Solution Set 2: Part B

Benjamin Bakker Mathematics 25a Prof. Berger

October 3, 2003

B.1. (1) Show that (ℚ, d_{p}) is a metric space and that d(x, z) ≤ max(d(x, y), d(y, z)).

Solution. We first have to verify the axioms for a metric, namely that ∀x, y, z ∈ ℚ

• d(x, y) ≥ 0

• d(x, y) = 0 ⇔ x = y

• d(x, y) = d(y, x)

• d(x, z) ≤ d(x, y) + d(y, z)

d(x, y) is defined as either a power of p or 0, and therefore satisfies the first condition. For any x, y, v(x − y) = v(y − x), since we can pull exactly the same number of factors out of x − y and y − x, so d(x, y) = d(y, x). d(x, y) is a power of p if x ≠ y, and zero otherwise, so it is nonnegative and zero if and only if x = y. Suppose x − y = p^{v_{1}} \frac{n_{1}}{d_{1}} , z − y = p^{v_{2}} \frac{n_{2}}{d_{2}} , with n_{i}, d_{i} relatively prime to p. We can assume v_{1} ≤ v_{2}, so that v(x − z) = v((x − y) − (z − y)) ≥ v_{1}, for

x − z = p^{v_{1}} \frac{n_{1}}{d_{1}} − p^{v_{2}} \frac{n_{2}}{d_{2}} = p^{v_{1}} (\frac{n_{1}d_{2} − n_{2}d_{1}p^{v_{2}−v_{1}}}{d_{1}d_{2}} )

and since d_{i} are relatively prime to p, we can’t pull out any more factors of p from the denominator, although we may be able to factor some out of the numerator (if v_{1} = v_{2}). Thus,

d(x, z) = p^{−v(x−z)} ≤ p^{−v_{1}} = max(p^{−v_{1}}, p^{−v_{2}}) = max(d(x, y), d(y, z))

The triangle inequality, d(x, z) ≤ d(x, y) + d(y, z), then follows.

(2) Show every element of B(a, r) is a center of B(a, r)

Solution. Note that every ball B(a, r) is nonempty since a ∈ B(a, r). For any x ∈ B(a, r), d(x, a) < r. Pick any y ∈ B(x, r), so that d(x, y) < r. By the above, d(y, a) ≤ max(d(x, a), d(x, y)) = r, so y ∈ B(a, r), and therefore B(x, r) ⊂ B(a, r). Certainly a ∈ B(x, r), so applying the same reasoning B(a, r) ⊂ B(x, r) ⇒ B(x, r) = B(a, r).

(3) Show that any given two balls are either disjoint or one is contained in the other

Solution. Suppose there is an element x ∈ B(a_{1}, r_{1}) ∩ B(a_{2}, r_{2}). From part (2), B(x, r_{1}) = B(a_{1}, r_{1}) and B(x, r_{2}) = B(a_{2}, r_{2}). Without loss of generality we may assume r_{1} ≤ r_{2}, so B(a_{1}, r_{1}) = B(x, r_{1}) ⊂ B(x, r_{2}) = B(a_{2}, r_{2}). Thus, either the two balls intersect, and therefore one is contained in the other, or they are disjoint.

B.2. If u = {u_{n}}_{n≥1} is a sequence of elements in a metric space (E, d), show that the set A(u) of accumulation points is closed.

Solution. To show that A(u) is closed, we will prove that the complement of A(u) is open. Take any point x ∉ A(u). x is not an accumulation point of {u_{n}}, so for some ϵ there can be no points of the sequence closer to x than ϵ, i.e., B(x, ϵ) ∩ {u_{n}} = ∅; otherwise, x would be an accumulation point. There can be no accumulation point u ∈ B(x, ϵ). To see this, suppose there was, and let δ = d(x, u). B(u, ϵ − δ) is contained in B(x, ϵ), since if y is in the former, d(y, u) < ϵ − δ ⇒ d(x, y) ≤ d(y, u) + δ < ϵ. But B(u, ϵ − δ) doesn’t contain any elements of the sequence {u_{n}}, so u is not an accumulation point. We have therefore produced a ball around an arbitrary x ∈ X \ A(u) that is contained in X \ A(u), and thus X \ A(u) is open. . Alternatively, we can show that for any sequence x_{n} in A(u), if x_{n} converges to x then x ∈ A(u). For every ball B(x, \frac{1}{2m}), there exists an N such that x_{n} ∈ B(x, \frac{1}{2m}) for n > N. x_{N+1} is an accumulation point, and therefore there must be an element of u, say u′_{m}, that is in B(x_{N+1}, \frac{1}{2m}). We can pick the u′_{m} to be different from our previous choices since there are always infinitely many elements of the sequence in B(x_{N+1}, \frac{1}{2m}), and therefore the u′_{m} may be rearranged into a subsequence of u_{n}. For any m, u′_{m} ∈ B(x, \frac{1}{m}) by the triangle inequality, and since \frac{1}{m} converges to 0, u′_{m} converges to x. Hence, x ∈ A(u), and A(u) is closed.

B.3. Let f : [0, 1] → ℝ be defined by f(x) = \sqrt{x}. Show that f is uniformly continuous.

Solution. If we know f(x) = \sqrt{x} is continuous, then this problem is really easy. We proved in class that every continuous function on a compact space is uniformly continuous, and that [0, 1] is compact, so f must be uniformly continuous. . To make this problem more fun, we will now show that f is continuous. Let x_{0} ∈ [0, 1], and let ϵ > 0 be given. If we take δ = ϵ^{2}, then by the triangle inequality

|\sqrt{x} − \sqrt{x′}| ≤ \sqrt{x} + \sqrt{x′}

and therefore

|x − x′| = |\sqrt{x} − \sqrt{x′}|(\sqrt{x} + \sqrt{x′}) < δ = ϵ^{2} ⇒ |\sqrt{x} − \sqrt{x′}|^{2} ≤ |x − x′| < ϵ^{2} ⇒ |\sqrt{x} − \sqrt{x′}| < ϵ

B.4. Check that a K-Lipshitz function is uniformly continuous. Show that for every 0 < a < 1, there is a K_{a} such that f(x) = \sqrt{x} is K_{a}-Lipshitz on [a, 1]. What happens at a = 0?

Solution. Given that f : X → Y is K-Lipshitz, then for all ϵ > 0, take δ = ϵ/K. For any x, x′ ∈ X, we then have

d_{X}(x, x′) < δ = ϵ/K ⇒ d_{Y} (f(x), f(x′)) ≤ Kd_{Y} (x, x′) < ϵ

and f is uniformly continuous. . For 0 < a < 1, take K_{a} = \frac{1}{2\sqrt{a}} (we might guess this because we can think of K_{a} as a bound on the derivative—see below—but we shouldn’t use this intuition in the formal proof!). For all x, x′ ∈ [a, 1], a ≤ x ≤ 1 so that \sqrt{a} ≤ \sqrt{x}, and similarly for x′. Thus, 1/K_{a} ≤ \sqrt{x} + \sqrt{x′} ⇒ 1 ≤ K_{a}(\sqrt{x} + \sqrt{y}), and

|\sqrt{x} − \sqrt{x′}| ≤ K_{a}|\sqrt{x} − \sqrt{x′}|(\sqrt{x} + \sqrt{x′}) ≤ K_{a}|x − x′|

The above does not work if a = 0, for then K_{a} is not defined. Heuristically, you can think of the constant K as a bound on the “derivative,”

\frac{|f(x) − f(x′)|}{|x − x′|} ≤ K

But the derivative of \sqrt{x}, which is −x^{−1/2}, is unbounded at the origin.