Answer
a) No
b) No
c) $A=B$
Work Step by Step
a) $A \cup C= B \cup C$
A and B do not necessarily have to be the same set.
For example:
$A=${$0$}
$B=${$1$}
$C=${$0,1$}
By the definition of the union,
$A \cup C=${$0,1$}
$B \cup C=${$0,1$}
Note that the equality $A \cup B=B \cup C$ holds, while the two sets $A$ and $B$ are not equal.
b)$A \cap C= B \cap C$
A and B do not necessarily have to be the same set.
For example:
$A=${$0,1$}
$B=${$0,2$}
$C=${$0$}
By the definition of the intersection,
$A \cap C=${$0$}
$B\cap C=${$0$}
Note that the equality $A \cap B=B \cap C$ holds, while the two sets $A$ and $B$ are not equal.
c) Given: $A \cup C=B \cup C$ and $A \cap C=B \cap C$
To prove: $A=B$
Proof:
First case: Let $x$ be an element of $A$$$x \in A$$
By the definition of union, $$x \in A \cup C$$
By using $A \cup C=B \cup C$, $$x \in B \cup C$$
then $x$ has to either in $B$ ($A \subseteq B$) or in $C$ or in both.
Let us assume $x \in C$
By the definition of intersection and $x \in A$ and $x \in C$ $$x \in A \cap C$$.
By $A \cap C=B\cap C$:
$\implies x\in B\cap C$
By the definition of intersection, $x$ is also an element of $B$.
$\implies x \in B$
Since all elements of $A$ are in $B$
$\therefore$ By the definition of sets $A\subseteq B$
Second case:Let $x$ be an element of $B$$$x \in B$$
By the definition of union, $$x \in B \cup C$$
By using $A \cup C=B \cup C$, $$x \in A \cup C$$
then $x$ has to either in $A$ ($B \subseteq A$) or in $C$ or in both.
Let us assume $x \in C$
By the definition of intersection and $x \in B$ and $x \in C$ $$x \in B \cap C$$.
By $A \cap C=B\cap C$:
$\implies x\in A\cap C$
By the definition of intersection, $x$ is also an element of $A$.
$\implies x \in A$
Since all elements of $B$ are in $A$
$\therefore$ By the definition of sets $B\subseteq A$
Conclusion: $A\subseteq B$ and $B\subseteq A$, then the two sets have to be equal.
$\implies A=B$
