Definition Let X be a metric space with distance function d. A subset A of X is said to be bounded if there exists a non-negative real number K such that d(x, y) ≤ K for all x, y ∈ A. The smallest real number K with this property is referred to as the diameter of A, and is denoted by diam A.

(Note that diam A is the supremum of the values of d(x, y) as x and y range over all points of A.)

Lemma 1.1 (Lebesgue Lemma) Let (X, d) be a compact metric space. Let U be an open cover of X. Then there exists a positive real number δ such that every subset of X whose diameter is less than δ is contained wholly within one of the open sets belonging to the open cover U.

Proof Every point of X is contained in at least one of the open sets belonging to the open cover U. It follows from this that, for each point x of X, there exists some δ_x > 0 such that the open ball B(x, 2δ_x) of radius 2δ_x about the point x is contained wholly within one of the open sets belonging to the open cover U. But then the collection consisting of the open balls B(x, δ_x)  of radius δ_x about the points x of X forms an open cover of the compact space X. Therefore there exists a finite set x₁, x₂, . . . , x_r of points of X such that

B(x1, δ1) ∪ B(x2, δ2) ∪ · · · ∪ B(xr, δr) = X,

where δi = δxi

for i = 1, 2, . . . , r. Let δ > 0 be given by

         δ = minimum(δ₁, δ₂, . . . , δ_r).

Suppose that A is a subset of X whose diameter is less than δ. Let u be a point of A. Then u belongs to B(x_i , δ_i) for some integer i between 1 and r.

But then it follows that A ⊂ B(x_i, 2δ_i), since, for each point v of A,

 d(v, x_i) ≤ d(v, u) + d(u, x_i) < δ + δ_i ≤ 2δ_i.

But B(x_i, 2δ_i) is contained wholly within one of the open sets belonging to the open cover U. Thus A is contained wholly within one of the open sets belonging to U, as required.

Let U be an open cover of a compact metric space X. A Lebesgue number for the open cover U is a positive real number δ such that every subset of X whose diameter is less than δ is contained wholly within one of the open sets belonging to the open cover U. The Lebesgue Lemma thus states that there exists a Lebesgue number for every open cover of a compact metric space.

Let X and Y be metric spaces with distance functions d_X and d_Y respectively, and let f: X → Y be a function from X to Y . The function f is said to be uniformly continuous on X if and only if, given ε > 0, there exists some δ > 0 such that d_Y (f(x), f(x΄)) < ε for all points x and x΄ of X satisfying d_X(x, x΄) < δ. (The value of δ should be independent of both x and   x΄.)

Theorem 1.2 Let X and Y be metric spaces. Suppose that X is compact.

Then every continuous function from X to Y is uniformly continuous.

 Proof Let d_X and d_Y denote the distance functions for the metric spaces X and Y respectively. Let f: X → Y be a continuous function from X to Y .

We must show that f is uniformly continuous.

Let ε > 0 be given. For each y ∈ Y , define

            V_y = {x ∈ X : d_Y (f(x), y) <1/2ε}.

Note that V_y = f ^-1(BY (y,1/2 ε) , where BY (y,1/2 ε) denotes the open ball of radius 1/2 ε about y in Y . Now the open ball BY (y, 1/2ε) is an open set in Y , and f is continuous. Therefore V_y is open in X for all y ∈ Y . Note that x ∈ V_f(x) for all x ∈X.

Now {V_y : y ∈ Y } is an open cover of the compact metric space X. It follows from the Lebesgue Lemma (Lemma 1.27) that there exists some δ > 0 such that every subset of X whose diameter is less than δ is a subset of some set V_y. Let x and x΄ be points of X satisfying d_X(x, x΄) < δ. The diameter of the set {x, x΄} is d_X(x, x΄), which is less than δ.  Therefore there exists some y ∈ Y such that x ∈ V_y and x΄ ∈ V_y. But then d_Y (f(x), y) <1/2ε and d_Y (f(x΄), y) <1/2ε, and hence

                                         d_Y (f(x), f(x΄)) ≤ d_Y (f(x), y) + d_Y (y, f(x΄)) < ε.

This shows that f: X → Y is uniformly continuous, as required.

Let K be a closed bounded subset of R_n . It follows that any continuous function f: K → R^k is uniformly continuous.

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