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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.

  • $S_1: f(E \cup F) = f(E) \cup f(F)$
  • $S_2: f(E \cap F)=f(E) \cap f(F)$

Which of the following is true about S1 and S2?

  1. Only $S_1$ is correct
  2. Only $S_2$ is correct
  3. Both $S_1$ and $S_2$ are correct
  4. None of $S_1$ and $S_2$ is correct
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5 Answers

Best answer
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98 votes

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, $S_2$ is not true. $S_1$ is always true (no counter example exists)

Correct Answer: $A$

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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.
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21 votes

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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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

Answer:

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