Help with Cartesian product subsets I want to prove that if  $A \subseteq C\,$ and $\,B \subseteq D,\,$ then $\,A \times B \subseteq C \times D.$  
I know that $A \subseteq C \iff a \in A \rightarrow a \in C$ and that $B\subseteq D\iff b \in B \rightarrow  b \in D$  I also know that $A \times B = \{(a, b)\mid a\in A, b\in B\}$ and that $C\times D = \{(c, d) \mid c \in C, d \in D\}$.
So keep that in mind, how do I connect the dots?
 A: Let $(x,y)$ be an arbitrary point in $A \times B$.you need to show that $(x,y)$ is in $C \times D$.
now $(x,y)$ be an arbitrary point in $A \times B \implies x \in A \subseteq C ,y \in B \subseteq D $
A: You have all of the pieces.
You want to show that every element of $A\times B$ is an element of $C\times D$, so start with an arbitrary element of $A\times B$. What does such an element look like? It’s $\langle a,b\rangle$ for some $a\in A$ and $b\in B$. Since $a\in A$, you know that $a\in C$, and since $b\in B$, you know that $b\in D$. Now what can you conclude about $\langle a,b\rangle$?
A: Here is a slightly more formal way to connect the dots: starting with the most complex expression, let's try to expand the definitions and simplify:
\begin{align}
& A \times B \subseteq C \times D \\
\equiv & \;\;\;\;\;\text{"definition of $\subseteq$"} \\
& \langle \forall p : p \in A \times B : p \in C \times D \rangle \\
\equiv & \;\;\;\;\;\text{"basic property of $\times$, twice: split $p$ into its components $x$ and $y$"} \\
& \langle \forall x,y : x \in A \land y \in B : x \in C \land y \in D \rangle \\
\equiv & \;\;\;\;\;\text{"logic: split range -- to try and separate $x$ and $y$"} \\
& \langle \forall x,y : x \in A \land y \in B : x \in C \rangle \;\land\; \langle \forall x,y : x \in A \land y \in B : y \in D \rangle \\
\equiv & \;\;\;\;\;\text{"logic: in LHS, bring $y$ to the only place where it is used; same for $x$ in RHS"} \\
& \langle \forall x : x \in A \land \langle \exists y :: y \in B \rangle : x \in C \rangle \;\land\; \langle \forall y : \langle \exists x :: x \in A \rangle \land y \in B : y \in D \rangle \\
\equiv & \;\;\;\;\;\text{"logic: in LHS, extract $x$-less part from $\forall x$; same for $y$ in RHS"} \\
& (\langle \exists y :: y \in B \rangle \Rightarrow \langle \forall x : x \in A : x \in C \rangle) \;\land\; (\langle \exists x :: x \in A \rangle \Rightarrow \langle \forall y : y \in B : y \in D \rangle) \\
\equiv & \;\;\;\;\;\text{"definition of $\subseteq$, twice; definition of $\emptyset$, twice; write $P \Rightarrow Q$ as $\lnot P \lor Q$, twice"} \\
& (B = \emptyset \lor A \subseteq C) \;\land\; (A = \emptyset \lor B \subseteq D) \\
\equiv & \;\;\;\;\;\text{"add $\lor A = \emptyset$ to LHS since $A = \emptyset$ implies $A \subseteq C$; same for RHS; simplify"} \\
& A = \emptyset \lor B = \emptyset \lor (A \subseteq C \land B \subseteq D) \\
\end{align}
This immediately implies the original statement.
Yes, this is longer than the other answers.  However, note how at most points there is really only one thing we could do to make progress: this kind of proof is easy to (re)construct.  And also note that we have not lost any information, so we proved something more general than was originally asked, viz., an equivalence which correctly accounts for the handling of the special case $A \times B = \emptyset$.
