# Intersection with Subset is Subset

## Theorem

$S \subseteq T \iff S \cap T = S$

where:

$S \subseteq T$ denotes that $S$ is a subset of $T$
$S \cap T$ denotes the intersection of $S$ and $T$.

## Proof

Let $S \cap T = S$.

Then by the definition of set equality, $S \subseteq S \cap T$.

Thus:

 $\displaystyle S \cap T$ $\subseteq$ $\displaystyle T$ Intersection is Subset $\displaystyle \leadsto \ \$ $\displaystyle S$ $\subseteq$ $\displaystyle T$ Subset Relation is Transitive

Now let $S \subseteq T$.

From Intersection is Subset we have $S \supseteq S \cap T$.

We also have:

 $\displaystyle S$ $\subseteq$ $\displaystyle T$ $\displaystyle \leadsto \ \$ $\displaystyle S \cap S$ $\subseteq$ $\displaystyle T \cap S$ Set Intersection Preserves Subsets $\displaystyle \leadsto \ \$ $\displaystyle S$ $\subseteq$ $\displaystyle S \cap T$ Intersection is Idempotent and Intersection is Commutative

So as we have:

 $\displaystyle S \cap T$ $\subseteq$ $\displaystyle S$ $\displaystyle S \cap T$ $\supseteq$ $\displaystyle S$

it follows from the definition of set equality that:

$S \cap T = S$

So we have:

 $\displaystyle S \cap T = S$ $\implies$ $\displaystyle S \subseteq T$ $\displaystyle S \subseteq T$ $\implies$ $\displaystyle S \cap T = S$

and so:

$S \subseteq T \iff S \cap T = S$

from the definition of equivalence.

$\blacksquare$