Satisfying the Definition of a Metric Let $S$ be any set. Let $X$ be the collection of all finite subsets of $S$. Show that
$d: X \times X \to [0,\infty), \quad (A,B) \mapsto |(A \setminus B) \cup (B \setminus A)|$ defines a metric on $X$.
My book gives the following definition of a metric in terms of a mapping.
A metric on a set $\Omega$ is a function $d: \Omega \times \Omega \to \mathbb{R}_+$ satisfying three properties:


*

*$d(f,g)=0$ if and only if $f=g$.

*$d(f,g) = d(g,f)$

*$d(f,g) \leq d(f,h) + d(h,g)$.


Here is what I have worked out:
$d(A,B) = |(A \setminus B) \cup (B \setminus A)| = 0 \Leftrightarrow A \setminus B = B \setminus A$. This means that $x \in A$ and $x \notin B$ or $x \in B$ and $ x \notin A$.
$d(A,B) = |(A \setminus B) \cup (B \setminus A)| = |(B \setminus A) \cup (A \setminus B)| = d(B,A).$
$d(A,B) = |(A \setminus C) \cup (C \setminus A)| + |(B \setminus C) \cup (C \setminus B)| = d(A,C) + d(C,B)$.
I do not think this is convincing enough and would like some feedback on how to properly show this is indeed a metric on $X$.
Thanks in advance.
 A: $d(A,B)=0$:
Remember, you are trying to show that $A=B$. Your $\iff$ is false, and your concluded a different statement entirely.
Instead, observe that for a union of two sets to be empty, each of the sets has to be empty. Thus $A\setminus B=\varnothing$ and also $B\setminus A=\varnothing$. Hence $A\subset B$ and $B\subset A$, whereupon $A=B$ as desired.
$d(A,B)=d(B,A)$:
Your proof is fine.
$d(A,B)\leq d(A,C)+d(C,B)$:
This is actually the "meat" of what you're trying to prove. It is called the triangle inequality and is usually the hardest property of a metric space to check. In this case, you're going to need to do some set theory to show it. The key fact you will use is that $(B\setminus C)\cup (C\setminus A)=B\setminus A$.
A: Well,the first big problem with the first part of the "proof" is that you seem to assume what you're trying to prove. It assumes axiom (1) is true when that's what we're trying to show. Consider the union of the relative complements and d(A,B) =0. $|(A \setminus B) \cup (B \setminus A)| = 0$ implies $(A \setminus B) \cup (B \setminus A)$ = $\emptyset$ 
Consider $(A \setminus B) \cup (B \setminus A)$ refers to the union of the set of all elements that is in B,but not in A and the set of all elements in A.but not in B. Therefore the set is asking for the set of all elements which is both in A and B and not in A or B. Clearly, such as set is empty! More importantly, this is a set where the complements of each A and B with respect to the other set are the same set. This can only be if A=B. 
Since the unions are symmetric-it doesn't matter what order you write the sets in-the second part of your proof is fine. 
The third part is the triangle inequality and it's going to a little tricky to prove. The key fact you're going to need from set theory is the following: $(B \setminus C) \cup (C \setminus A) = B \setminus A$ . The proof will be somewhat tedious, but it's worth doing in detail. 
Good luck! 
