Union with Superclass is Superclass

Theorem

Let $A$ and $B$ be classes.

Then:

$A \subseteq B \iff A \cup B = B$

where:

$A \subseteq B$ denotes that $A$ is a subclass of $B$

$A \cup B$ denotes the union of $A$ and $B$.

Proof

Let $A \cup B = B$.

Then by definition of class equality:

$A \cup B \subseteq B$

Thus:

\(\ds x\)

\(\in\)

\(\ds A\)

\(\ds \leadsto \ \ \)

\(\ds x\)

\(\in\)

\(\ds A\)

Rule of Addition

\(\, \ds \lor \, \)

\(\ds x\)

\(\in\)

\(\ds B\)

\(\ds \leadsto \ \ \)

\(\ds x\)

\(\in\)

\(\ds A \cup B\)

Definition of Class Union

\(\ds \leadsto \ \ \)

\(\ds x\)

\(\in\)

\(\ds B\)

Definition of Subclass: $A \cup B \subseteq B$

\(\ds \leadsto \ \ \)

\(\ds A\)

\(\subseteq\)

\(\ds B\)

Definition of Subclass

$\Box$

Now let $A \subseteq B$.

We have:

\(\ds x\)

\(\in\)

\(\ds B\)

\(\ds \leadsto \ \ \)

\(\ds x\)

\(\in\)

\(\ds A\)

Rule of Addition

\(\, \ds \lor \, \)

\(\ds x\)

\(\in\)

\(\ds B\)

\(\ds \leadsto \ \ \)

\(\ds x\)

\(\in\)

\(\ds A \cup B\)

Definition of Class Union

\(\ds \leadsto \ \ \)

\(\ds B\)

\(\subseteq\)

\(\ds A \cup B\)

Definition of Subclass

From Subset of Union, we have $S \cup T \supseteq T$.

We also have:

\(\ds x\)

\(\in\)

\(\ds A \cup B\)

\(\ds \leadsto \ \ \)

\(\ds x\)

\(\in\)

\(\ds A\)

Rule of Addition

\(\, \ds \lor \, \)

\(\ds x\)

\(\in\)

\(\ds B\)

\(\ds \leadsto \ \ \)

\(\ds x\)

\(\in\)

\(\ds B\)

as $x \in A \implies x \in B$

\(\ds \leadsto \ \ \)

\(\ds A \cup B\)

\(\subseteq\)

\(\ds B\)

Definition of Subclass

Thus:

\(\ds A \cup B\)

\(\subseteq\)

\(\ds B\)

\(\ds B\)

\(\subseteq\)

\(\ds A \cup B\)

By definition of class equality:

$A \cup B = B$

$\blacksquare$

Also see

• Intersection with Subclass is Subclass

Sources

• 2010: Raymond M. Smullyan and Melvin Fitting: Set Theory and the Continuum Problem (revised ed.) ... (previous) ... (next): Chapter $2$: Some Basics of Class-Set Theory: $\S 5$ The union axiom: Exercise $5.6. \ \text {(f)}$