Homework 6 <me@tylerlogic.com>
October 27, 2014
http://coursework.tylerlogic.com/courses/upenn/math508/homework06

### 1 Prove cosx and sinx are continuous for all x ∈ ℝ

cos x is continuous: Let x For any |h| we have

where the last inequality comes from the fact that -1 ≤-sinx 1 and -1 cosx 1 for all values of x. Now, as h 0, cosh 1 and sinh 0. This implies the last line in the equation above approaches zero as h 0. Thus, for any ε > 0 we can find a small enough δ so that for |h| < δ |sin(x + h) - sinx| < ε. Hence sinx is continuous at x.

sin x is continuous: Let x For any |h| we have

where the last inequality comes from the fact that -1 sinx 1 and -1 cosx 1 for all values of x. Like we saw in the previous paragraph, the last line of the above equation can be made arbitrarily close to zero as h 0. Hence, similar to what we saw in the previous paragraph, cosx is continuous at x.

### 2

Let f(x) = x2 + 4x and 0 < ε < 4. Define δ = . Then 0 < δ < 1, given the constraints on ε. Assume that |x| < δ, so that 0 < |x| < δ < 1, in particular 0 < |x|2 < |x| < 1. With this and the triangle inequality, we then have

as desired.

### 3 Prove the existence of an x ∈ [1, 2] such that x5+ 2x + 5 = x5+ 10

Put f(x) = x5 + 2x + 5 and g(x) = x5 + 10. Being polynomials, f and g are both continuous. Therefore the intermediate value theorem implies that on the interval [1,2] f will take on every value between f(1) = 8 and f(2) = 41 and g will take on every value between g(1) = 11 and g(2) = 26. Since f(1) = 8 < 11 = g(1) and g(2) = 26 < 41 = f(2), f and g must therefore intersect on the interval [1,2]. Hence there’s a point x on [1,2] with f(x) = g(x), i.e. where x5 + 2x + 5 = x5 + 10.

### 4 Show there exists two diametrically opposite points on the Eath’s equator that have the same temperature.

We’ll use the 3-dimensional cartesian coordinate system and orient the earth so that the z-axis contains the earth’s axis of rotation, the north pole has a positive z coordinate, and the center of the earth is at the orgin. Then the equator will be contained in the xy-plane. Finally, we will denote, by pθ for θ 0, the point on the equator that is intersected by the line through the origin and (cosθ,sinθ,0).

Now let T : 0 be the function whose value at θ 0 is the temperature of pθ. We will assume that this temperature function is continuous. Define the function D : 0 by D(θ) = T(θ) -T(θ + π), i.e. it is the difference in temperature of pθ and its diametrically opposite point, pθ+π. Since D is made up of the composition and difference of continuous functions, then it too is continuous.

Let pθ be a point on the equator that doesn’t have the same temperature as its diametric opposite (there must be such a point, otherwise there would be nothing to prove as all points diametrically opposite would have the same temperature). Then D(θ) = -D(θ + π) and neither values are zero. Without loss of generality, assume that D(θ) < 0 < D(θ + π). Since D is continuous and it’s domain, 0, is connected, then this and the intermediate value theorem implies that there is a θ′∈ (θ,θ + π) such that D(θ) = 0. But this implies that T(θ) = T(θ+ π), i.e. the diametrically opposed points pθ and pθ+π have the same temperature.

### 5

Define {an} and {bn} by an = -1∕n and bn = 1∕n. Then of course both sequences converge to zero. Then the function f : defined as

has the desired property of f(an) 0 and f(bn) being unbounded.

Does there exist such a function that’s continuous at x = 0 This is not possible as it would contradict Rudin’s Theorem 4.2.

### 6

Let f(a,n) = (1 + a)n where a and n are positive.

#### (a) Behavior of f(a,n) for constant a or n

For constant a how does f(a,n) behave as n →∞? When a is constant, 1 + a > 1 since a was assumed positive. So f(a,n) = (1 + a)n →∞ as n →∞.

For constant n how does f(a,n) behave as a 0? As a 0, 1 + a 1. So for constant n, f(a,n) = (1 + a)n 1 as a 0.

#### (b)

Defining an = - 1, then an is a sequence of positive values, and furthermore

so that f(an,n) L as n →∞.

### 7 Which of the following functions are uniformly continuous on [0,∞)

#### (a) f(x) = xsinx

For any x [0,) and h we have that
For a fixed h, hsin(x + h) is bounded, but x(sinxcosh + sinhcosx - sinx) gets arbitrarily large as x gets large. Thus no matter how small h is, we can find x large enough so that |hsin(x + h) + x(sinxcosh + sinhcosx - sinx)|, and therefore |(x + h)sin(x + h) - xsinx|, can be made arbitrarily large. Hence there is no way for f(x) = xsinx to be uniformly continuous on [0,).

#### (b) f(x) = ex

For any x [0,) and h we have that

so no matter the value of h, |ex(eh - 1)| can be made arbitrarily large, and therefore so can |ex+h -ex|. Hence ex cannot be uniformly continuous on [0,)

#### (c) f(x) =

For any ε > 0, set δ = ε. Thus for any x,y [0,when |y - x| < δ we have

Since x,y [0,) then 1 + y 1 and 1 + x 1 which implies

so that f(x) is uniformly continuous on [0,)

### 8 Show that f(x) = is continuous ∀x ≥ 0

Let ε > 0 and x [0,). Define δ = ε so that in particular ε = . Then for any y 0 such that |x - y| < δ we have

indicating that f is continuous at x.

Is the function uniformly continuous? Yes. First note that because f is continous, as per above, then because [0,1] is compact in f is uniformly continuous on [0,1]. Furthermore, for any x,y [1,), + 1. Therefore for any ε > 0, setting δ = ε means that when |y - x| < δ we have

so that f is uniformly continuous on [1,).

Finally for any ε > 0, define δ = min(δ-+) where δ- is such that |x-y| < δ-⇒|-| < for all x,y [0,1] and δ+ is such that |x - y| < δ+ ⇒| -| < for all x,y [1,). Thus with this definition, whether x,y [0,1] or x,y [1,), |x - y| < δ will imply | -| < ϵ.

It remains to be seen that |x-y| < δ implies |-| < ϵ when x [0,1] and y [1,). This nevertheless holds, since for x [0,1] and y [1,), |x - y| < δ implies both |x - 1| < δ < δ- and |y - 1| < δ < δ+. These two equations in turn imply | -| < and | -| < . Hence we obtain

Therefore f is uniformly continous on all of [0,).

### 9

#### (a) Prove [0, 1] and ℝ are not homeomorphic.

Since [0,1] is compact but is not, there exists no continuous function from [0,1] to . Thus there exists no homeomorphism between them either.

#### (b) Prove ℝ and (0,∞) are homeomorphic.

The function f : (0,) defined by f(x) = ex is both bijective and continuous, and thus is a homeomorphism between the two sets.

#### (c) Prove ℝ2 and {(x,y) ∈ ℝ2: y > 0} are homeomorphic.

Since f(x) = ex is a homeomorphism from (0,) then the function g : 2 →{(x,y) 2 : y > 0} defined by

will also be a homeomorphism.

#### (d) Prove ℝ and (-1, 1) are homeomorphic.

The map f : (-1,1) is continuous since its composed of the quotient, product, and difference of continuous functions. It’s injective, because
It’s surjective because for any y , if it were to be true that

then we’d have

as one of the roots. Thus because

then we have a solution for x < 1. Thus f is a homeomorphism.

### 10

Let

#### (a) Show that f is continuous on ℝ

The function f is the composition and product of functions x, sinx, and 1∕x, which are all continuous at nonzero reals. Therefore f is continuous at all nonzero reals. It remains to be shown that f is continuous at zero.

Let ε > 0 and set δ = ε. The for |x| < δ we have

so that f is continuous at zero.

#### (b) Is f uniformly continous on [0, 2π]?

Yes, f is continous on [0,2π] and [0,2π] is compact.

### 12 Show a continuous real-valued function f that has f(x + y) = f(x) + f(y) for all x,y ∈ ℝ must be f(x) = cx for some c.

Let f : be continuous and have that f(x + y) = f(x) + f(y) for all x,y . Put c = f(1). Then for integer n, f(n) = f(1) + f(n- 1) = f(1) + + f(1) = nf(1) = nc. Thus for any rational n∕m we have

so that f(n∕m) = (n∕m)c. Thus any rational q has f(q) = qc. With this, we finally have that for x and a rational sequence {an} with an r, the continuity of f yields

as desired.

### 13

Let E be a set and f : E be uniformly continuous.

#### (a) Show that if E is bounded, then so is f(E)

Lemma 13.1. If E X is a bounded set, then so is its closure, E.

Proof. Let E be a bounded subset of a metric space X, and x,y two points of its closure, E. If x,y E, then there’s nothing to prove since E is already bounded. We must address two remaining case:

1. when one of x and y is a limit point not in E and the other a point of E
2. when both x and y are limit points not in E

Assume the first, and without loss of generality, let x E and y be a limit point of E. Then any open ball centered at y, say B1(y), contains some point of x E. Thus d(x,y) d(x,x) + d(x,y) < M + 1 where M is a bound for the set E. Hence the distance between any point of E and a limit point is bounded.

Now assume the second case above. By the previous paragraph we have d(x,y) d(x,z) + d(z,y) < 2(M + 1) for any z E. So in this case, the distance is bounded.

Hence we have that the closure of E is bounded. __

Let E be a bounded set. Then the closure, E, is also bounded by the above lemma. Hence it’s closed and bounded, implying that it’s compact as a subset of . Therefore, the set f(E) is also compact since f is continuous. Thus f(E) is bounded since it’s a subset of . But because E E, then f(E) f(E), and so f(E) must also be bounded.

#### (b) If E is not bounded, give an example showing f(E) might not be bounded.

Let E = (0,) and f(x) = x. Then f is uniformly continuous, E is not bounded, and f(E) = (0,) which is also not bounded.