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## Here's the Solution to this Question

Q#1.

LetΒ $x\in A\cap C$

Case 1,

SinceΒ $A\sube B, x\in A\implies x\in B$

$x\in B\implies x\in B\cap C$

Case 2,

$x\in C$

$x\in C\implies x\in B\cap C$Β sinceΒ $x\in B$

Thus, in either cases,Β $x\in A\cap C\implies x\in B\cap C.$

So,

$\forall x(x\in A\cap C\implies x\in B\cap C)$

We now have that,

$A\cap C\sube B\cap C$Β $\blacksquare.$

Q#2.

LetΒ $(x,y)\in(A\cap B)\times(C\cap D)$

$\iff x\in (A\cap B)\times(C\cap D)\times y$

$\iff x\in A\cap B\land y\in C\cap D$

$\iff x\in A\land x\in B\land y\in C\land y\in D$

$\iff x(A\times C)y\land x(B\times D)y$

$\iff x(A\times C)\cap(B\times D)y$

Therefore,

$(A\cap B)\times(C\cap D)=(A\times C )\cap (B\times D)$Β as requiredΒ $\blacksquare$Β .

Q#3.

To begin with, we will first show thatΒ $A\times (B\cap C)\sube (A\times B)\cap(A\times C )$. SupposeΒ $(x,y)\in A\times(B\cap C)$. This means that,Β $x\in A$Β andΒ $y\in B\cap C$Β by definition of Cartesian product. By definition of intersection, it follows thatΒ $y\in B$Β andΒ $y\in C$.SinceΒ $x\in A$Β andΒ $y\in B$, it follows thatΒ $(x,y)\in A\times B$Β by definition of Cartesian product. Also, sinceΒ $x\in A$Β andΒ $y\in C$Β it follows thatΒ $(x,y)\in A\times C$Β by definition ofΒ $\times$. Now, we haveΒ $(x,y)\in A\times B$Β andΒ $(x,y)\in A\times C$Β and so,Β $(x,y)\in (A\times B)\cap(A\times C)$. Having shown thatΒ $(x,y)\in A\times(B\cap C)\implies (x,y)\in (A\times B)\cap(A\times C)$, we have,Β $A\times(B\cap C)\sube(A\times B)\cap(A\times C)$.

Next, we show thatΒ $(A\times B)\cap(A\times C)\sube A \times(B\cap C)$,

SupposeΒ $(x,y)\in (A\times B)\cap(A\times C)$. By definition of intersection, it means that,Β $(x,y)\in A\times B$Β andΒ $(x,y)\in A\times C$. By definition of Cartesian product,Β $(x,y)\in A\times B$Β meansΒ $x\in A$Β andΒ $y\in B$Β also,Β $(x,y)\in A\times C$Β meansΒ $x\in A$Β andΒ $y\in C$. We have that,Β $y\in B$Β andΒ $y\in C$Β so,Β $y\in B\cap C$Β by definition of intersection. Thus, we have deduced thatΒ $x\in A$Β andΒ $y\in B\cap C$, soΒ $(x,y)\in A\times (B\cap C)$. Hence,

$(A\times B)\cap(A\times C)\sube A\times (B\cap C)$.

In summary, we have shown thatΒ $A\times (B\cap C)\sube (A\times B)\cap(A\times C)$Β andΒ $(A\times B)\cap(A\times C)\sube A\times (B\cap C)$. It follows that,Β $A\times(B\cap C)=(A\times B)\cap(A\times C)$Β as required$\blacksquare.$

Q#4.

SupposeΒ $x\in \mathcal{p}(A)\cup \mathcal{p}(B)$,

Then,Β $x\in\mathcal{p}(A) \space or\space x\isin\mathcal{p}(B)$

Hence,

$x\sube A\space or\space x\sube B$

We need to show thatΒ $x\in\mathcal{p}( A\cup B),$Β that is, we need to show thatΒ $x\sube A\cup B$

SupposeΒ $y\in x$Β then,Β $y\in A\space or\space y\in B$Β sinceΒ $x\sube A \space or \space x\sube B$

Thus,Β $y\in x\implies y\in (A\cup B)$Β and so,Β $x\sube (A\cup B)$.

Hence,

$x\in\mathcal {p}(A\cup B)$

$\therefore x\in \mathcal{p}(A)\cup \mathcal{p}(B)\implies x\in\mathcal {p}(A\cup B)$

and so,

$\mathcal{p}(A)\cup\mathcal{p}(B)\sube \mathcal{p}(A\cup B)$Β as requiredΒ $\blacksquare$Β .

Q#5.

Suppose thatΒ $A$Β andΒ $B$Β are sets andΒ $\mathcal p(A)\sube\mathcal {p})(B)$

By definition of a power set,Β $A\sube \mathcal {p}(A)$.SinceΒ $A\in \mathcal{p}(A)$Β andΒ $\mathcal{p}(A)\sube \mathcal{p}(B)$, we know thatΒ $A\in \mathcal{p}(B)$Β by definition of subset.

So, by definition of power set,Β $A\sube B$Β $\blacksquare$.