Final Exam
代考分析导论 In the induction step, the original proof wrote “x and y were arbitrary pens in P so all pens in P are the same color”.
Problem 1
In the induction step, the original proof wrote “x and y were arbitrary pens in P so all pens in P are the same color”. Being arbitrary pens in P doesn’t imply x and y are of same color because P is an arbitrary set of n+1 pens, which is not proved to be of same color yet. The flaw in this proof provided is that it used the conclusion that we need to prove as an evidence.
Problem 2
(a) Suppose sets 𝑆1, 𝑆2,… ∈ 𝒪, by definition then its complement is finite, and its . Note that if any one of them is not ℝ or ∅,union with another set whose complement is also finite would generate a new set whose complement is finite.
Therefore if 𝑆1, 𝑆2 ∈ 𝒪, it follows that 𝑆1∪𝑆2∈𝒪.For a countable union of sets in 𝒪, we have, by induction, the union is also in 𝒪
(b) Suppose 𝑆1, 𝑆2 ∈ 𝒪, we have that:
The third scenario can be proved by showing:
(S1 ∩ S2 )c = S1 c∪S2 c
Is a union of two finite sets. It’s obvious (S1 ∩ S2 )c is also finite. By induction, a union of finite number of finite sets is also finite.Therefore a finite intersection of sets in 𝒪 is also in 𝒪.
(c) Proof: for every element 𝑂 ∈ 𝒪 that contains 0, then 𝑂 = ℝ or that 𝑂c is finite by definition of 𝒪. If 𝑂=ℝ,thenthere exists 100 ∈ ℕ that 
∈𝑂=ℝ for all 𝑛≥100.
If 𝑂c is finite and 𝑂≠ℝ,then there exists a finite set of points{𝑑1,𝑑2,… ,𝑑𝑘|𝑑𝑖∈ℝ, 𝑑𝑖 ∉𝑂} that is not in 𝑂. Without loss of generality assume 𝑑1 is the minimum of these points with a positive value. Then we have 𝑁 = 
+ 1 ∈ ℕ, such that 
< 𝑑1. For any 𝑛 ≥ 𝑁,0 << 𝑑1 and ∈ 𝑂. If such positive 𝑑1 doesn’t exist, then let N = 100.
So () 𝒪-converges to 0. 代考分析导论
(d) Following the routine in part (c). Note that if 𝑂 = ℝ that contains 1, then 𝑁 = 100, for all 𝑛 ≥100,∈ 𝑂.If 𝑂c is finite and 1∈𝑂. Because 𝑂c is finite, so 𝑂 must contain an interval that has 0 in it. We can still have 𝑁= + 1∈ℕ where 𝑑1 is the smallest positive value in the point set that is not in 𝑂. And for all 𝑛 ≥ 𝑁, 0 << 𝑑1 and ∈ 𝑂.
The above argument can be further expanded to prove that sequence () 𝒪-converges to any 𝐿 ∈ ℝ.
(e) Prove: for any 𝐿∈ℝ,if 𝑂∈𝒪 and 𝐿∈𝑂,then 𝑂=ℝor 𝑂c is finite
If 𝑂=ℝ, let 𝑁=100, for any 𝑛≥100, 𝑥n=𝑛∈ 𝑂.If 𝑂≠ℝ, 𝐿 ∈𝑂, note that only a finite set of k points is not in 𝑂,let 𝑑𝑘 be the maximum value of such points, and 𝑁=⌈𝑑𝑘⌉ +1∈ℕ. For any 𝑛≥𝑁,𝑥n=𝑛 >𝑑𝑘, 𝑥𝑛 ∈ 𝑂.
Therefore (𝑛) 𝒪-converges to any 𝐿 ∈ ℝ.
Problem 3 代考分析导论
(a) With 𝑥 ≠ 0, the derivative of f is:
When x = 0, the derivative is:
Because cos(𝑥) ∈ [−1,1] for any 𝑥 ∈ ℝ, so:
𝑓 ′ (0) = 
Therefore, f(x) is differentiable on ℝ, and that 𝑓 ′ (0) = < 0
(b) In an open interval that contains 0:
𝑓(−0.01) = − 
+ 0.0001 ∗ cos(100) ≤ −0.005 + 0.0001 < 0
And 𝑓(0) = 0
Therefore there must be an open interval containing both 𝑥 = −0.01 and 𝑥 = 0 that f(x) is not decreasing.
Problem 4
(a) Regardless of the enumeration scheme of ℚ,there is always a one-to-one mapping between ℕ and ℚ.Without loss of generality, assume that there is a subset XN⊂ ℕ such that if 𝑛∈ XN,then 𝑞n < 𝑥. Now for any 𝜖>0:
Note that 
, so we can find a smallest N such that 
, and:
The first part is finite, the second part ≤𝜖.So it’s obvious that 𝑓(𝑥) is finite and well defined for all 𝑥 ∈ ℝ.
(b) Suppose 𝑦 > 𝑥 and 𝑥, 𝑦∈ ℝ, then there must be a subset 𝑌N ⊂ ℕ and a subset XN ⊂ ℕ such that if 𝑛∈XN, 𝑞n <𝑥, and if 𝑛 ∈ 𝑌N, 𝑞n < 𝑦.
Since if 𝑞n <𝑥,then 𝑞𝑛 < 𝑦.
Rationales are dense, so in the interval(𝑥, 𝑦) ≠ ∅,there must be at least one 𝑁∗that 𝑥< 𝑞N∗ <𝑦. And 𝑁∗ ∈ 𝑌N, 𝑁∗ ∉ 𝑋N
We have:
𝑋N⊂𝑌N
So f(x) is monotone.
(c) Suppose 𝑥∈ ℚ, then there exists 𝑁∗ that
𝑞𝑁∗ = 𝑥. For any 𝜖 > 0, in the interval of (𝑥 − 𝜖, 𝑥),there must be a subset 𝜖𝑁 ⊂ℕ that if 𝑛 ∈𝜖𝑁,𝑞n ∈(𝑥 −𝜖, 𝑥).
Therefore 𝑓(𝑥) is discontinuous at all 𝑥 ∈ ℚ.
Problem 5 代考分析导论
(a) The antiderivative of the Weierstrass function is:
Both 𝐹(𝑥) and 𝑓(𝑥) is continuous on ℝ and 𝐹(𝑥) is differentiable with 𝐹 ′ (𝑥) = 𝑓(𝑥).
By the fundamental theorem of calculus, 𝑓(𝑥) is integrable.
(b) The function 𝐹(𝑥) is found in part (a):
(c) By definite integral:
Problem 6
(a) Prove:
When 𝑥 ′ = 𝑥” ∈ [𝑥]𝑑([𝑥],[𝑥])=inf{|𝑥′−𝑥”|:𝑥′∈[𝑥],𝑥”∈[𝑥]}=0
And for any 𝑥, 𝑦 ∈ 𝑋:
𝑑([𝑥],[𝑦])=inf{|𝑥′−𝑦′|: 𝑥′∈[𝑥], 𝑦′∈[𝑦]}
=inf{|𝑦′−𝑥′|:𝑥′∈[𝑥],𝑦′∈[𝑦]}
=𝑑([𝑦],[𝑥])
For any 𝑥, 𝑦, 𝑧∈𝑋:
𝑑([𝑥],[𝑧])=inf{|𝑥′−𝑧′|:𝑥′∈[𝑥], 𝑧′∈[𝑧]} 代考分析导论
≤inf{|𝑥′−𝑦′|+|𝑦′−𝑧′|: 𝑥′∈[𝑥], 𝑦′ ∈[𝑦], 𝑧′∈[𝑧]}
≤inf{|𝑥′−𝑦′|:𝑥′∈[𝑥],𝑦′ ∈[𝑦]} +inf{|𝑦′−𝑧′|: 𝑦′ ∈[𝑦], 𝑧 ′ ∈ [𝑧]}
≤𝑑([𝑥],[𝑦])+𝑑([𝑦],[𝑧])
therefore 𝑑([𝑥],[𝑦])
is a valid metric on the set of
equivalence classes of this relation.
(b) The value of 𝑑([𝜋],[1000]) can be calculated by definition:
𝑑([𝜋],[1000]) = inf{|𝑥′ −𝑦 ′|: 𝑥′ ∈ [𝜋], 𝑦 ′∈ [1000]}
Now the equivalence relation is given by𝑥~𝑥 ′if𝑥− 𝑥 ′ ∈ℤ
So [𝜋] = {𝑥 + 𝜋|𝑥 ∈ ℤ}, [1000]=ℤ, and it’s easy to verify that:
𝑑([𝜋],[1000]) = 3.14 − 3 = 0.14
Problem 7 代考分析导论
(a) By definition:
𝑑∞((𝑥𝑛 ), (𝑦𝑛)) =sup{||sin(𝑛 −1)| − |cos(𝑛 − 1)||: 𝑛 ∈ ℕ}
Now to maximize |sin(𝑛 −1)|− |cos(𝑛 − 1)|, we can assume that for a proper n,sin(𝑛 − 1) → 1 and cos(𝑛 − 1) → 0 at the same time, i.e. 𝑛 − 1 → 2𝑘𝜋 + 𝜋2 for some 𝑘 ∈ ℤ.
And 𝑑∞((𝑥𝑛 ), (𝑦𝑛 )) = 1
(b) A sequence (𝑥𝑛 )𝑖i nℓ ∞ is Cauchy if for every 𝜖 > 0, there exists 𝑁 ∈ ℕ such that whenever 𝑎, 𝑏≥𝑁,then𝑑∞((𝑥𝑛 )𝑎, (𝑥𝑛 )𝑏 ) <𝜖
(c) Assume that sequence (𝑥𝑛)𝑖 is Cauchy inℓ∞, thenfor every 𝜖>0, there exists 𝑁 ∈ ℕ such that whenever 𝑎,𝑏≥𝑁,then: 代考分析导论
𝑑∞((𝑥𝑛)𝑎, (𝑥𝑛 )𝑏 ) =sup{|𝑥𝑛𝑎 −𝑥𝑛𝑏 |: 𝑛 ∈ ℕ}≤, 𝑥𝑛𝑎 ∈ (𝑥𝑛 )𝑎,𝑥𝑛𝑏∈(𝑥𝑛 )𝑏} < ∈
Now that(𝑥𝑛)𝑖 is bounded for any i, then we must have a converging subsequence of (𝑥𝑛 )𝑖 . Let (𝑥𝑛)𝑝𝑖 denote the subsequence, and 
be the limit. For any 𝜖 > 0,
we
=2𝜖
With 𝑎, 𝑏 ≥𝑁, and (𝑥𝑛)𝑏 is in the converging subsequence.So Cauchy sequences in ℓ ∞ converges. And (ℓ∞,𝑑∞((𝑥𝑛),(𝑦𝑛)))is a complete metric space.
