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.