Problem:

B-1

[Rudin, p. 98 # 3]. Let be a metric space and f : → ℝ a continuous function. Denote by Z(f) the zero set of f. These are the points p ∈ where f is zero, f(p) = 0.

a)

Show that Z(f) is a closed set.

b)

[See also Rudin, p. 101 #20] Given any set E ∈, the distance of a point p to E is defined by

Show that h is a uniformly continuous function.

c)

Use the previous part to show that given any closed set E ∈, there is a continuous function that is zero on E and positive elsewhere.

B-2

[Rudin, p. 99 # 13 or #11, see also p. 98 #4] extension by continuity Let X be a metric space, E ⊂ X a dense subset, and f : E → ℝ a uniformly continuous function. Show that f has a unique continuous extension to all of X. That is, there is a unique continuous function g : X → ℝ with the property that g(p) = f(p) for all p ∈ E. [Remark: One generalize this by replacing ℝ by any complete metric space.]

Solution:

B-1

Let , f, and Z(f) be as in the statement of the problem.

a)

By definition of Z(f) we have Z(f) = f^-1({0}). Since {0} = (ℝ -{0})^c = (-∞,0) ∪ (0,∞)^c and both (-∞,0) and (0,∞) are open subsets of ℝ, then {0} is closed as it is the complement of the union of two open sets. Because f is continuous, this implies that f^-1({0}) = Z(f) is also closed.

b)

Let E and h be defined as in the problem statement. For any p,q ∈ and z ∈ E we have

by the definition of h and the triangle inequality. However, we can take the inf of the right-hand side of the inequality to obtain h(p) ≤ d(p,q) + h(q), which implies h(p) - h(q) ≤ d(p,q). Thus for any ε > 0 if we define δ > 0 so that δ = ε, then d(p,q) < δ for any p,q ∈ implies

as needed for the uniform continuity of h.

c)

Being the infimum of distances, we necessarily have that h(p) ≥ 0 for all p ∈ . Furthermore, because d(z,z) = 0 for any point of , then h(z) = 0 for any point of E. We are left only to prove that h is positive on points outside of E.

Since E was assumed closed, then it’s complement is open. Thus for any point p in its complement, we can find an open ball of radius r > 0 around p that’s disjoint from E. However, this implies that r ≤ d(p,z) for all z ∈ E, in other words r ≤ h(p) so that h(p) is positive as desired.

B-2

Let p ∈ X, and define V _n(p) to be the open ball of radius centered at p. Furthermore let U_n(p) = f(V _n(p)) be the closure of f’s image of V _n(p) in ℝ, and define U_p to be the set ∩_n=1^∞U_n(p). With these definitions, we obtain the following lemmas.

Lemma 1. Any finite intersection of {U_n(p)} is nonempty.

Proof. Since E is dense in X, then each V _n(p) has elements of E, i.e. each U_n(p) is nonempty. Furthermore, we have V _n+1(p) ⊂ V _n(p) which in turn means f ⊂ f and therefore f ⊂f so that finally,

(1)

Therefore, since each U_n(p) is nonempty, any finite intersection of {U_n(p)} is nonempty __

Lemma 2. The set U_p is nonempty.

Proof. Equation 1 implies one of two things:

a)

All U_n(p) are unbounded.

b)

There exists an integer N such that U_N(p),U_N+1(p),… are bounded.

If the former is true, equation 1 implies U_p is nonempty. If the latter is true, then for some integer N each U_n(p) for n ≥ N is closed (being the closure of a set) and bounded, i.e. each U_n(p) for n ≥ N is compact as they are subsets of ℝ. This fact and Lemma 1 tell us, via Rudin’s Theorem 2.36, that ∩_n≥NU_n(p) is nonempty. However, due to equation 1, U_p = ∩_n≥NU_n(p), and so U_p must be nonempty in this case as well. __

Lemma 3. The set U_p contains only a single point.

Proof. For later contradiction, assume that there are distinct points x,y ∈ U_p. Set ε = |y - x| so that, in particular, the closure of the open ball of radius 2ε does not contain y. Since f is uniformly continuous, there exists a δ > 0 such that any two points in E within a distance of δ of each other have that their images under f are within a distance of ε of each other. Therefore, for integer m with < , any two points q,q′∈ V _m(p)∩E have that f(q)-f(q′) < ε. This implies, since x ∈ U_p, that U_m(p) ⊂B_2ε(x). However, in light of our definition of ε, this implies y ⁄∈ U_m(p), which contradicts the assumption that y ∈ U_p. Hence U_p contains no such distinct points x and y. Given that Lemma 2, this then implies U_p must contain a single point. __

Now in light of Lemma 3, we may define a function g : X → ℝ by

From here, I would prove continuity and uniqueness of g, but was unable to do so.