Relative Complement of Cartesian Product/Proof 1

Theorem

Let $A$ and $B$ be sets.

Let $X \subseteq A$ and $Y \subseteq B$.

Then:

$\relcomp {A \mathop \times B} {X \times Y} = \paren {A \times \relcomp B Y} \cup \paren {\relcomp A X \times B}$

Proof

From Set with Relative Complement forms Partition:

$A = \set {X \mid \relcomp A X}$

$B = \set {Y \mid \relcomp B Y}$

and so by definition of partition:

$A = X \cup \relcomp A X$

$B = Y \cup \relcomp B Y$

By Cartesian Product of Unions:

$A \times B = \paren {X \times Y} \cup \paren {\relcomp A X \times \relcomp B Y} \cup \paren {X \times \relcomp B Y} \cup \paren {\relcomp A X \times Y}$

and so:

\(\ds \paren {A \times B} \setminus \paren {X \times Y}\)

\(=\)

\(\ds \paren {\relcomp A X \times \relcomp B Y} \cup \paren {X \times \relcomp B Y} \cup \paren {\relcomp A X \times Y}\)

\(\ds \)

\(=\)

\(\ds \paren {X \times \relcomp B Y} \cup \paren {\relcomp A X \times \relcomp B Y} \cup \paren {\relcomp A X \times Y} \cup \paren {\relcomp A X \times \relcomp B Y}\)

\(\ds \)

\(=\)

\(\ds \paren {\paren {X \cup \relcomp A X} \times \relcomp B Y} \cup \paren {\relcomp A X \times \paren {Y \cup \relcomp B Y} }\)

\(\ds \)

\(=\)

\(\ds \paren {A \times \relcomp B Y} \cup \paren {\relcomp A X \times B}\)

$\blacksquare$

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