Homework 6 - Bonus Problem <me@tylerlogic.com>
October 27, 2014
http://coursework.tylerlogic.com/courses/upenn/math508/homework06/bonusproblem

### Problem:

B-1
[Rudin, p. 98 # 3]. Let be a metric space and f : 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 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 with the property that g(p) = f(p) for all p E. [Remark: One generalize this by replacing 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} = (-{0})c = (-∞,0) (0,)c and both (-∞,0) and (0,) are open subsets of , 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 Un(p) = f(V n(p)) be the closure of f’s image of V n(p) in , and define Up to be the set n=1Un(p). With these definitions, we obtain the following lemmas.

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

Proof. Since E is dense in X, then each V n(p) has elements of E, i.e. each Un(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 Un(p) is nonempty, any finite intersection of {Un(p)} is nonempty __

Lemma 2. The set Up is nonempty.

Proof. Equation 1 implies one of two things:

a)
All Un(p) are unbounded.
b)
There exists an integer N such that UN(p),UN+1(p), are bounded.

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

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

Proof. For later contradiction, assume that there are distinct points x,y Up. 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 Up, that Um(p) B2ε(x). However, in light of our definition of ε, this implies y ⁄∈ Um(p), which contradicts the assumption that y Up. Hence Up contains no such distinct points x and y. Given that Lemma 2, this then implies Up must contain a single point. __

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

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