Intersection with Subclass is Subclass
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Theorem
Let $A$ and $B$ be classes.
Then:
- $A \subseteq B \iff A \cap B = A$
where:
- $A \subseteq B$ denotes that $A$ is a subclass of $B$
- $A \cap B$ denotes the intersection of $A$ and $B$.
Proof
Let $A \cap B = A$.
Then by the definition of class equality:
- $A \subseteq A \cap B$
Thus:
\(\ds x\) | \(\in\) | \(\ds A\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds x\) | \(\in\) | \(\ds A \cap B\) | Definition of Subclass: $A \subseteq A \cap B$ | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds x\) | \(\in\) | \(\ds A\) | Definition of Class Intersection | ||||||||||
\(\, \ds \land \, \) | \(\ds x\) | \(\in\) | \(\ds B\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds x\) | \(\in\) | \(\ds B\) | Definition of Conjunction | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds A\) | \(\subseteq\) | \(\ds B\) | Definition of Subclass |
$\Box$
Now let $A \subseteq B$.
\(\ds x\) | \(\in\) | \(\ds A \cap B\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds x\) | \(\in\) | \(\ds A\) | Definition of Class Intersection | ||||||||||
\(\, \ds \land \, \) | \(\ds x\) | \(\in\) | \(\ds B\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds x\) | \(\in\) | \(\ds A\) | Definition of Conjunction | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds A \cap B\) | \(\subseteq\) | \(\ds A\) | Definition of Subclass |
We also have:
\(\ds x\) | \(\in\) | \(\ds A\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds x\) | \(\in\) | \(\ds A\) | |||||||||||
\(\, \ds \land \, \) | \(\ds x\) | \(\in\) | \(\ds B\) | Definition of Subclass | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds x\) | \(\in\) | \(\ds A \cap B\) | Definition of Class Intersection |
So as we have:
\(\ds A \cap B\) | \(\subseteq\) | \(\ds A\) | ||||||||||||
\(\ds A\) | \(\subseteq\) | \(\ds A \cap B\) |
it follows from the definition of class equality that:
- $A \cap B = A$
$\blacksquare$
Also see
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 $5.6. \ \text {(e)}$