Real Analysis Homework 1 Solutions 1. Show that Rn with the usual ...

Real Analysis Homework 1 Solutions1. Show that R n with the usual euclidean distance is a metric space. Itemsa-c will guide you through the proof.a. Define a function called an inner product on pairs of points x =(x 1 , x 2 , . . . , x n ) and y = (y 1 , y 2 , . . . , y n ) in R n by〈x, y〉 = x 1 y 1 + x 2 y 2 + · · · + x n y n .Show that for any x ∈ R n , 〈x, x〉 ≥ 0 and is equal to 0 if and only ifx = 0 (i.e. the inner product is positive definite).Also show that for any x, y ∈ R n , 〈x, y〉 = 〈y, x〉 (i.e. the innerproduct is symmetric). Third, show that the inner product is linearin each component, that is for α, β ∈ R and x, y, z ∈ R n show〈αx + βy, z〉 = α〈x, z〉 + β〈y, z〉to establish linearity in the first coordinate, then use the symmetryof the inner product to give linearity in the second coordinate.Proof. Since 〈x, x〉 = x 2 1 + x 2 2 + · · · + x 2 n and each coordinate is real,this sum is always nonnegative. If x = 0, then this sum is 0 2 + 0 2 +· · · + 0 2 = 0. If this sum is zero, then every single coordinate mustalso be zero, thus we have a function so that 〈x, x〉 ≥ 0 and is equalto zero if and only if x = 0. Therefore 〈·, ·〉 is positive definite. [Notethat when we say x = 0, this means that x is the origin in R n , notthat x is the real number 0.] √From the commutativity of multiplication,〈x, y〉 = x 1 y 1 + · · · + x n y n = y 1 x 1 + · · · y n x n = 〈y, x〉. √With z = (z 1 , z 2 , . . . , z n ), thenThereforeαx + βy = α(x 1 , . . . , x n ) + β(y 1 , . . . , y n )= (αx 1 , . . . , αx n ) + (βy 1 , . . . , βy n )= (αx 1 + βy 1 , . . . , αx n + βy n ).〈αx + βy, z〉 = (αx 1 + βy 1 )z 1 + · · · + (αx n + βy n )z n= α(x 1 z 1 + · · · + x n z n ) + β(y 1 z 1 + · · · y n z n )= α〈x, z〉 + β〈y, z〉,1

which establishes linearity in the first variable. To get linearity inthe second, let α, β ∈ R and x, y, z ∈ R n . Then using the symmetryof 〈·, ·〉 and that it’s linear in the first variable,〈x, αy + βz〉 = 〈αy + βz, x〉= α〈y, x〉 + β〈z, x〉= α〈x, y〉 + β〈x, z〉.Therefore it is linear in the second variable, too.b. For x ∈ R n define the norm of x by‖x‖ = 〈x, x〉 1 2 .Note that this gives the standard euclidean distance in R n of x fromthe origin.We now establish the Cauchy-Schwarz-Bunyakovsky inequality:|〈x, y〉| ≤ ‖x‖‖y‖.This is a significant inequality, and is essential in the study of innerproduct spaces.Let λ ∈ R and x, y ∈ R n . Consider the inequality from positivedefiniteness〈λx + y, λx + y〉 ≥ 0.Use linearity and symmetry to rewrite this as〈x, x〉λ 2 + 2〈x, y〉λ + 〈y, y〉 ≥ 0.Think of this as a quadratic in the variable λ, and since the left sideis always nonnegative, the discriminant in the quadratic formula isnonpositive, so thatFrom this deduce thatProof. For λ ∈ R and x, y ∈ R n ,4〈x, y〉 2 − 4〈x, x〉〈y, y〉 ≤ 0.|〈x, y〉| ≤ ‖x‖‖y‖.0 ≤ 〈λx + y, λx + y〉= λ〈x, λx + y〉 + 〈y, λx + y〉= λ 2 〈x, x〉 + λ〈x, y〉 + λ〈y, x〉 + 〈y, y〉= λ 2 〈x, x〉 + 2λ〈x, y〉 + 〈y, y〉. √2

By considering this as a quadratic in the variable λ and that it isalways nonnegative, it either has a repeated real root or no real rootsat all. Therefore the discriminant in the quadratic formula must beat most 0, i.e. b 2 − 4ac ≤ 0. This translates toThis leads to(2〈x, y〉) 2 − 4〈x, x〉〈y, y〉 ≤ 0.4|〈x, y〉| 2 ≤ 4〈x, x〉〈y, y〉,which by taking square roots and substituting in ‖ · ‖ where appropriate,|〈x, y〉| ≤ ‖x‖‖y‖.c. Define the function d(x, y) = 〈x − y, x − y〉 1 2 . We show that d is ametric on R n .i. Show that d is positive definite.ii. Show that d is symmetric.iii. Consider three points x, y, z ∈ R n . Since R n is a linear space andthat in the function d we are considering differences of points, wemay assume y = 0 for simplicity, so x − y = x and y − z = −z.With this simplification, we have d(x, y) = ‖x‖ and d(y, z) = ‖z‖.Show thatd(x, z) 2 = ‖x‖ 2 − 2〈x, z〉 + ‖z‖ 2 .Using the Cauchy-Schwarz inequality, show thatd(x, z) 2 ≤ ‖x‖ 2 + 2‖x‖‖z‖ + ‖z‖ 2 .Taking square roots, conclude thatd(x, z) ≤ ‖x‖ + ‖z‖ = d(x, y) + d(y, z).Hence, R n is a metric space using the metric derived from the innerproduct.Proof. We prove the positive definiteness of d since 〈w, w〉 ≥ 0 andis equal to zero if and only if w = 0. The form 〈w, w〉 1 2 also has thisproperty. By replacing w with x − y establishes that d is positivedefinite. √Symmetry is attained by using linearity in both components andobservingd(y, x) 2 = 〈y − x, y − x〉= 〈−(x − y), −(x − y)〉= − − 〈x − y, x − y〉= 〈x − y, x − y〉= d(x, y) 2 √3

For the triangle inequality, let x, y, z ∈ R n . If we define x ′ = x − y,y ′ = 0, z ′ = z −y, by definition, d(x ′ , y ′ ) = d(x, y), d(y ′ , z ′ ) = d(y, z),and d(x ′ , z ′ ) = d(x, z), then we are able to use x ′ , y ′ , z ′ in place ofx, y, z. Therefore we may assume without loss of generality thaty = 0. Then d(x, y) = ‖x‖ and d(y, z) = ‖z‖.Consider the following calculation:d(x, z) 2 = 〈x − z, x − z〉= 〈x, x〉 − 〈x, z〉 − 〈z, x〉 + 〈z, z〉= ‖x‖ 2 − 2〈x, z〉 + ‖z‖ 2≤ ‖x‖ 2 + 2|〈x, z〉| + ‖z‖ 2≤ ‖x‖ 2 + 2‖x‖‖z‖ + ‖z‖ 2= (‖x‖ + ‖z‖) 2 .= (d(x, y) + d(y, z)) 2So we arrive at d(x, z) 2 ≤ (d(x, y) + d(y, z)) 2 , and by taking squareroots we get d(x, z) ≤ d(x, y) + d(y, z).d. Given any real vector space V , if there is a function 〈·, ·〉 which satisfiesi. 〈x, x〉 ≥ 0, and 〈x, x〉 = 0 if and only if x = 0 (positive definiteness),ii. 〈x, y〉 = 〈y, x〉 (symmetry), andiii. 〈αx + βy, z〉 = α〈x, z〉 + β〈y, z〉 for all α, β ∈ R and x, y, z ∈ V ,then we say V is a real inner product space. The Cauchy-Schwarzinequality follows from these axioms, so V becomes a metric spaceunder the metric d(x, y) = 〈x − y, x − y〉 1 2 .If we want to consider complex vector spaces we have to change thingsa slight bit. First, we have to make sure the function respects complexlinearity, so we must consider α, β ∈ C instead of the reals. Second,the symmetric property must account for the complex variables aswell. So we replace ii. withii. 〈x, y〉 = 〈y, x〉.so V equipped with such a function is a complex inner productspace.e. There are a lot of inner product spaces out there. For example C[0, 1]is a real (complex) inner product space with inner product〈f, g〉 =∫ 10(f(x)g(x) dx =∫ 10)f(x)g(x) dx .4

f. There are a lot of spaces which are not inner product spaces. Forexample, the metrics d 1 and d ∞ on R 2 do not come from inner products.Similarly, the metrics d 1 and d ∞ on C[0, 1] don’t come from innerproducts.2. Given a metric space (X, d) define a function d ′ : X × X → R byd ′ (x, y) =d(x, y)1 + d(x, y) .a. Show that (X, d ′ ) is a metric space. Hint: When showing the triangleinequality, at some point you will have something like d(x, y)+d(y, z)in the denominator, and you replace this with d(x, z) which will makethe fraction larger (but that’s ok since you’re showing an inequality).Proof. The function d ′ is positive definite since the function f(t) =t1+tis nonnegative on the nonnegative real numbers and is equal tozero exactly when t = 0.The function d ′ is symmetric sinced ′ (x, y) =d(x, y) d(y, x)=1 + d(x, y) 1 + d(y, x) = d′ (y, x).We show the triangle inequality for d ′ by starting with the triangleinequality for d. For simplicity, let a = d(x, z), b = d(x, y) andc = d(y, z). Thena ≤ b + ca + (ab + ac + abc) ≤ b + c + (ab + ac + abc)a + (ab + ac + abc) ≤ b + c + (ab + ac + abc) + (bc + abc + bc)a(1 + b)(1 + c) ≤ b(1 + a)(1 + c) + c(1 + a)(1 + b)a1 + a ≤ b1 + b + c1 + c ,where the second line arises from that adding equal amounts to eachside of an inequality preserves the inequality, the third line arises fromthat bc + abc + bc ≥ 0, the fourth line is algebra, and the fifth lineresults from division by the positive number (1 + a)(1 + b)(1 + c).3. Show that an arbitrary union of open sets is open.Proof. Let U α be open for every α ∈ A, where A is an index set. Letx ∈ U = ⋃ α∈A U α. Since U is a union, there is some β ∈ A so thatx ∈ U β . Thus there is some r > 0 so that B r (x) ⊆ U β . Since U β ⊆ U,this implies B r (x) ⊆ U. Since x was an arbitrary element of U, this showsthat U is open.5

4. Let X be a metric space and S ⊆ X. Prove that ext(S) = (X\S) ◦ .Proof. We show ext(S) = (X\S) ◦ by double containment.(⊆). Let x ∈ ext(S). Then there some r > 0 so that B r (x) ∩ S = ∅. Usingthe same r, x is also a point in X\S so that B r (x) ⊆ X\S. Thereforex ∈ (X\S) ◦ .(⊇). Let x ∈ (X\S) ◦ . Then there is some r > 0 so that B r (x) ⊆ X\S.Thus B r (x) ∩ S = ∅. But this shows that x ∈ ext(S).5. Show that R\Z is open in R.Proof. Let x /∈ Z be a real number. Then ⌊x⌋ < x⌊x⌋ + 1, where ⌊·⌋ isthe greatest integer function. Let r = min{x − ⌊x⌋, ⌊x⌋ + 1 − x}. Thenfor y ∈ R, if |x − y| < r, then y > ⌊x⌋ and y < ⌊x⌋ + 1, so y /∈ Z. HenceB r (x) ⊆ R\Z, so R\Z is open.Implicit in this argument is that the integers ⌊x⌋ and ⌊x⌋ + 1 are thetwo closest integers to x. The archimedean principle will show this to betrue.6. Prove that in R, ∂Q = R. Hint: between any two real numbers, there isa rational.Proof. We want to show that for any x ∈ R and r > 0, then B r (x) containsa point in Q and a point in R\Q.Case 1. Let x ∈ R\Q and r > 0. Then there is a rational q between x andx + r. Thus B r (x) contains a rational. This ball automatically containsx ∈ R\Q. Therefore x ∈ ∂Q.Case 2. Let x ∈ Q and r > 0. There is some rational q so that 0 < q < r.Then 0 < q π < r as well, and the number q πis irrational. Thus theirrational number x + q πis contained in B r(x). Thus any neighborhoodof x contains a rational (namely x) and an irrational. Therefore x ∈ ∂Q.Thus ∂Q = R in R.This proof also contains the idea that between any two reals there is anirrational as well. This and the hint are consequences of the archimedeanprinciple.K&F 6.1. Give an example of a metric space R and two open spheres B r1 (x) andB r2 (y) in R such that B r1 (x) ⊆ B r2 (y) although r 1 > r 2 .Proof. Your example may be different. Let R be a nonempty discretemetric space, x = y ∈ R, and 0 < r 2 < r 1 < 1. Then B r1 (x) = B r2 (x),and set containment follows from equality.For an example of strict containment, consider R = {(x, y) ∈ R 2 |x ≥0} ∪ {(−1, 0)}. Then B √ 2((−1, 0)) B 11 ((0, 0)). Note that you can’t get10strict containment if x = y.6