Proving Set Operations I'm trying to prove that if $A$ is a subset of $B$ then $A \cup B = B$, but I am having trouble trying to proves this mathematically. I know that since $A$ is a subset, then $A$ has an element $x$ which is in $B$. So when $A \cup B$, some of the elements in $A$ are already in $B$; so the result of the union would be $B$. How would I show this mathematically? What about if $A$ is a subset of $B$; then $A \cap B = A$?
 A: The first thing you need to do is make sure you know what the definitions of union, intersection, and subset really mean. 
Union: $A\cup B = \{x : x\in A \space\text{or}\space x\in B\}$.
Intersection: $A\cap B = \{x : x\in A \space\text{and}\space x\in B\}$.
Subset: We say that $A$ is a subset of $B$, written $A\subseteq B$, provided that for all $x$, if $x\in A$, then $x\in B$. That is
$$
(A\subseteq B) \Longleftrightarrow (\forall x)(x\in A\to x\in B) \Longleftrightarrow (\forall x\in A)(x\in B).
$$
The two problems you are considering are proved by showing mutual subset inclusion; that is, if you can show that, for your first problem, that $A\cup B \subseteq B$ and also that $B\subseteq A\cup B$, then you will have shown that $A\cup B = B$. Similarly, for your second problem, if you can show that $A\cap B \subseteq A$ and also that $A\subseteq A\cap B$, then you will have shown that $A\cap B = A$. 
Given these facts (refer back to them often while reading on), try to follow the two proofs below (and let me know if you have questions). 

Problem 1: If $A$ is a subset of $B$, then $A\cup B = B$. 
Proof. Suppose $A\subseteq B$. Then we have the following:
($\subseteq$): Pick $x\in A\cup B$. Thus, either $x\in A$ or $x\in B$. Now, if $x\in B$, we are done; thus, suppose $x\in A$. Then, since $A\subseteq B$, we have that $x\in B$ also. In either case, $x\in B$, so that $A\cup B \subseteq B$. 
($\supseteq$): Pick $x\in B$, which implies that $x\in A\cup B$. Thus, $B\subseteq A\cup B$.
Since we have shown that $A\cup B \subseteq B$ and also that $B\subseteq A\cup B$, we necessarily have that $A\cup B = B$.

Problem 2: If $A$ is a subset of $B$, then $A\cap B = A$. 
Proof. Suppose $A\subseteq B$. Then we have the following:
($\subseteq$): Pick $x\in A\cap B$. Then $x\in A$ and $x\in B$. Since $x\in A$, we have that $A\cap B\subseteq A$. 
($\supseteq$): Pick $x\in A$. Then, since $A\subseteq B$, we have that $x\in B$ also. Therefore, since $x\in A$ and $x\in B$, it follows that $x\in A\cap B$. Thus, $A\subseteq A\cap B$. 
Since we have shown that $A\cap B \subseteq A$ and also that $A\subseteq A\cap B$, we necessarily have that $A\cap B = A$.
