October 27, 2014

http://coursework.tylerlogic.com/courses/upenn/math508/homework06
cos x is continuous: Let x ∈ ℝ For any |h| we have

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

Let f(x) = x

as desired.

Put f(x) = x

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 p

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

Let p_{θ} 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, ℝ_{≥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 p_{θ′} and
p_{θ′+π} have the same temperature.

Define {a

has the desired property of f(a_{n}) → 0 and f(b_{n}) 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.

Let f(a,n) = (1 + a)

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.

Defining a

so that f(a_{n},n) → L as n →∞.

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

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

so no matter the value of h, |e^{x}(e^{h} - 1)| can be made arbitrarily large, and therefore so can |e^{x+h} -e^{x}|. Hence e^{x} cannot be
uniformly continuous on [0,∞)

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,∞)

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 ℝ 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^{-},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,∞).

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

The function f : ℝ → (0,∞) defined by f(x) = e

Since f(x) = e

will also be a homeomorphism.

The map f : (-1,1) → ℝ is continuous since its composed of the quotient, product, and difference of continuous functions. It’s injective, because

then we’d have

and the quadratic formula yields

as one of the roots. Thus because

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

Let

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.

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

Let f : ℝ → ℝ be continuous and have that f(x + y) = f(x) + f(y) for all x,y ∈ ℝ. 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 ∈ ℚ has f(q) = qc. With this, we finally have that for x ∈ ℝ and a rational
sequence {a_{n}} with a_{n} → r, the continuity of f yields

as desired.

Let E ⊂ ℝ be a set and f : E → ℝ be uniformly continuous.

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:

- when one of x and y is a limit point not in E and the other a point of E
- 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
B_{1}(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 ⊂ ℝ 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 ℝ. Therefore, the set f(E) is also compact since f is continuous. Thus f(E) is bounded since it’s a subset of ℝ. But because E ⊂E, then f(E) ⊂ f(E), and so f(E) must also 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.