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 where j=1V j 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
     {            }
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]


Interior points: (0,1)

Boundary points: {0,1}

Closure: [0,1]

(b) 2 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)


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
   {  sin-1 x ⁄= 0
y =   0 x  x = 0

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]


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

(b) {0}∪{1,1
2,,1-
n,}⊂


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

(c) X=[0, 1] \ as a subset of


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

so

  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 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
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 ⟨A,B⟩ for A,B Mk,n by ⟨A,B ⟩ = trace(ABt)

(a) Show this ⟨⋅,⋅⟩ is an inner product.


Positive Definite
Let A Mk×n with r1,r2,,rk being the k vectors that make up the rows of A. Then ⟨A,A ⟩ = 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 ⟨A,A ⟩0 with equality when A is the zero matrix.
Additivity
Let A,B,C Mk×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 Mk×n and α . By properties of the trace function we have
                  t             t
⟨αA,B ⟩ = trace(αAB ) = α trace(AB ) = α ⟨A,B ⟩

Symmetry
Let A,B Mk×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 = ⟨A,A ⟩ 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 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 + 1
2r 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
    ⋃
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


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
    ∑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 2 as × we have
d((0,0),(1,2)) = d(0,1)+ d(0,2) = 1 + 2 = 3

and the “disc” centered at the origin is

PIC

(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 r-d1(x,u1)
   2 will be contained in U1 since any y Br-d1(x,u1)
   2(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 Br-d2(2x,u2)(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

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 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 (,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

         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 ni or ki. 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(ν12). 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 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 = pνnk. Then y - x = pν-kn, 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
d(x - z) ≤ max (d(x,y),d(y,z)) ≤ d(x,y)+ d(y,z)

(b) For x,a 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
                ′       ′ ′
Nr(a) = Nr(x) ⊂ N r(x) = Nr(a )

as desired.