The Gateway to Computer Science Excellence
First time here? Checkout the FAQ!
x
+22 votes
2k views

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 $Y, f^{-1}(S \cap T) = f^{-1}(S) \cap f^{-1}(T)$
asked in Set Theory & Algebra by Veteran (106k points)
edited by | 2k views
Inclusion–exclusion principle.

2 Answers

+35 votes
Best answer

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)
LHS: $|f ( A \cup B )| = |f (\{a,b,c\})| = |\{1\}| = 1$

RHS: $|f(A)|+|f(B)| = 1 + 1 = 2$ ,

LHS != RHS.
B)
LHS: $f ( A  \cap B ) = f (\{\}) = \{\}$.

RHS: $f(A) \cap f(B) = \{ 1\} \cap \{ 1\} = \{ 1\}$

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

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
Can you please elaborate a little on d)...This is true that in a function a value can be mapped only to one value but how is d) true..?

(S ∩ T) contains common elements of S and T and f-1 of that will give the elements of X mapping to those common elements.

Now, f-1 (S) will give the elements of X mapping to the elements in S, and f-1 (T) will give the same for elements in T. Now, we take intersection and get elements of X which have a mapping to both S and T, or we only get elements of X for which a mapping exist to both S and T. As we know an element cannot have more than 1 mapping in a function. So, this would mean the mapped element must be present in both S and T. 

 

Thanks

aren't you assuming that function is bijective for the inverse(otherwise inverse doesn't exist) and so if we take bijective assumption for the option B then it is also true.

@srinath sri

i think in part A) |f({a,b,c})| would be 1 as f({a,b,c}) = {1} and this is same for |f(A)| and |f(B)|.
@srinath,

in option b , you have taken {1,1} and {1} as different sets, and concluded L.H.S != R.H.S , but in sets we dont allow same values again, then L.H.S will be equal to R.H.S.
yes. Those were wrong. Corrected the example now..
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. ?

Please explain.
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.
0 votes
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 Junior (559 points)
where hav they mentioned bijective function ???
If inverse exists for a function then it must be bijective.
Jst see Arjun sir's commet in the best answer ....
Answer:

Related questions



Quick search syntax
tags tag:apple
author user:martin
title title:apple
content content:apple
exclude -tag:apple
force match +apple
views views:100
score score:10
answers answers:2
is accepted isaccepted:true
is closed isclosed:true

32,732 questions
39,309 answers
110,280 comments
36,733 users