Note: If f is injective (one to one) then s2 will also be correct.

33 votes

Let $f: A \rightarrow B$ a function, and let E and F be subsets of $A$. Consider the following statements about images.

- $S1: f(E \cup F) = f(E) \cup f(F)$
- $S2: f(E \cap F)=f(E) \cap f(F)$

Which of the following is true about S1 and S2?

- Only S1 is correct
- Only S2 is correct
- Both S1 and S2 are correct
- None of S1 and S2 is correct

62 votes

Best answer

Say $E=\{1,2\}$ and $F=\{3,4\}.$

- $f(1)=a$
- $f(2)=b$
- $f(3)=b$
- $f(4)=d$

$f(E\cup F)=f(1,2,3,4)=\{a,b,d\}$

$f(E)\cup f( F)=f(1,2)\cup f(3,4)=\{a,b\}\cup \{b,d\}=\{a,b,d\}$

Now, $E\cap F=\emptyset$

$f(E\cap F)=f(\emptyset)=\emptyset$

But, $f(E)\cap f(F)=f(1,2)\cap f(3,4)=\{a,b\}\cap \{b,d\}=\{b\}$

So, S2 is not true. S1 is always true (no counter example exists)

Correct Answer: $A$

22 votes

Here Answer is A .

S1 is always True.

S2 is false Consider case where E & F do not intersect, i.e. Intersection is empty set. In that case , F(E) and F(F) might have some common elements.

S1 is always True.

S2 is false Consider case where E & F do not intersect, i.e. Intersection is empty set. In that case , F(E) and F(F) might have some common elements.

26

$S_2$ is **False**: Consider the case where $f$ is constant and $A$ and $B$ are disjoint.

But if function $f$ is injective, then $S_2$ is also true.

$ f \: \text{injective} \rightarrow f(A \cap B) = f(A) \cap f(B) $

http://www.blaetterundsterne.org/proofTaoAnalysisi1E3431.html

see option D of this question also- https://gateoverflow.in/2036/gate2014-3_2

16 votes

Answer is A , becouse ...

*S*1:*f*(*E*∪*F*)=*f*(*E*)∪*f*(*F*)*S*2:*f*(*E*∩*F*)<=*f*(*E*)∩*f*(*F*)

For S2, Consider no common elements between E and F but some element in E mapping to an element x, and some other element in F also mapping to that x. Here, LHS will be empty set while RHS will have x in it.

0 votes

**Correct Answer (A)**

$1.\ f(A_{1}\cup A_{2})=f(A_{1})\cup f(A_{2})$

$f(A_{1}\cup A_{2})=\{y\in B|y=f(x),x\in A_{1}\cup A_{2}\}=\{y\in B|y=f(x),x\in A_{1}\ or\ x\in A_{2}\}=\{y\in B|y=f(x),x\in A_{1}\}\cup \{y\in B|y=f(x),x\in A_{2}\}=f(A_{1})\cup f(A_{2}).$

$2.\ f(A_{1}\cap A_{2})\neq f(A_{1})\cap f(A_{2})$

$A=\{1,2\},B=\{3,4\}$ and $f=\{(1,3),(2,3)\}$

$A_{1}=\{1\},A_{2}=\{2\}$

$f(A_{1}\cap A_{2})=f(\phi)=\phi$ while $f(A_{1})\cap f(A_{2})=\{3\}\cap \{3\}=\{3\}$

**Note : **

$f(A_{1}\cap A_{2})\subseteq f(A_{1})\cap f(A_{2})$

$f(A_{1}\cap A_{2})=f(A_{1})\cap f(A_{2})$ if f is ONE-ONE