We show that the real line has a dense subset S, such that for any s, t in S, the translation f(x) = x + (t - s) is the unique automorphism S -> S that takes s to t. Almost certainly this has been done before, but who knows. Any ordered set with this unique-automorphism property will have a natural measure (on its intervals) and metric (on pairs of its points), which the real line, considered just as an open ordered continuum, does not. Lines like this could therefore be helpful in explaining the seemingly "built in" metrics of space and time. We assume the continuum hypothesis (i.e., 2^omega_0, the size of the set of real numbers, is taken to be omega_1, the next highest cardinal number after the size of the integers). The basic method is taken from a 1950 paper by Sierpinski, in which a form of diagonalization produces a subset of the real line whose only automorphism is the identity function.

Our diagonalization, like Sierpinski's, starts by assuming a well-ordering of certain functions on the reals. Let { f_alpha : alpha < omega_1 } be all the continuous monotone increasing functions from R onto R which, within at least one interval of R, do not have constant integer slope over any subinterval. This last clause rules out maps that are piecewise equivalent to maps of the form f(x) = nx + r, where n is an integer and r is a real number; the reason for the clause will emerge later. (Unlike the discontinuous functions, the continuous functions from R to R have the same cardinality as R, because each is determined by its values on just the rational numbers, and countably many real values can be encoded as a single real.)

We build up, in parallel, a set S and a "blacklist" of points that we bar ourselves from adding to S later. At each stage we ensure that S is closed under the translations that take any of its points to another, and we "rule out" one function f by blacklisting f(x) for some x just added to S.

Start with S and the blacklist empty. (Aside: we could start with S containing the rational numbers.) At each step alpha, 0 <= alpha < omega_1, we choose a new non-blacklisted x_alpha to add to S, using the procedure below. Then to ensure that S is closed under translations (addition and subtraction), we add to it all points in the set T = { n*(alpha_x) + s : n any integer, s either zero or any number previously in S }.

(To illustrate where T's definition comes from, imagine that you first pick x_0 = 1. T will then include n*(1) for all integers n, i.e. all the integers, which will be added to S. Next imagine that you pick x_1 = pi = 3.1415... and add it to S. You want to be able to translate the points in S so that 0 maps onto pi; this translation would take pi itself to 2pi. Thus you need to add 2pi, and 3pi, 4pi, etc., and since you then want to be able to translate -47 to 3pi (say), you need to add every point's image under f(x) = x + 47 + 3pi, etc. The definition of T takes care of all these cases at once.)

Once we have chosen x_alpha and added it along with its T to S, we blacklist f_alpha(x_alpha), ensuring that f_alpha will not be a function from S onto S. The difficulty comes in choosing x_alpha so that f_alpha is not in S, nor equal to any of the points in T that are now being added to S.

x_alpha will itself be chosen by diagonalization. It must be chosen from an interval I on none of whose subintervals is f_alpha(x) identically equal to nx + r, for any integer n and real r. The extra clause in the definition of the f_alpha was meant to ensure such an I. Let d_n be a decimal in the interior of I, terminating at exactly n decimal points, such that any extension of d_n with more digits is still in I.

We add digits to d_n to get d_n+1, d_n+2, etc., yielding a real number d (our x_alpha) not equal to any "bad" points. What points are bad? (a) Those that have been blacklisted; (b) those already in S; (c) those whose image under f_alpha (to be blacklisted) would be in T (needed by S to ensure closure under translations). In order to yield d not in any of these categories, we set up a hodgepodge enumeration { h_i : i >= 0}. Each h_i is either (a) a blacklisted number, (b) a number in S, or (c) an ordered pair <s,p> such that s is in S and p is an integer. { h_i } is to contain all possible elements of these three types, which can be arranged since at no stage alpha < omega_1 is either S or the blacklist uncountable.

Now add digits to d_n at each stage i >= 0 as follows: If h_i is a member of S or of the blacklist, add one digit to d_n so that d_n+1 differs from h_i at the n+1st decimal place; if h_i is an ordered pair <s,p>, add enough digits to d_n to ensure that no extension d of d_n+m can have f_alpha(d) = px + s. (If this cannot be ensured by adding any number of digits, then the points x on which f_alpha(x) = px + s are dense in the interval between d_n and d_n's extension by infinitely many 9's, and by f_alpha's continuity it must be identically equal to px + s on that subinterval, contradicting our initial choice of the interval I.)

Let us check that no point is ever added to both S and the blacklist. A point is added to S because it is in T at some stage alpha. (x_alpha itself is trivially in T at stage alpha.) A point is in T because it equals n*x_alpha + s for some integer n and some s added to S at a previous stage. But our diagonalization ensures that f_alpha(x_alpha) is not equal to any number of this form.

Now consider the set S obtained at the end of this transfinite process. S is closed under addition and subtraction, as is easily seen by induction on its construction process. Therefore, for any s and t in S, the translation mapping s to t (namely f(x) = x + t - s) is clearly an automorphism of S: for any x, x + t - s is in S, and for any y, y + s - t, its pre-image under f, is in S. Are there any other automorphisms of S? We explicitly robbed all others of at least one point in their range, except for functions piecewise equivalent to maps of the form f(x) = nx + r, where n is an integer and r is a real number. We have to see whether any functions of this form slipped in as valid automorphisms. First, if r is not in S for one of the piecewise definitions of f, then f is not an automorphism on S. S is closed under addition and subtraction, so if x and (nx + r) are both in S, so is x + x + ... + x = nx, and also (nx + r) - nx = r, contradiction. Second, no function that agrees on any nontrivial interval with f(x) = nx + s, n an integer > 1, can be an automorphism on S. f(x) = 2x, for example, maps S into S, but not onto S -- there are x in S such that x/2 is not in S; S is not closed under multiplication or division. Let f(x) be equal to nx + s on some interval [a,b], a < b, with n > 1. Consider f's inverse g(x) on the interval [f(a),f(b)]: g(x) = (x - s) / n on this interval. Now define g'(x) to agree with g on that interval, but having g'(x) = x + r_1 (r_1 a real number chosen to mage g' continuous) for x < f(a), and g'(x) = x + r_2 (chosen similarly) for x > f(b). g'(x) is one of the f_alpha encountered in our diagonalization process. Call it f_alpha_g'. The witness against its being an automorphism on S must be chosen from the interval [f(a),f(b)], because outside of it there are always subintervals on which g' has the proscribed behavior (being identically equal to nx + r for integer n). The upshot is that at stage alpha_g' we add an x_alpha_g' to S and blacklist (x_alpha_g' - s) / n, which is between a and b. Then f(x) cannot map S onto S, because the preimage under f of x_alpha_g' was blacklisted from S.

Despite not being closed under multiplication, S (or another set with the features S was designed to have) may provide a better model for physical lines in space or time, because although it is densely ordered it has a unique natural metric on pairs of its points. It would be interesting to see if a structure with similar properties could be designed in R2, R3, or R4.