proofs:
<h2>S subset S Union T</h2>
We want to prove 
Let
. By definition
is the set that contains the elements of
and the elements of
. Then
must be in
. As
was arbitrary, we conclude that
.
<h2>T Subset S Union T</h2>
This proof is analogous to the previous one. In fact, this result is the same result as the previous one.
<h2>S Intersection T subset S </h2>
We want to prove 
Let
. By definition of the intersection
should be in
and also in
. Then, we already saw that
. As
was arbitrary we can conclude that
.
<h2>S Intersection T subset T</h2>
This is the same result as the previous one. There is no need to prove it anymore, but if you wish, you can reply the exact same proof.