elementary-set-theory
We want to show:
$$|S_1 \times S_2| = |S_1| \cdot |S_2|$$
I am not sure how to go about showing this in general terms. I believe that we will need to use the definition of the cartesian product during this, so that is where I am at right now.
$$S_1 \times S_2 = \{(a,b): a \in S_1 \; \text{and} \; b \in S_2\}$$
These are finite sets.
Let the elements we remove be denoted by $s^j=(s_1^j,s_2^j,\ldots,s_k^j)$ for $j=1,2,\ldots,n$. We are interested in the set $$(S_1\times S_2\times\ldots\times S_k)\setminus\{s^1,s^2,\ldots,s^n\}.$$
Claim. A (possibly very crude) upper bound is given by $\frac{n^k-1}{n-1}$.
Proof. We will prove this by induction on $k$.
For $k=1$, there is no problem. The set $S_1\setminus\{s^1,s^2,\ldots,s^n\}$ is a Cartesian product with one factor, so we are done: the upper bound is $1$ in this case.
For $k\to k+1$, note that we can rewrite the set as follows: $$\begin{align}&(S_1\times S_2\times\ldots\times S_{k+1})\setminus\{s^1,s^2,\ldots,s^n\}=\\&=(S_1\times S_2\times\ldots\times S_{k+1})\setminus\{s^1,s^2,\ldots,s^n\}\\&=\bigcup_{x\in S_{k+1}}(S_1\times S_2\times\ldots\times S_k\times\{x\})\setminus\{s^1,s^2,\ldots,s^n\}\\&=\left(\bigcup_{j=1}^n(S_1\times S_2\times\ldots\times S_k\times\{s_{k+1}^j\})\setminus\{s^1,s^2,\ldots,s^n\}\right)\cup(S_1\times S_2\times\ldots\times (S_{k+1}\setminus\{s_{k+1}^1,\ldots,s_{k+1}^n\})),\end{align}$$ which is (if we omit the sets that appear more than once in the first union) a disjoint union of at most $n$ sets of the form $$(S_1\times S_2\times\ldots\times S_k\times\{s_{k+1}^j\})\setminus\{s^1,s^2,\ldots,s^n\},$$ and a single Cartesian product of $k+1$ sets. By the induction hypothesis, each set $$(S_1\times S_2\times\ldots\times S_k\times\{s_{k+1}^j\})\setminus\{s^1,s^2,\ldots,s^n\}$$ can be written as a disjoint union of at most $\frac{n^k-1}{n-1}$ Cartesian products. Therefore our set can be written as a disjoint union of at most $n\frac{n^k-1}{n-1}+1 = \frac{n^{k+1}-1}{n-1}$ Cartesian products (with $k+1$ factors each, of course). $\square$
Note that finding better bounds might require more sophisticated combinatorial methods.
Take the set $D$ and make all possible pairs out of that set (since you are asked $D$ $\times$ $D$). In general, if $B$ is another set, and you want $D$ $\times$ $B$, then $D$ $\times$ $B$ = { (x,y) such that x $\epsilon$ $D$ and y $\epsilon$ $B$ } where if you replace $B$ by $D$, you get what you asked. Note that $D$ $\times$ $B$ and $B$ $\times$ $D$ aren't equal if $B$ is not he same as $D$. The first set (conventionally) comes on x-axis and the second on y-axis and so on if there are more than two sets.
Alternatively, for ease of understanding, fix one element of $D$ on x-axis and make pair of it with elements of $D$ on y-axis. This will give you points on top of the fixed point and then repeat it with another fixed element until you exhaust $D$ on x-axis. For example, fix point {2} from $D$ on x-axis and make pairs (2,y) where y varies in $D$ on y-axis. This will give you {2} $\times$ $D$. Now change the fixed point to another point in $D$ on x-axis and repeat the process until every element of $D$ is used as a fixed point, which wil give $D$ $\times$ $D$.
Best Answer
Say $S_1 = \{a,b\}$ and $S_2=\{x,y,z\}$. Then
$$ S_1\times S_2= \big\{ (a,x),\ (a,y),\ (a,z),\ (b,x),\ (b,y),\ (b,z) \big\}. $$ So you can look at it in either of two ways: $$ \big\{ \underbrace{(a,x),\ (a,y),\ (a,z)},\ \underbrace{(b,x),\ (b,y),\ (b,z)} \big\}. $$ $$ \big\{ \underbrace{(a,x),\ (b,x)},\ \underbrace{(a,y),\ (b,y)},\ \underbrace{(a,z),\ (b,z)} \big\} $$ The first is a sum of two $3$s; the second is a sum of three $2$s. So the first is $2\times3$ and the second is $3\times 2$. That's one way of knowing that $2\times3$ is the same number as $3\times2$.
$$ |S_1|\times|S_2| = \sum_{a\in S_1}\ \sum_{x\in S_2}\ 1 = \sum_{(a,x)\in S_1\times S_2} \ 1. $$