Cartesian Product of Subsets

Theorem
Let $A, B, S, T$ be sets such that $A \subseteq B$ and $S \subseteq T$.

Then:
 * $A \times S \subseteq B \times T$

In addition, if $A, S \ne \O$, then:
 * $A \times S \subseteq B \times T \iff A \subseteq B \land S \subseteq T$

Proof
First we show that $A \subseteq B \land S \subseteq T \implies A \times S \subseteq B \times T$.

First, let $A = \O$ or $S = \O$.

Then from Cartesian Product is Empty iff Factor is Empty:
 * $A \times S = \O \subseteq B \times T$

so the result holds.

Next, let $A, S \ne \O$.

Then from Cartesian Product is Empty iff Factor is Empty:
 * $A \times S \ne \O$

and we can use the following argument:

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

Now we show that if $A, S \ne \O$, then:
 * $A \times S \subseteq B \times T \implies A \subseteq B \land S \subseteq T$

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

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

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

So that explains the restriction $A, S \ne \O$.

Now, as $A, S \ne \O$, $\exists x \in A, y \in S$.

Thus:

So when $A, S \ne \O$, we have:
 * $A \subseteq S \land B \subseteq T \implies A \times S \subseteq B \times T$
 * $A \times S \subseteq B \times T \implies A \subseteq B \land S \subseteq T$

from which:
 * $A \times S \subseteq B \times T \iff A \subseteq B \land S \subseteq T$