# Set Intersection Preserves Subsets/Corollary/Proof 1

## 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.

$A \subseteq B, \ S \subseteq T \implies A \cap S \subseteq B \cap T$

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:

$A \subseteq B \implies A \cap S \subseteq B \cap S$

$\blacksquare$