Final
分析导论代考 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 分析导论代考
(a)Suppose sets 
,by definition. Note that if any one of them is not 
or ∅, then its complement is finite, and its union with another set whose complement is also finite would generate a new set whose complement is finite.
Therefore if , it follows that 
. For a countable union of sets in 
, we have, by induction, the union is also in
(b)Suppose ,we have that:
The third scenario can be proved by showing:
Is a union of two finite sets. It’s obvious 
is also finite. By induction, a union of finite number of finite sets is also finite. Therefore a finite intersection of sets in is also in
(c)Proof: for every element 
that contains 0, then 
or that 
is finite by definition of . If , then there exists 
for all n ≥100.
If is finite and 
, then there exists a finite set of points 
that is not in 0. Without loss of generality assume d1 is the minimum of these points with a positive value. Then we have 
, such that 
. For any 
, and 
. If such positive doesn’t exist, then let N = 100.d1
So 
-converges to 0.
(d)Following the routine in part (c). Note that if that contains 1, then N = 100, for all n≥100, 
. 分析导论代考
If is finite and 1∈0 . Because is finite, so 0 must contain an interval that has 0 in it. We can still have 
where d1 is the smallest positive value in the point set that is not in 0. And for all 
.
The above argument can be further expanded to prove that sequence -converges to any 
.
(e)Prove: for any ,if 
, then 
is finite
If , let N = 100, for any n = 100, 
.
If ,L∈0 note that only a finite set of k points is not in , let dk be the maximum value of such points, and 
. For any
.
Therefore -converges to any.
Problem 3
(a)With x≠0, the derivative of f is:
When x = 0, the derivative is:
Because cos(x)∈[-1,1] for any 
, so:
Therefore, f(x) is differentiable on , and that 
(b)In an open interval that contains 0:
And f(0)＝0
Therefore there must be an open interval containing both and 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 
such that if 
, then 
. Now for any 
:
<
Note that 
, so we can find a smallest N such that 
, and:
The first part is finite, the second part ≤∈. So it’s obvious that f(x) is finite and well-defined for all .
(b)Suppose y>xand y,x∈R, then there must be a subset YN ⊂N and a subset XN⊂N
such that if 
, and if 
. Since if , then qn<y.
Rationales are dense, so in the interval (x,y)≠Ø, there must be at least one N*that 
We have:
So f(x) is monotone.
(c)Suppose 
,then there exists N* that 
. For any 
, in the interval of 
, there must be a subset 
that if 
.
Therefore f(x) is discontinuous at all x∈R.
Problem 5 分析导论代考
(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 is found in part (a):
(c)By definite integral:
Problem 6
(a)Prove:
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:
Now the equivalence relation is given by if
So 
and it’s easy to verify that:
Problem 7 分析导论代考
(a)By definition:
Now to maximize 
, we can assume that for a proper n, sin(n-1)→1 and sin(n-1)→0 at the same time, i.e. 
for some 
.
And 
(b)A sequence 
in 
is Cauchy if for every 
, there exists 
such that whenever 
, then 
(c)Assume that sequence is Cauchy in , then for every , there exists 
such that whenever 
, then:
Now that is bounded for any i, then we must have a converging subsequence of . Let 
denote the subsequence, and 
be the limit. For any ∈>0, we
With a,b<N and 
is in the converging subsequence. So Cauchy sequences in converges. And 
is a complete metric space.
