Direct proof set theory. Okay, I am in a first year Discrete Math course in university and struggling to keep up, and need some additional help. We are learning about proving set relations and I just can't keep up with my professor's pace. I'll just get right to the problem I can't seem to understand.
(I don't know how to show the proper notation but I think this explanation in words will be clear enough, it's pretty simple)
"Let A and B be sets. Prove that A union B equals A if and only if B is a subset of A."
The way my professor chose to solve this was by splitting the biconditional into its two implications, and proving them separately. This splitting up is straightforward; I understand that. The two implications are "if A union B equals A, then B is a subset of A", and "if B is a subset of A, then A union B equals A".
Then my professor says that to prove each implication, we will assume that the antecedent (hope that's the right word) is true and then see if the consequent is true. She describes this as a direct proof. This is what I don't understand. I can't seem to reason out in my mind why this method works, or how to actually implement it. I'm sure it's simple, but my professor just talks too fast for me to follow what she's saying, and I can't seem to figure this out on my own.
So if what little I understand is correct, then I should first assume that "A union B equals A" is true, and see if it follows that "B is a subset of A" is also true. But to put it simply, I haven't got a clue what to do from here.
I understand this sounds like "Please do my homework for me" but really I'm asking "Please teach me how to tackle problems like this so I can do it on my own in future."
 A: Suppose that $A \cup B = A$. We will show that $B \subseteq A$. Select $b \in B$. Notice that $b \in B \subseteq A \cup B = A$, where the equality comes from the first sentence. We have $b \in A \cup B$. Since $A \cup B = A$ we have $b \in A$. Try to do the other half of the problem yourself.
A: The general strategy that you can use to solve such problems is that although problem statements are about sets, the proofs usually involve the elements of the sets, as in Jay's excellent approach.  You need to get under the hood and follow the spinning and thumping bits.
A: In general, when we want to prove $P\implies Q$, we suppose $P$ is true and then try to prove $Q$. If $P$ were false, then the implication is vacuously true.
The statement is: $A\cup B =A\iff B\subseteq A$ or equivalently, $(A\cup B =A\implies B\subseteq A)\land (B\subseteq A\implies A\cup B =A)$.
So, we try to separately prove $A\cup B =A\implies B\subseteq A$ and $B\subseteq A\implies A\cup B =A$.
To prove the first statement, suppose $A\cup B=A$ is true. To show $B\subseteq A$, we suppose $x$ to be an arbitrary element of $B$ and then try to prove $x\in A$. (Note that if $B$ was the null set, the statement becomes $\emptyset\subseteq A$ which is always true). Now since $x\in B$, $x\in A\cup B$ and since $A\cup B=A$, $x\in A$ which proves this statement.
To prove the second, again suppose the antecedent, i.e. $B\subseteq A$. To prove $A\cup B=A$, we need to prove $A\cup B\subseteq A$ and $A\subseteq A\cup B$. To prove $A\cup B\subseteq A$, suppose $x\in A\cup B$. Then either $x\in A$ or $x\in B$. If $x\in B$, then from $B\subseteq A$, we conclude $x\in A$. Since both cases imply $x\in A$, $A\cup B\subseteq A$ is true. The second statement, $A\subseteq A\cup B$ follows immediately since if $x\in A$ then $x\in A\cup B$.
So we have proved $A\cup B\subseteq A$ and $A\subseteq A\cup B$ is true, and thus $A\cup B=A$ is true.
