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
With this definition, x is the only limit point of V _{x} and is bounded since it is contained within the ball B_{1}(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) ℝ^{2} ⊂ ℝ^{3} (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
as a subset of ℝ^{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 ℝ, so it’s compact by
HeineBorel.
(b) {0}∪{1,,…,,…}⊂ ℝ
Again, this set is a closed and bounded subset of ℝ, so it’s compact by
HeineBorel.
(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
HeineBorel.
5
For any element x_{k} ∈ S we have that
so
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
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 r_{1},r_{2},…,r_{k} being the k vectors that make up the rows of A. Then
= trace(AA^{t}) = r_{1}r_{1}^{t} +r_{2}r_{2}^{t} ++r_{k}r_{k}^{t}. Since each r_{i}r_{i}^{t} is simply the dot product of r_{i} with itself,
then each r_{i}r_{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

Homogeneity
 Let A,B ∈_{k×n} and α ∈ ℝ. 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.

Nonnegativity


Symmetry


Triangle Inequality

(c)
(d)
7
Let K be compact in ℝ^{n} with x ⁄∈ K. Further let d = inf{d(x,y)  y ∈ K}. There
must be a limit point z ∈ ℝ^{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 B_{r}(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 HeineBorel. 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 ℝ^{2} that contains every point with rational
coordinates. Let {r_{n}} be a sequence of positive real numbers such that ∑
_{n}r_{n} = 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
ℝ^{2} because ℚ^{2} is dense in ℝ^{2} and U contains ℚ^{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 ∈ 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_{nk},t_{nk}). However, this implies that
and we thus arrive at a contradiction. Hence L is not completely contained in U.
10 Product Topology
Let (E_{1},d_{1}) and (E_{2},d_{2}) be two metric spaces and define the distance on E_{1} ×E_{2} to
be
for each (x_{1},y_{1}),(x_{2},y_{2}) ∈ E_{1} × E_{2}.
(a) Warmup
Viewing ℝ^{2} as ℝ × ℝ we have
and the “disc” centered at the origin is
(b)
Let U ⊂ E_{1} × E_{2} be an open subset. Let (u_{1},u_{2}) ∈ U. We can thus
find an open ball B_{r}(u_{1},u_{2}) ⊂ U. Define U_{1} = {x ∈ E_{1}  d((x,u_{2}),(u_{1},u_{2})) < r}. With this definition for
any x ∈ U_{1}, the open ball of radius will be contained in U_{1} since any y ∈ B_{}(x) has that
so that d((y,u_{2}),(u_{1},u_{2})) < r. Thus U_{1} is open in E_{1}. Similarly, we also have that the set U_{2} = {x ∈ E_{2}  d((u_{1},x),(u_{1},u_{2})) < r}
is open in E_{2} because each x ∈ U_{2} has the open ball B_{}(x) around it which is completely contained in U_{2}. Finally,
with the simple definitions of U_{1} and U_{2}, we see (u_{1},u_{2}) ∈ U_{1} × U_{2} and U_{1} × U_{2} ⊂ B_{r}(u_{1},u_{2}) ⊂ U, as
desired.
Conversely, assume that for each point (u_{1},u_{2}) ∈ U there exist U_{1}, U_{2} open in E_{1}, E_{2}, respectively, with
(u_{1},u_{2}) ∈ U_{1} × U_{2} and U_{1} × U_{2} ⊂ U. So fix (u_{1},u_{2}) ∈ U. As a consequence of the hypothesis at the opening of this
paragraph, we can find open balls B_{r1}(u_{1}) ⊂ U_{1} and B_{r2}(u_{2}) ⊂ U_{2}. Define r = min(r_{1},r_{2}). Then for any point
(x,y) ∈ B_{r}((u_{1},u_{2})) we have
and
which implies x ∈ B_{r1}(u_{1}) and y ∈ B_{r2}(u_{2}), respectively. In turn, this means (x,y) ∈ U_{1} ×U_{2} and thus (x,y) is also in U.
Hence the open ball B_{r}((u_{1},u_{2})) is contained in U, i.e. U is open.
11 The padic 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
where p does not divide any n_{i} or k_{i}. With this in place we assume without loss of generality that ν_{1} ≤ ν_{2} so that
Now with both the above and below forms of x  z, we see ν_{3} ≥ ν_{1}, which given our assumption that ν_{1} ≤ ν_{2},
implies ν_{3} ≥ min(ν_{1},ν_{2}). Hence we have
as desired. __

Nonnegativity
 Let x,y ∈ ℚ be distinct. Then d(x,y) = xy_{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
(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
as desired.