Proof 8.3.17
Suppose that f and g are isometries. Prove that the composition f(g(x)) is also an isometry.
Let the functions g and f be isometries such that the outputs of g belong to the domain of f. That is, we can compute the composition f(g(x)).
Because the function f is an isometry, it is onto. That is for any element, F, in the target of f, there exists an element, G, of the domain of f such that f(G)=F. Because the function g is an isometry, it is also onto. That is for any element, G, in the target of g, there exists an element, x, of the domain of g such that g(x)=G. By the property of transitivity, this implies that for any element, F, in the target of f(g(x), there exists an element, x, of the domain such that f(g(x))=F. Therefore, f(g(x) is onto.
Similarly, because the functions f and g are isometries, they are one-to-one. That is, for the elements G1 and G2 of the domain of f, f(G1)=f(G2) implies G1=G2. And, for the elements x1 and x2 of the domain of g, g(x1)=g(x2) implies x1=x2. Thus, for the elements x1 and x2 of the domain of the composition f(g(x)), f(g(x1))=f(g(x2)) implies x1=x2 by transitivity.
Lastly, because f and g are isomtries they are both distance preserving. Thus, |f(g(x2))-f(g(x1))|=|g(x2)-g(x1)|=|x2-x1|. Therefore, f(g(x) is distance preserving.
Because f(g(x) is onto, one-to-one, and distance preserving. It is an isometry.