Homework 3 <me@tylerlogic.com>
September 19, 2014
http://coursework.tylerlogic.com/courses/upenn/math508/homework03

### 1 Give an example of closed sets Vj⊂ ℝ where ∪j=1∞Vj is open.

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

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

Define

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 .

Boundary points:

Closure:

#### (d)

The graph of the function

as a subset of 2. 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 , so it’s compact by Heine-Borel.

#### (b) {0}∪{1,,…,,…}⊂ ℝ

Again, this set is a closed and bounded subset of , 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 then it is not compact, again by Heine-Borel.

### 5

For any element xk S we have that

so

which gives us a closed form for each xk 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

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

### 6

Define for A,B k,n by = trace(ABt)

#### (a) Show this is an inner product.

Positive Definite
Let A k×n with r1,r2,,rk being the k vectors that make up the rows of A. Then = trace(AAt) = r1r1t +r2r2t ++rkrkt. Since each ririt is simply the dot product of ri with itself, then each ririt 0 with equality when ri is the zero vector. Thus 0 with equality when A is the zero matrix.
Let A,B,C k×n. By properties of the trace function and matrix multiplication we have

Homogeneity
Let A,B k×n and α . By properties of the trace function we have

Symmetry
Let A,B k×n. By properties of the trace function we have

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

Non-negativity

Symmetry

Triangle Inequality

### 7

Let K be compact in n with x ⁄∈ K. Further let d = inf{d(x,y) | y K}. There must be a limit point z n 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 Br(p) K = {} for p with d(x,p) = d. This would imply that, say, d + 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 = -{0} with the normal metric on 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 {xn} be a sequence of points in 2 that contains every point with rational coordinates. Let {rn} be a sequence of positive real numbers such that nrn = 1. Define

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 2 because 2 is dense in 2 and U contains 2 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 . By way of contradiction, assume that L is completely contained in U. Then for any n , we have that D(xn,rn) L = {u + c(v - u) | c (sn,tn)} for some sn,tn . Thus by defining the function f : L by f(u + c(v - u)) = c we see that f(D(xn,rn) L) = (sn,tn). So for all 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 (sn1,tn1),,(snk,tnk). However, this implies that

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

for each (x1,y1),(x2,y2) E1 × E2.

#### (a) Warmup

Viewing 2 as × we have

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 Br(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 will be contained in U1 since any y B(x) has that
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 Br(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 Br1(u1) U1 and Br2(u2) U2. Define r = min(r1,r2). Then for any point (x,y) Br((u1,u2)) we have

and

which implies x Br1(u1) and y Br2(u2), respectively. In turn, this means (x,y) U1 ×U2 and thus (x,y) is also in U. Hence the open ball Br((u1,u2)) is contained in U, i.e. U is open.

### 11 The p-adic topology on the rational numbers.

#### (a) Show that (ℚ,dp) is a metric space.

Fix a prime p.

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

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

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

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(ν12). Hence we have

as desired. __

Non-negativity
Let x,y 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 with x - y = pν. Then y - x = pν, implying that |x - y|p = |y - x|p = p-ν. Hence d(x,y) = d(y,x).
Triangle Inequality
Let x,y,z . Making use of the lemma above, we obtain

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

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

So for any y Nr(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 Nr(a), i.e. Nr(x) Nr(a).

On the other hand, we similarly have that for any y Nr(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 Nr(x), meaning Nr(a) Nr(x).

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

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

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

as desired.