| = 0$
$\min\{|f(A)|,|f(B)|\} = \min(1,1) = 1$
LHS!=RHS

D) Its easy to see that this is true because in a function a value can be mapped only to one value. The option assumes inverse of function $f$ exists.
answered by Loyal (3.7k points)
edited by
if  f(a)=1,f(b)=2,f(c)=1 and
A={a},B={b,c}
A)
what would have been the value of    |f(A∪B)|??

what is this | f(x) |  in case of set ?
@Arjun sir.. For option D , we must assume that function is bijective. And if function is bijective then option B should also be true. ?

In option D, the inverse is used -- so it is safe to assume that it exists.

When I solve this question with the example in d pic I Iet the answer as option B.

Plz correct me if I am wrong anywhere

Here you have made a mistake .. according to your diagram .. f inv (s) = 1,3,4 .. and f inv (t)= 3,2 .. hence their intersection will be 3 .. which is equal to LHS .

Therefore option d is correct.
What question wants to ask is for One-One Functions option B) always holds.

B) For One to One functions, whatever is common in A and B will also give the same image in Y else it will give different images. Eg F(a)=1, F(b)=2, and F(c)=3

A={a,b} and B={b,c}

In A /\ B = {b} then images of A and B will also give common element which is the image of b= 2. This is in agreement with One-One property if F(a)= F(b) then a=b only.

In D) option they have given a Bijective Function, which is One-One from both sides. So This property will hold in D)

A) and C) are easy to prove by taking counter examples. However, i can see that A) is definitely true if Function is One-One and A and B are disjoint sets(A /\ B = Null).

In C) This will be valid always only if either A is a subset of B or vice-versa((A /\ B) =A or B).
answered by (269 points)