Math 508: Advanced Analysis

Homework 3

Lawrence Tyler Rush

<me@tylerlogic.com>

September 19, 2014

http://coursework.tylerlogic.com/courses/upenn/math508/homework03

1 Give an example of closed sets V _j ⊂ ℝ where ∪_j=1^∞V _j is open.

Define V _j = [-j,j]. Then ∪_j=1^∞V _j = ℝ and ℝ is open. But this example leaves a bad taste in one’s mouth since the union of the {V _j} is the whole space, ℝ. As for a more slick example, define V _j = [-(1 - 1∕j),1 - 1∕j]. In this case ∪_j=1^∞V _J = (-1,1), which is open in ℝ.

2 Construct a bounded set of real numbers with exactly three limit points

Define

{ } Vx = x+ 1|n ∈ ℤ+ n

With this definition, x is the only limit point of V _x and is bounded since it is contained within the ball B₁(x). Therefore V ₁ ∪ V ₂ ∪ V ₃ is a bounded set with exactly three limit points.

3 Find the interior points and boundary points of each set, and describe the sets closure

(a) (0, 1] ⊂ ℝ

Interior points: (0,1)

Boundary points: {0,1}

Closure: [0,1]

(b) ℝ² ⊂ ℝ³ (the coordinate plane z = 0)

Interior points: There are no interior points because there is no ball of any point completely contained in the plane.

Boundary points: The set itself is the set of boundary points.

Closure: This set is closed, so the closure is itself.

(c) ℚ ⊂ ℝ

Interior points: There are no interior points. The set is countable, and therefore discrete, so no neighborhood of any point can be completely contained within ℝ.

Boundary points: ℝ

Closure: ℝ

(d)

The graph of the function

{ sin-1 x ⁄= 0 y = 0 x x = 0

as a subset of ℝ². Call the set G.

Interior points: There are no interior points.

Boundary points: G ∪{(0,y) | |y|≤ 1}

Closure: Same as the boundard points.

4 Which sets are compact? Why?

(a) [0, 1] ⊂ ℝ

This set is a closed and bounded subset of ℝ, so it’s compact by Heine-Borel.

(b) {0}∪{1,,…,,…}⊂ ℝ

Again, this set is a closed and bounded subset of ℝ, so it’s compact by Heine-Borel.

(c) X=[0, 1] \ ℚ as a subset of ℝ

Since X contains no rational numbers then 0 ⁄∈ X, however, every open ball of zero will contain a point of X since it will contain an irrational in [0,1]. Therefore 0 is a limit point of X, but since it’s not contained in X, X is not closed. Because X is not closed as a subset of ℝ then it is not compact, again by Heine-Borel.

5

For any element x_k ∈ S we have that

1 9 9 9 9 10xk = 102 + 103 + 104 + ⋅⋅⋅+ 10k+1

xk - -1xk = 9-- --9k+1 ( 1)0 10 10 1- -1 xk = 9-- --9-- 10 10 10k+1 (10 - 1)x = 9- -9- k 10k 9x = 9- -9- k 10k x = 1- -1- k 10k

which gives us a closed form for each x_k ∈ S. With this formula, it’s obvious that 1 is an upper bound. Assume for later contradiction that there exists an a < 1 where a is an upper bound. However, for any integer k > -log ₁₀(1 - a) we have

xk = 1- -1- > 1- ----lo1g-(1--a) = 1- 10log10(1-a) = 1- (1- a) = a 10k 10 10

which contradicts our assumption that a is an upper bound. Thus 1 is the supS.

6

Define

for A,B ∈

_k,n by

= trace(AB^t)

(a) Show this is an inner product.

Positive Definite: Let A ∈ _k×n with r₁,r₂,…,r_k being the k vectors that make up the rows of A. Then = trace(AA^t) = r₁r₁^t +r₂r₂^t ++r_kr_k^t. Since each r_ir_i^t is simply the dot product of r_i with itself, then each r_ir_i^t ≥ 0 with equality when r_i is the zero vector. Thus ≥ 0 with equality when A is the zero matrix.
Additivity: Let A,B,C ∈_k×n. By properties of the trace function and matrix multiplication we have $( ) ( ) ⟨A + B,C ⟩ = trace (A + B)Ct = trace ACt + BCt = trace(ACt) +trace(BCt ) = ⟨A,C ⟩+ ⟨B,C⟩$
Homogeneity: Let A,B ∈_k×n and α ∈ ℝ. By properties of the trace function we have $t t ⟨αA,B ⟩ = trace(αAB ) = α trace(AB ) = α ⟨A,B ⟩$
Symmetry: Let A,B ∈_k×n. By properties of the trace function we have $( t) ( t t) ( t) ⟨A,B ⟩ = trace AB = trace (AB ) = trace BA = ⟨B,A ⟩$

(b) Let |A|² = and define d(A,B) := |A - B|. Show d is a metric.

Non-negativity: $∘ ------ √ - d(A,B ) = |A - B| = ⟨A,B ⟩ ≥ 0 = 0$
Symmetry: $∘ ------ ∘ ------ d(A,B ) = |A - B | = ⟨A, B⟩ = ⟨B, A⟩ = |B - A| = d(B,A )$
Triangle Inequality: $d(A,B ) = |A - B | = |A - C + C - B| = |(A- C )+ (C - B)| ≤ |A - C |+ |C - B| = d(A,C )+ d(C,B )$

(c)

(d)

7

Let K be compact in ℝⁿ with x ⁄∈ K. Further let d = inf{d(x,y) | y ∈ K}. There must be a limit point z ∈ ℝⁿ of K such that d(x,z) = d, If there were not such a point, then we’d be able to find some r > 0 such that B_r(p) ∩ K = {} for p with d(x,p) = d. This would imply that, say, d + 1
2

r would be a lower bound on the set {d(x,y) | y ∈ K}, which is a contradiction since d is the infimum of that set. Now because z is a limit point of K and K is compact, then it is also closed by Heine-Borel. Hence z ∈ K, as desired.

8

Let X = ℝ -{0} with the normal metric on ℝ of the absolute value of the difference between two points. Then the sets A = [-1,0) and B = (0,1] are closed sets since zero is not in X. They are obviously dijsoint. Furthermore dist(A,B) = 0 since each has points arbitrarily close to zero.

9

Let {x_n} be a sequence of points in ℝ² that contains every point with rational coordinates. Let {r_n} be a sequence of positive real numbers such that ∑ _nr_n = 1. Define

⋃ U = D(xn,rn) n

where D(x,r) is the disc of radius r centered at x.

(a) Show U is open and dense in ℝ

The set U is open because it is the union of open sets. Also, the set U is dense in ℝ² because ℚ² is dense in ℝ² and U contains ℚ² as a subset.

(b) Show that no straight line L is completely contained in U

Let L be a line in the plane. A line is the linear combination of two specified points, so let those points be u and v. Then L is the set of points (1 - c)u + cv = u + c(v - u) for all c ∈ ℝ. By way of contradiction, assume that L is completely contained in U. Then for any n ∈ ℕ, we have that D(x_n,r_n) ∩ L = {u + c(v - u) | c ∈ (s_n,t_n)} for some s_n,t_n ∈ ℝ. Thus by defining the function f : L → ℝ by f(u + c(v - u)) = c we see that f(D(x_n,r_n) ∩ L) = (s_n,t_n). So for all n ∈ ℕ, we get an open cover of the real line. For instance, we can cover the closed interval [0,3], but since [0,3] is compact, we can find a finite open subcover (s_n₁,t_n₁),…,(s_{n_k},t_{n_k}). However, this implies that

∑k ∞∑ ∞∑ ∑∞ 3 ≤ (tni - sni) ≤ (tn - sn) ≤ 2rn = 2 rn = 2 i=1 n=1 i=1 i=1

and we thus arrive at a contradiction. Hence L is not completely contained in U.

10 Product Topology

Let (E₁,d₁) and (E₂,d₂) be two metric spaces and define the distance on E₁ ×E₂ to be

d((x1,y1),(x2,y2)) = d1(x1,x2)+ d2(y1,y2)

for each (x₁,y₁),(x₂,y₂) ∈ E₁ × E₂.

(a) Warmup

Viewing ℝ² as ℝ × ℝ we have

d((0,0),(1,2)) = d(0,1)+ d(0,2) = 1 + 2 = 3

and the “disc” centered at the origin is

(b)

Let U ⊂ E₁ × E₂ be an open subset. Let (u₁,u₂) ∈ U. We can thus find an open ball B_r(u₁,u₂) ⊂ U. Define U₁ = {x ∈ E₁ | d((x,u₂),(u₁,u₂)) < r}. With this definition for any x ∈ U₁, the open ball of radius r-d1(x,u1)
2

will be contained in U₁ since any y ∈ B(x) has that

d((y,u ),(u ,u )) = d (y,u )+ d (u ,u ) 2 1 2 1 1 2 2 2 = d1(y,u1) ≤ d1(y,x)+ d1(x,u1) r--d1(x,u1) < 2 + d1(x,u1) r--d1(x,u1)+-2d1(x,u1) = 2 r+ d1(x,u1) = -----2----- r+ r < --2-- = r

so that d((y,u₂),(u₁,u₂)) < r. Thus U₁ is open in E₁. Similarly, we also have that the set U₂ = {x ∈ E₂ | d((u₁,x),(u₁,u₂)) < r} is open in E₂ because each x ∈ U₂ has the open ball B(x) around it which is completely contained in U₂. Finally, with the simple definitions of U₁ and U₂, we see (u₁,u₂) ∈ U₁ × U₂ and U₁ × U₂ ⊂ B_r(u₁,u₂) ⊂ U, as desired.

Conversely, assume that for each point (u₁,u₂) ∈ U there exist U₁, U₂ open in E₁, E₂, respectively, with (u₁,u₂) ∈ U₁ × U₂ and U₁ × U₂ ⊂ U. So fix (u₁,u₂) ∈ U. As a consequence of the hypothesis at the opening of this paragraph, we can find open balls B_r₁(u₁) ⊂ U₁ and B_r₂(u₂) ⊂ U₂. Define r = min(r₁,r₂). Then for any point (x,y) ∈ B_r((u₁,u₂)) we have

d1(u1,x) ≤ d1(u1,x) +d2(u2,y) = d((u1,x),(u2,y)) < r ≤ r1

and

d2(u2,y) ≤ d1(u1,x)+ d2(u2,y) = d((u1,x),(u2,y)) < r ≤ r2

which implies x ∈ B_r₁(u₁) and y ∈ B_r₂(u₂), respectively. In turn, this means (x,y) ∈ U₁ ×U₂ and thus (x,y) is also in U. Hence the open ball B_r((u₁,u₂)) is contained in U, i.e. U is open.

11 The p-adic topology on the rational numbers.

(a) Show that (ℚ,d_p) is a metric space.

Fix a prime p.

Lemma 11.1. For x,y,z ∈ ℚ, d(x,z) ≤ max(d(x,y),d(y,z)).

Proof. Let x,y,z be rational numbers. Then we have

n n n x- y = pν1-1 , y- z = pν2-2 and x - z = pν3-3 k1 k3 k2

where p does not divide any n_i or k_i. With this in place we assume without loss of generality that ν₁ ≤ ν₂ so that

( ) ( ) ν1n1 ν2n2 ν1 n1 ν2-ν1n2 ν1 n1k2 +-pν2--ν1n2k1- x - z = (x- y)+ (y - z) = p k1 + p k3 = p k1 + p k3 = p k1k2

Now with both the above and below forms of x - z, we see ν₃ ≥ ν₁, which given our assumption that ν₁ ≤ ν₂, implies ν₃ ≥ min(ν₁,ν₂). Hence we have

( ) d(x,z) = |x- z|p = p-ν3 ≤ p- min(ν1,ν2) = max p- ν1,p-ν2 = max (|x - y|p,|y- z|p) = max(d(x,y),d(y,z))

as desired. __

Non-negativity: Let x,y ∈ ℚ be distinct. Then d(x,y) = |x-y|_p = p^-ν for some ν. Since p is positive, then d(x,y) will always be positive. On the other hand d(x,x) = |x - x|_p = 0.
Symmetry: Let x,y ∈ ℚ with x - y = p^ν. Then y - x = p^ν, implying that |x - y|_p = |y - x|_p = p^-ν. Hence d(x,y) = d(y,x).
Triangle Inequality: Let x,y,z ∈ ℚ. Making use of the lemma above, we obtain $d(x - z) ≤ max (d(x,y),d(y,z)) ≤ d(x,y)+ d(y,z)$

(b) For x,a ∈ ℚ show that x ∈ N_r(a) implies N_r(x) = N_r(a).

Let x,y,a ∈ ℚ with r > 0 and x ∈ N_r(a). Then d(x,a) < r.

So for any y ∈ N_r(x), d(x,y) < r. In particular, d(y,a) ≤ max(d(x,a),d(x,y)) < max(r,r) = r, by the above lemma. So y ∈ N_r(a), i.e. N_r(x) ⊂ N_r(a).

On the other hand, we similarly have that for any y ∈ N_r(a), d(y,a) < r and therefore d(x,y) ≤ max(d(x,a),d(a,y)) < max(r,r) = r again by the above lemma. Thus y ∈ N_r(x), meaning N_r(a) ⊂ N_r(x).

Combining these two results leaves us with N_r(x) = N_r(a).

(c) Show that any two neighborhoods are disjoint, or one is contained in the other.

Let a,a′∈ ℚ and r,r′ be positive reals. If N_r(a) and N_r′(a′) are disjoint, we are done. So assume not. Then there is an x ∈ N_r(a) ∩ N_r′(a′). By the previous part of the problem, we then have N_r(x) = N_r(a) and N_r′(x) = N_r′(a′). Thus by assuming without loss of generality that r < r′ we obtain

′ ′ ′ Nr(a) = Nr(x) ⊂ N r(x) = Nr(a )

as desired.