in Set Theory & Algebra edited by
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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
in Set Theory & Algebra edited by
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Note: If f is injective (one to one) then s2 will also be correct.

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4 Answers

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

Correct Answer: $A$

edited by

2 Comments

Well, you can never claim no counter example exists without formal proof.
17
This explanation is very good. But if you have drawn the diagrams then it would have been more easy to understand,

Still the best explanation
0
27 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.
edited by

5 Comments

what is the meaning of f(E Union F) .please explain
0
edited by

$S_2$ 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

32

Nice point  Sachin Mittal 1 .Thanks:-)

0

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

13

in the proof you shared, what’s the meaning of this?

 

f 1-1→f(A∩B)=f(A)∩f(B).

what does 1-1 imply here? sorry if this is a dumb question but I see no notation like this used in Rosen.

0
20 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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4 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

Answer:

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