Metric Spaces, Extending a Uniform Function

The Image of a Cauchy Sequence

The function 1/x is continuous on (0,1), yet the cauchy sequence 1/n does not map to a cauchy sequence in the range. A cauchy sequence does map to a cauchy sequence, however, when its limit point p is in the domain. This is a simple consequence of continuity at p.

Let f be a uniformly continuous function from one metric space into another, and let s be a cauchy sequence in the domain, with image sequence t in the range. Select an ε and find the corresponding δ, such that points within δ of x map to points within ε of f(x), for every x in the domain. Find n such that points in s beyond s_n are never more than δ apart. In the image sequence, points beyond t_n are never more than ε apart. A uniform map carries cauchy sequences to cauchy sequences.

Extending f to a Complete Metric Space

We would like to extend this map to g(), from the completion of the domain to the completion of the range. The first task is to show that g is well defined. Different cauchy sequences could represent the same point in the complete metric space, as when 3.4999999… = 3.5000000…

Let the cauchy sequences s and t in the domain map to u and v in the range, respectively, and assume the distance sequence |s_n,t_n| approaches 0, whence s and t are equivalent. Select ε and δ as before, and find n such that the distances |s_j,t_j| are bounded below δ for j > n. Apply f to build the image sequences u and v, and the distance sequence |u_j,v_j| is bounded by ε, for j > n. The image sequences approach each other, and represent the same point in the completion of the range. The map is well defined.

If x is in the domain of f, then the constant sequence s_n = x is in the domain of g, and maps to the constant sequence f(x) in the range. In other words, g is backward compatible with f. The action of g on x is simply f(x), when x lies in the domain of f.

Finally, we would like to show that g is uniformly continuous. Start with a sequence s in the domain, which maps to u in the range. Given ε, select δ according to the dictates of f. Consider any other sequence t such that |s,t| < δ. Beyond some n, the terms of s are within δ of the terms of t. Moving to the image space, the terms of u are within ε of the terms of v, and |u,v| ≤ ε. (Technically, we're looking for strict inequality, but we can always choose δ for ½ε, and that is strictly less than ε.) The selection of δ does not depend on s, so we have uniform continuity.

Extending f to the Boundary

If S and T are metric spaces, and f is a uniform map from U, a subset of S, into T, and T is complete, we can extend f to a unique continuous function on the closure of U.

Let p be a limit point of U that is not in U. find a sequence in U that converges to p. This sequence is cauchy, and its image in T is cauchy, with a limit point y. Declare f(p) = y.

But is this well defined? If two different sequences approach p, the points in these sequences must approach each other by the triangular inequality. The distances between corresponding terms goes to 0, and when we apply the uniform map f, distances still go to 0. In other words, the two image sequences are 0 distance apart, and represent the same point in the complete metric space T. The value of y is well defined.

Note that f(p) really can't be anything else, if f is to remain continuous at p. The extension of f at p is well defined and unique. Make a similar extension of f for all the points in the closure of U.

We still need to show f is uniform. We will demonstrate a viable δ for an arbitrary ε, which applies across all of U closure. Find δ for ½ε across U, which we can do since f is uniform on U, then set γ = ½δ.

Let p be a limit point of U. Picture a cauchy sequence, drawn from U, approaching p. Remember, the limit of the image of this sequence in T is y, and it doesn't matter which sequence we use. Any sequence will do.

Restrict attention to a neighborhood in U about p, bounded by γ. Use f to map this neighborhood, and the end of the cauchy sequence, over to T. Terms of the image sequence are now separated by ½ε, and they must be within ½ε of their limmit point y, else the sequence would not converge to y.

Any point in the neighborhood of radius γ about p can participate in a cauchy sequence approaching p, hence the entire neighborhood maps to a region in T that is within ½ε of y.

Let p lie in U. Using γ, points close to p map to image points that are within ½ε of y. This certainly holds for nearby points in U, and it also holds for nearby points in U closure.

If p is in the closure of U, but not in U, the points of U within γ of p map to points within ½ε of y. But what if nearby points are in U closure, and not in U? Let q be in U closure, with f(q) = z. The distance from p to q is strictly less than γ, so there is some r in U, quite close to p, that is still within γ of q. The image of r is within ½ε of both y and z, hence y and z are within ε of each other. That completes the proof.

The uniform function f can be extended to a uniform function on the closure of its domain, and the extension is unique. If the range space T is not complete, we can always embed T in its completion, which does not change f, and then apply this theorem.