Let $P$ be the set of all subsets of a given set $X$. Show that $\cup: P \times P \rightarrow P$ given by $(A, B) \rightarrow A \cup B$ and $\cap: P \times P \rightarrow P$ given by $(A, B) \rightarrow A \cap B$ are binary operations on the set $P$.
(N/A) binary operation $*$ on a set $S$ is a function $*: S \times S \rightarrow S$.
For the union operation $\cup: P \times P \rightarrow P$,for any two subsets $A, B \in P$,their union $A \cup B$ is also a subset of $X$,which means $A \cup B \in P$. Since every pair $(A, B)$ maps to a unique element $A \cup B$ in $P$,$\cup$ is a binary operation on $P$.
Similarly,for the intersection operation $\cap: P \times P \rightarrow P$,for any two subsets $A, B \in P$,their intersection $A \cap B$ is also a subset of $X$,which means $A \cap B \in P$. Since every pair $(A, B)$ maps to a unique element $A \cap B$ in $P$,$\cap$ is a binary operation on $P$.
For the union operation $\cup: P \times P \rightarrow P$,for any two subsets $A, B \in P$,their union $A \cup B$ is also a subset of $X$,which means $A \cup B \in P$. Since every pair $(A, B)$ maps to a unique element $A \cup B$ in $P$,$\cup$ is a binary operation on $P$.
Similarly,for the intersection operation $\cap: P \times P \rightarrow P$,for any two subsets $A, B \in P$,their intersection $A \cap B$ is also a subset of $X$,which means $A \cap B \in P$. Since every pair $(A, B)$ maps to a unique element $A \cap B$ in $P$,$\cap$ is a binary operation on $P$.
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(Hint: use the following table)
$^*$ $1$ $2$ $3$ $4$ $5$ $1$ $1$ $1$ $1$ $1$ $1$ $2$ $1$ $2$ $2$ $2$ $2$ $3$ $1$ $2$ $3$ $3$ $3$ $4$ $1$ $2$ $3$ $4$ $4$ $5$ $1$ $2$ $3$ $4$ $5$
(Hint: use the following table)
| $^*$ | $1$ | $2$ | $3$ | $4$ | $5$ |
| $1$ | $1$ | $1$ | $1$ | $1$ | $1$ |
| $2$ | $1$ | $2$ | $2$ | $2$ | $2$ |
| $3$ | $1$ | $2$ | $3$ | $3$ | $3$ |
| $4$ | $1$ | $2$ | $3$ | $4$ | $4$ |
| $5$ | $1$ | $2$ | $3$ | $4$ | $5$ |
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