Cartesian Product of Subsets

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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:

$A \times S \subseteq B \times T \iff A \subseteq B \land S \subseteq T$


Corollary 1

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


Then:

$A \times S \subseteq B \times S$


Corollary 2

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


Then:

$A \times S \subseteq A \times T$


Corollary 3

Let $A, B, C$ be sets such that $B \ne \O$.

Let $A \times B \subseteq C \times C$.


Then:

$A \subseteq C$


Family of Subsets

Let $\family {S_i}_{i \mathop \in I}$ be a family of sets where $I$ is an arbitrary index set.

Let $S = \ds \prod_{i \mathop \in I} S_i$ be the Cartesian product of $\family {S_i}_{i \mathop \in I}$.


Let $\family {T_i}_{i \mathop \in I}$ be a family of sets.

Let $T = \ds \prod_{i \mathop \in I} T_i$ be the Cartesian product of $\family {T_i}_{i \mathop \in I}$.


Then:

$\paren {\forall i \in I: T_i \subseteq S_i} \implies T \subseteq S$.


Nonempty Subsets

Let $T_i \ne \O$ for all $i \in I$.


Then:

$T \subseteq S \iff \forall i \in I: T_i \subseteq S_i$.


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:

\(\ds \) \(\) \(\ds \tuple {x, y} \in A \times S\)
\(\ds \) \(\leadsto\) \(\ds x \in A, y \in S\) Definition of Cartesian Product
\(\ds \) \(\leadsto\) \(\ds x \in B, y \in T\) Definition of Subset
\(\ds \) \(\leadsto\) \(\ds \tuple {x, y} \in B \times T\) Definition of Cartesian Product


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:

\(\ds \) \(\) \(\ds x \in A, y \in S\)
\(\ds \) \(\leadsto\) \(\ds \tuple {x, y} \in A \times S\) Definition of Cartesian Product
\(\ds \) \(\leadsto\) \(\ds \tuple {x, y} \in B \times T\) Definition of Subset
\(\ds \) \(\leadsto\) \(\ds x \in B, y \in T\) Definition of Cartesian Product


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:

$A \times S \subseteq B \times T \iff A \subseteq B \land S \subseteq T$

$\blacksquare$


Sources