Compact Sets, Continuous Becomes Uniform

Continuous Becomes Uniform

Let S be a compact metric space and let T be a metric space. Let f be a continuous function from S into T. We will show that f is uniform.

Given ε, and a point x in S, there is some δ satisfying continuity at x, at the level of ½ε. Build a ball of radius ½δ about x. Do this for every x, and we have an open cover for S. Select a finite subcover and set γ to the smallest radius. This is the value that will satisfy uniformity.

Let y and z be points in S, with |y,z| < γ. Refer to the open cover; y is contained in a ball of radius ½δ, centered at a point we will call x. (The value of δ may depend on x, but it is at least as large as γ.) By the triangular inequality, |x,z| < δ. That means |f(x),f(z)| < ½ε. The same is true of |f(x),f(y)|, hence f(y) and f(z) are within ε of each other. The value of γ satisfies continuity for every x, and f is uniform.

Since compact = closed and bounded in Rⁿ, this is a generalization of an earlier theorem.

Bounded

Continuing the above, the image of S is bounded in T. Fix an origin in T and consider the distance between f(S) and the origin. This is a continuous map from S into the reals. Since S is compact the image is compact, thus closed and bounded. Let b be the least upper bound. Being closed, the image attains its upper bound. Hence there is some x with |f(x)| = b. Once again this is a generalization of an earlier theorem.

Distance Between Functions

Let f and g be continuous functions from S into T, and let h(x) be the distance from f(x) to g(x). Choose an open interval in the reals, and let h(x) lie inside this interval. In other words, x is in the preimage of an open set. Choose an open set in S that keeps f(y) close to f(x), and g(y) close to g(x). Within this open set, h(y) is close to h(x), and lies inside our interval. (This is the triangular inequality.) The preimage of every open interval is covered by open sets, and h is continuous.

By the above, h is bounded. Every two functions differ by an upper bound b, and b = 0 iff the functions are equal.

Shrinking Diameters

Let S be a compact region in Rⁿ. In other words, S is closed and bounded. Let C be a finite open cover for S. Remember that any open cover of S admits a finite subcover. There is then a distance d, such that every open ball in S, of radius d or less, lies inside at least one of the open sets of C.

Suppose this is not the case. Build a sequence of balls, with decreasing diameters, such that none of these balls is contained in an open set in C. The centers imply a cluster point p. Restrict attention to a subsequence of balls whose centers approach p.

Since p is contained in an open set U in C, place p at the center of an open ball of radius d, contained in U. If you go down far enough, the centers of our balls are within d/2 of p. Go down further, if necessary, so that the radius of each ball is less than d/2. This places the entire ball inside the set U, which is a contradiction. Therefore a distance d exists, such that every ball of radius d or less lies in some open set U in our finite cover C.