A metric space on Turing machines
The map $f$ is not a contraction.  To see why, let's take for concreteness $R = 100$.  Suppose $w$ is a word and $A,B$ are two machines such that $t_A(w) = 200$ and $t_B(w) = 400$.  Then $d(A,B) = 100$.  Now $t_{f(A)}(w) = 100 + |w| + 2$ and $t_{f(B)}(w) = 200 + |w| + 2$, so $|t_{f(A)}(w) - t_{f(B)}(w)| = 100$ and $d(f(A),f(B)) = 100$ also.  But to have a contraction in the sense of the Banach fixed point theorem (also called a strict contraction) you need to have a constant $c &lt; 1$ such that $d(f(A), f(B)) \le c d(A,B)$, so this can't be achieved.