Set Intersection Preserves Subsets/Corollary/Proof 1
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Theorem
Let $A, B, S$ be sets.
Then:
- $A \subseteq B \implies A \cap S \subseteq B \cap S$
Proof
Let $A \subseteq B$, and let $S$ be any set.
From Set Intersection Preserves Subsets:
- $A \subseteq B, \ S \subseteq T \implies A \cap S \subseteq B \cap T$
for arbitrary sets $S$ and $T$.
Substituting $S$ for $T$:
- $A \subseteq B, \ S \subseteq S \implies A \cap S \subseteq B \cap S$
From Set is Subset of Itself, $S \subseteq S$ for all sets $S$.
Hence the result:
- $A \subseteq B \implies A \cap S \subseteq B \cap S$
$\blacksquare$