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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?

1. Only S1 is correct
2. Only S2 is correct
3. Both S1 and S2 are correct
4. None of S1 and S2 is correct
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+1

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

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)

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+1
Well, you can never claim no counter example exists without formal proof.

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.
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what is the meaning of f(E Union F) .please explain
+19

$S_2$ 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

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Nice point  Sachin Mittal 1 .Thanks:-)

+7

S2 is false consider an example f(x)=x2   and  A={Set of integers} B={set of integers(greater than zeros)}

Let E={3} F={-3}

f(E∩F)=f(E)∩f(F)

Taking LHS

E∩F= {3}∩{-3} =Empty

f(Empty)=Empty Set

Taking RHS

f(E)=9  f(F)=9

f(E)∩f(F) =9

Clearly LHS is not equal to RHS

Answer is A , becouse ...

• S1:f(EF)=f(E)f(F)
• S2:f(EF)<=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.
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