Let E={a,b,c,b} and F={1,2,3} Show that no pair of p(E×F) is included in {A×B,A€E,B€F}

To show that no pair of P(E \times F) is included in {A \times B, A \in E, B \in F} , we need to consider all possible pairs of P(E \times F) and show that none of them can be expressed as A \times B where A \in E and B \in F .

Given E=\{a,b,c,b\} and F=\{1,2,3\} ,

E \times F = \{(a,1), (a,2), (a,3), (b,1), (b,2), (b,3), (c,1), (c,2), (c,3), (b,1), (b,2), (b,3)\}

Let's list all possible pairs in P(E \times F) :

1. \{(a,1)\}
2. \{(a,2)\}
3. \{(a,3)\}
4. \{(b,1)\}
5. \{(b,2)\}
6. \{(b,3)\}
7. \{(c,1)\}
8. \{(c,2)\}
9. \{(c,3)\}
10. \{(b,1)\}
11. \{(b,2)\}
12. \{(b,3)\}

Now, let's consider A \in E and B \in F :

A=\{a,b,c\} and B=\{1,2,3\}

A \times B does not cover all possible pairs in P(E \times F) :

1. Missing pair: \{(b,1)\}
2. Missing pair: \{(b,2)\}
3. Missing pair: \{(b,3)\}

Therefore, no pair of P(E \times F) is included in {A \times B, A \in E, B \in F} .

\textbf{Answer:} No pair of P(E \times F) is included in {A \times B, A \in E, B \in F} .