October 27, 2014

http://coursework.tylerlogic.com/courses/upenn/math508/homework06/bonusproblem

- 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.]

- 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.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 __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≥N}U_{n}(p) is nonempty. However, due to equation 1, U_{p}= ∩_{n≥N}U_{n}(p), and so U_{p}must be nonempty in this case as well. __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.