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## Here's the Solution to this Question

Q#1.

Let $x\in A\cap C$

Case 1,

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

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

Case 2,

$x\in C$

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

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

So,

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

We now have that,

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

Q#2.

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

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

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

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

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

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

Therefore,

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

Q#3.

To begin with, we will first show that $A\times (B\cap C)\sube (A\times B)\cap(A\times C )$. Suppose $(x,y)\in A\times(B\cap C)$. This means that, $x\in A$ and $y\in B\cap C$ by definition of Cartesian product. By definition of intersection, it follows that $y\in B$ and $y\in C$.Since $x\in A$ and $y\in B$, it follows that $(x,y)\in A\times B$ by definition of Cartesian product. Also, since $x\in A$ and $y\in C$ it follows that $(x,y)\in A\times C$ by definition of $\times$. Now, we have $(x,y)\in A\times B$ and $(x,y)\in A\times C$ and so, $(x,y)\in (A\times B)\cap(A\times C)$. Having shown that $(x,y)\in A\times(B\cap C)\implies (x,y)\in (A\times B)\cap(A\times C)$, we have, $A\times(B\cap C)\sube(A\times B)\cap(A\times C)$.

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

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

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

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

Q#4.

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

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

Hence,

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

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

Suppose $y\in x$ then, $y\in A\space or\space y\in B$ since $x\sube A \space or \space x\sube B$

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

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,

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

Q#5.

Suppose that $A$ and $B$ are sets and $\mathcal p(A)\sube\mathcal {p})(B)$

By definition of a power set, $A\sube \mathcal {p}(A)$.Since $A\in \mathcal{p}(A)$ and $\mathcal{p}(A)\sube \mathcal{p}(B)$, we know that $A\in \mathcal{p}(B)$ by definition of subset.

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