If $A \subseteq C$ and $B \subseteq D$ then $A \times B \subseteq C \times D$ Show that:
if $A \subseteq C\,$ and $\,B \subseteq D,\,$ then $\,A \times B \subseteq C \times D.$  
Can anyone help me with this?
 A: Here, it's a matter of spelling out that


*

*$A \subseteq C \iff a \in A \rightarrow a \in C$ and 

*$B\subseteq D\iff b \in B \rightarrow  b \in D$ 

*knowing that $A \times B = \{(a, b)\mid a\in A, b\in B\}$, and 

*knowing that $C\times D = \{(c, d) \mid c \in C, d \in D\}$


Using the above: show that 
$$A\times B \subseteq C \times D \;\;\text{ if and only if}\;\;\; (a, b) \in A\times B \rightarrow (a, b) \in C \times D$$
A: Take any element of $A \times B$.  It has the form $(a,b)$ where $a \in A$ and $b \in B$.  We want to show that $(a,b) \in C \times D$.  We have $a \in A$ and $A \subseteq C$, so $a \in C$.  Perhaps you can take it from here.
A: The first step is recognizing that you need to prove an implication ($⇒$). 
$$A⊆C,B⊆D \quad ⇒ \quad A \times B \subseteq C \times D \tag{1}$$
How do you prove such an implication ? You assume the left-hand side to be true. And you try to deduce the right-hand side from this assumption. If you succeed in proving the right-hand side, then you are done.
So we assume $A⊆C$ and $B⊆D$. Now we need to proof $A \times B \subseteq C \times D$. By definition this means, that we need to prove (another!) implication:
$$x \in A \times B \quad ⇒ \quad x\in C×B \tag{2}$$
Again, we assume the left-hand side to be true. And we try to deduce the right-hand side. So assume $x \in A × B$. We can write such an element $x$ as a pair $(a,b)$ with $a\in A$ and $b\in B$. Now notice that we assumed $A⊆C$ and $B⊆D$. So we have (by definition!) $a \in C$ and $b\in D$. Therefore $x=(a,b) \in C ×B$. So we succesfully proved the second implication. This means that  we proved that $A \times B \subseteq C \times D$. We proved this from the assumption $A⊆C,B⊆D$. Therefore we also succesfully proved the first implication ! So we are done. $\blacksquare$
