- #1

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Let a function ##f:X \to X## be defined.

Let A and B be sets such that ##A \subseteq X## and ##B \subseteq X##.

Then which of the following are correct ?

a) ##f(A \cup B) = f(A) \cup f(B)##

b) ##f(A \cap B) = f(A) \cap f(B)##

c) ##f^{-1}(A \cup B) = f^{-1}(A) \cup f^{-1}(B)##

d) ##f^{-1}(A \cap B) = f^{-1}(A) \cap f^{-1}(B)##

My attempt-

For option a, let an element ##x \in f(A \cup B)##

##\Leftrightarrow## ##f^{-1}(x) \in A \cup B##

##\Leftrightarrow## ##f^{-1} (x)\in A## or ##f^{-1}(x) \in B##

##\Leftrightarrow## ##x \in f(A)## or ##x \in f(B)##

##\Leftrightarrow## ##x \in f(A) \cup f(B)##

##\Rightarrow## ##f(A \cup B) = f(A) \cup f(B)##

A similar analogy can be applied to options c and d as well.

However, option b doesn't seem to fit into this argument. Even though this approach seems rationale, a counter example can be given to disprove option b. It goes as follows :

Let ##X=\left\{1,2,3 \right\}##, ##A=\left\{1 \right\}## and ##B=\left\{2 \right\}##

Let function ##f## be defined as ##f(1)=3## and ##f(2)=3##

Clearly ##A \cap B = \phi## and hence ##f(A \cap B)## becomes undefined.

Therefore disproving option b.

But option d is correct even though option b is incorrect. Can somebody clarify this for me ?

In simple terms, if option a is right then why not option b (surely there must be some flaw in the above proof when applied for option b but what is it) ? And since option b is incorrect how can option d be correct ?

NOTE: The above question appeared in an exam and the correct answers are options a,c,d.

Let A and B be sets such that ##A \subseteq X## and ##B \subseteq X##.

Then which of the following are correct ?

a) ##f(A \cup B) = f(A) \cup f(B)##

b) ##f(A \cap B) = f(A) \cap f(B)##

c) ##f^{-1}(A \cup B) = f^{-1}(A) \cup f^{-1}(B)##

d) ##f^{-1}(A \cap B) = f^{-1}(A) \cap f^{-1}(B)##

My attempt-

For option a, let an element ##x \in f(A \cup B)##

##\Leftrightarrow## ##f^{-1}(x) \in A \cup B##

##\Leftrightarrow## ##f^{-1} (x)\in A## or ##f^{-1}(x) \in B##

##\Leftrightarrow## ##x \in f(A)## or ##x \in f(B)##

##\Leftrightarrow## ##x \in f(A) \cup f(B)##

##\Rightarrow## ##f(A \cup B) = f(A) \cup f(B)##

A similar analogy can be applied to options c and d as well.

However, option b doesn't seem to fit into this argument. Even though this approach seems rationale, a counter example can be given to disprove option b. It goes as follows :

Let ##X=\left\{1,2,3 \right\}##, ##A=\left\{1 \right\}## and ##B=\left\{2 \right\}##

Let function ##f## be defined as ##f(1)=3## and ##f(2)=3##

Clearly ##A \cap B = \phi## and hence ##f(A \cap B)## becomes undefined.

Therefore disproving option b.

But option d is correct even though option b is incorrect. Can somebody clarify this for me ?

In simple terms, if option a is right then why not option b (surely there must be some flaw in the above proof when applied for option b but what is it) ? And since option b is incorrect how can option d be correct ?

NOTE: The above question appeared in an exam and the correct answers are options a,c,d.

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