# 这是一篇来自美国的关于探讨下列一些问题逻辑等价性的**数学代写**最终测验，作业具体详情可扫码咨询客服

Instructions: You may use textbooks and course notes on this exam. You may also exchange ideas with others in the class. However, all submitted solutions must reflflect your own understanding and evidence to the contrary (including having a solution which is identical to a solution in another source, or which is identical to another student’s solution) will be considered cheating and will result in a automatic failure and possible disciplinary instructions.

All solutions must include rigorous proof for full credit unless stated otherwise. Any result covered so far in the course may be used freely (with citation) unless stated otherwise.

Throughout this course, we have proved various results related to the nonexistence of “gaps” in R including the Nested Interval Property, the Monotone Convergence Theorem, the Bolzano-Weierstrass Theorem, and the (original) Cauchy Criterion. Each of these was proved, either directly or indirectly, by assuming the Axiom of Completeness. However, it turns out that each of these results (including AoC) is logically equivalent in that if you assume any one of these statements as an axiom, you can prove all of the others as theorems. In Problems 1 and 2, we will explore this logical equivalence by taking one of these theorems as an axiom and proving the Axiom of Completeness.

For these problems, you may assume the Algebraic Axioms and Order Axioms of R stated in Class Notes 1 as well as any immediate consequences of these axioms. However, you may not assume the Axiom of Completeness, nor any result which uses the Axiom of Completeness, since that would make your argument circular.

- (10 points)

Assume the Nested Compact set Property holds and prove the Axiom of Completeness. In other words, suppose given any nested sequence of nonempty compact sets *K*1 *⊇ **K*2 *⊇ **K*3 *⊇ **K*4 *⊇ · · · *, the intersection T *∞ **n*=1 *K**n **6 *=∅.

Prove that any set *A **⊂ *R which is bounded above has a least upper bound.

- (10 points)

Assume the Cauchy Criterion for sequences of real numbers and prove the Axiom of Completeness. In other words, suppose that for all sequences of real numbers,the sequence is Cauchy if and only if it is convergent. Prove that any set *A **⊂ *R which is bounded above has a least upper bound.

- (10 points)

Let *F*0 = [0*, *1]. We obtain *F*1 by removing the middle 1 4 from *F*0, so *F*1 = 0*, *38 *∪ *58* *1 *. *

To form *F*2, we remove the middle116from each of the two intervals, resulting in the set *F*2 = 0*, *532 *∪ *732*, *38 *∪ *58*, *2532 *∪ *2732*, *1 *. *

Continue in this fashion: Once we have constructed *F**n *which will consist of 2*n *intervals, we construct *F**n*+1 by removing the middle

❼ (5 points) Prove that a set *G **⊆ *R is dense if and only if for every nonempty open set *U **⊆ *R, the intersection *G **∩ **U **6 *=∅.

❼ (5 points) Let *f *: R *→ *R be a continuous function and let *G **⊆ *R be dense.

Is the image *f*[*G*] necessarily dense? Is the premiage *f **−*1 [*G*] necessarily dense? For each, give a proof or a counterexample.

- (2 points each) Determine if each of the following statements is true of false. If true, give a brief (approximately one sentence) argument. If false,provide an example.

❼ There is no continuous surjection R *→ *Q.

❼ There is no continuous surjection R *→ *[0*, *1].

❼ If *F*1 *⊇ **F*2 *⊇ **F*3 *⊇ · · · *is a nested sequence of nonempty closed sets, then T *∞ **n*=1*F**n **6 *=∅.

❼ If*G*1 *⊇ **G*2 *⊇ **G*3 *⊇ · · · *is a nested sequence of dense sets, then T *∞ **n*=1 *G**n **6 *=∅.

❼ If *E*1 *⊇ **E*2 *⊇ **E*3 *⊇ · · · *is a nested sequence of nonempty connected sets,then T *∞ **n*=1 *E**n **6 *=∅.

- (10 points) Prove or disprove the following:

The function *f*(*x*) =3*√ **x *is uniformly continuous on R.

- (10 points)

Let *f *: [0*, *1] *→ *[0*, *1] be a continuous function. Prove that *f *has a fifixed point, i.e, there is some*c **∈ *[0*, *1] such that *f*(*c*) = *c*.

In fact, this can be generalized as follows: if *K **⊂ *R is any compact set and *f *: *K **→ **K *is a continuous map, then *f *has a fifixed point. But let’s just stick with the case where *K *= [0*, *1] for now.

- (10 points)

Suppose *f *: *A **→ **B *is a difffferentiable bijection and let *a **∈ **A *such that *f*(*a*)*6 *= 0.

Prove using the defifinition of difffferentiation that the inverse function *f **− *1 is difffferentiable at the point *b *= *f*(*a*) and