Cartesian Product of Subsets

Theorem
Let $$A \subseteq S$$ and $$B \subseteq T$$.

Then $$A \times B \subseteq S \times T$$.

In addition, if $$A, B \ne \varnothing$$, then $$A \times B \subseteq S \times T \iff A \subseteq S \and B \subseteq T$$.

Proof

 * First we show that $$A \subseteq S \and B \subseteq T \implies A \times B \subseteq S \times T$$.

First, let $$A = \varnothing$$ or $$B = \varnothing$$.

Then from Cartesian Product Null, $$A \times B = \varnothing \subseteq S \times T$$, so the result holds.

Next, let $$A, B \ne \varnothing$$. Then from Cartesian Product Null, $$A \times B \ne \varnothing$$ and we can use the following argument:

$$ $$ $$ $$

Thus $$A \times B \subseteq S \times T$$ as we were to prove.


 * Now we show that if $$A, B \ne \varnothing$$, then $$A \times B \subseteq S \times T \implies A \subseteq S \and B \subseteq T$$.

So suppose that $$A \times B \subseteq S \times T$$.

First note that if $$A = \varnothing$$, then $$A \times B = \varnothing \subseteq S \times T$$, whatever $$B$$ is, so it is not necessarily the case that $$B \subseteq T$$.

Similarly if $$B = \varnothing$$; it is not necessarily the case that $$A \subseteq S$$.

So that explains the restriction $$A, B \ne \varnothing$$.

Now, as $$A, B \ne \varnothing$$, $$\exists x \in A, y \in B$$. Thus:

$$ $$ $$ $$

So when $$A, B \ne \varnothing$$, we have:
 * $$A \subseteq S \and B \subseteq T \implies A \times B \subseteq S \times T$$;
 * $$A \times B \subseteq S \times T \implies A \subseteq S \and B \subseteq T$$,

from which $$A \times B \subseteq S \times T \iff A \subseteq S \and B \subseteq T$$.