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\vspace*{.3in}
\noindent{\bf Beals page 11, \#7:}
The function $d(A,B)$ is symmetric in $A$ and $B$, because 
$A\triangle B = B\triangle A$.

For transitivity, consider sets $A,B,C\subset\R$ and let 
$x\in A\triangle C$ be any element.  Say for instance (and with no loss of
generality) that $x\in A$ but $x\notin C$.  Then if $x\in B$,
it follows that $x\in B\triangle C$; and if $x\notin B$, it follows
that $x\in A\triangle B$.  Either way, $x\in A\triangle B \cup B\triangle C$.
This proves that 
$$
A\triangle C \subset A\triangle B \cup B\triangle C,
$$
Consequently,
$$
m(A\triangle C) \leq m(A\triangle B) + m(B\triangle C).
$$
It follows that $d$ is transitive and a semi-metric.
\qed 

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