I came across this question in my assignment:
Are the following true for all functions f and for all subsets A, B and C of the domain f?
(i) $(\{f(x): x\in A\} \cup \{f(x):x\in B \}) \cap \{f(x):x\in C\}$ is a subset of $(\{f(x):x \in A\}\cap \{f(x):x\in B \}) \cup \{f(x):x\in C\}$
(ii) $(\{f(x):x \in A \} \cap \{f(x):x \in B\}) \cup \{f(x):x \in C\}$ is a subset of $(\{f(x):x \in A\} \cup \{f(x):x \in B\}) \cap \{f(x):x \in C\}$
The way I reasoned this was I assume some random elements for A,B,C and then I calculated the results. And the answer I came up with: True for (i) and False for (ii).
Can anyone tell me if the answer is correct, or I am not supposed to do it this way?
$\endgroup$ 121 Answer
$\begingroup$Let me consider $A$ in place of $f(A)$ and hope, that you can translate results to functions.
Accordingly to comments, let me understand $(i)$ as$$(A \cup B) \cap C \subset (A \cap B) \cup C \quad (i)$$Classically, in sets, for prove subset by definition we take any element from left side and try to prove, that it is in right. Here I would like suggest following consideration: left side is subset of $\boldsymbol{C}$, when right side is $\boldsymbol{C}$ union with some extra elements, so left is subset of right.
Accordingly, we write $(ii)$ as$$(A \cap B) \cup C \subset (A \cup B) \cap C \quad (ii)$$Here is enough counterexample: lets consider $C=\emptyset$. Then right side, obviously is $\emptyset$, when left side is $A \cap B$. To finish you need consider such functions, for which $A \cap B \ne \emptyset$.
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