# 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 1

Let $S \cap T = S$.

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

Thus:

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

Now let $S \subseteq T$.

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

We also have:

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

So as we have:

 $\ds S \cap T$ $\subseteq$ $\ds S$ $\ds S \cap T$ $\supseteq$ $\ds S$

it follows from the definition of set equality that:

$S \cap T = S$

So we have:

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

and so:

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

from the definition of equivalence.

$\blacksquare$

## Proof 2

 $\ds$  $\ds S \cap T = S$ $\ds \leadstoandfrom \ \$ $\ds$  $\ds \paren {x \in S \land x \in T \iff x \in S}$ Definition of Set Equality $\ds \leadstoandfrom \ \$ $\ds$  $\ds \paren {x \in S \implies x \in T}$ Conditional iff Biconditional of Antecedent with Conjunction $\ds \leadstoandfrom \ \$ $\ds$  $\ds S \subseteq T$ Definition of Subset

$\blacksquare$