I have an mathmatic exam, you can choose any time from now to monday,thanks

I have an mathmatic exam,you can choose any time from now to monday,The files are my course note, almost 1 hours,Attached file are example of past year examthanks

Math 321 Exam 2 November 25, 2019

MATH 321 Test 2 Solutions

Problem 1 [15 points]

State whether the following statements are true or false.[Correct answer is enough to

get full credit. If you are not sure about your answer, provide a brief explanation – if you answer

is wrong, your solution may still earn partial credit. ]

(a) *E *is the intersection of all closed sets that contain *E*

Let *W *= *∩*

*V *:*V *=*V , E**⊆**V **V *Notice that if *E **⊆ **V *, *V *-closed implies *E **⊆ **V *= *V *, so *E *is contained

in any closed set containing *E*, so *E **⊆ **W*. On the other hand, *E *is itself a closed set that

contains *E*, thus, must contain the intersection of all such sets: *E **⊇ **W*. Hence, *E *= *W*

TRUE

(b) Union of any compact sets is compact.

Counterexample: R = *∪**n**∈*Z[*n, n *+ 1] is a noncompact union of compact sets

FALSE

(c) *f **0*(*x*0) is not defined if *x*0 is not in the domain of *f*

By definition. In applications though, some functions can be redefined at one point to become

differentiable. We are not talking about those cases here.

TRUE

(d) *f*(*x*) = *x *is uniformly continuous on R

By definition, since we can pick *δ *= .

TRUE

(e) for any function *f *defined everywhere on R and any constant *a > *0

lim sup

*x**→**x*0

*f*(*x*) = inf

*δ*:*δ>*0

sup

*x*:0*<**|**x**−**x*0*|**<δ*

*f*(*x*) = inf

*δ*:*a>δ>*0

sup

*x*:0*<**|**x**−**x*0*|**<δ*

*f*(*x*)

True, since sup*x*:0*<**|**x**−**x*0*|**<δ **f*(*x*) is non-increasing with respect to *δ*.

TRUE

1

Math 321 Exam 2 November 25, 2019

Problem 2 [20 points]

(a) Show that if a set *E **⊆ *R is bounded, then *E *is compact.

(b) Show that any bounded non-empty open set does not have a minimum(i.e. its

infimum is not contained in the set) (Hint: assume, by contradiction, that the greatest

lower bound is contained in the set. Since the set is open, every point in it is an interior point.

Recall the definition of an interior point and the definition of infimum and lead the argument

to a contradiction.)

Solution(a) Assume *E *is bounded, i.e. *∃**M > *0 s.t. *∀**x **∈ **E **|**x**| ≤ **M*. Let us show that *E**0 *is

bounded as well: if *x **∈ **E**0 **∩ **E*, *|**x**| ≤ **M*, if *x **∈ **E**0 **\ **E*, we have proven(in class + homework)

that *∃{**x**n**} ⊆ **E *s.t. *x**n **→ **x*. [Optional explanation: *∀**n **∈ *N by definition of an accumulation

point *∃**x**n **∈ **E *s.t. *x**n **∈ *(*x **− *1*/n, x *+ 1*/n*).]

Since for each *n **|**x**n**| ≤ **M*, the limit value, by the properties of sequence limits must satisfy

*−**M **≤ **x **≤ **M*, i.e. *|**x**| ≤ **M*. As show above, if *E *is bounded, *E**0 *is also bounded by the same

constant *M*, and so is *E *= *E **∪ **E**0*.

We have also proved that closure of any set is closed.[Optional explanation: consider *x **∈ **E**0*,

by definition *∀**r > *0 (*x **− **r, x *+ *r*) *∩ **E *contains infinitely many points from *E *= *E **∪ **E**0*. Fix

any *r > *0. If (*x **− **r, x *+ *r*) *∩ **E *contains infinitely many points from *E*, *x **∈ **E**0 **⊆ **E*, If not, it

contains at least one point *y **∈ **E**0*, then letting *ρ *= min*{|**x **− **y**|**, r**} *we observe that the interval

(*y **− **ρ, y *+ *ρ*) must contain infinitely many points from *E*, and thus (*y **− **ρ, y *+ *ρ*) *⊆*, (x-r, x+r).

Thus, by definition, hence, *x **∈ **E*.]

*E *was proven to be closed and bounded, hence it compact.

(b) Assume that an open set *U **⊆ *R is bounded and not empty, then there exists a finite

*m *= inf *U*. If *m **∈ **U *= R *U *(as *U *is open) then *∃**r > *0 s.t. (*m **− **r, m *+ *r*) *⊂ **U*. In particular, it

implies that *∃**x **∈ **U *s.t. *x < m *(any *x **∈ *(*m**−**r, m*), for instance, *x *= *m**−**r/*2). This contradicts

the fact that *m *= min *U*.

2

Math 321 Exam 2 November 25, 2019

Problem 3 [20 points]

(a) Give an example of a function *f*(*x*) that is defined everywhere on R and is not

continuous at infinitely many points. Prove your statement.

(b) Assume that *f *is a monotone increasing continuous function defined on R

with the range not bounded from either above or below. Carefully show that

lim

*x**→*+*∞ **f*(*x*) = +*∞*, lim*x**→−∞ **f*(*x*) = *−∞ *and the range of *f *is equal to R, i.e. it

attains every real value. Cite or state all definitions and theorems you use in your

argument.

Solution (a) for example, the characteristic function of rational numbers *f*(*x*) = *χ*Q(*x*) is discontinuous at every point *a **∈ *R. Indeed, take a sequence *{**x**n**} ⊂ *Q s.t. *x**n **→ **a*, then *f*(*x**n*) = 1

for all *n*, so *f*(*x**n*) *→ *1. On the other hand, if we take a sequence *{**x**0**n**} ⊂ *R *\ *Q s.t. *x**0**n **→ **a*,

then *f*(*x**0**n*) = 0 for all *n*, so *f*(*x**0**n*) *→ *0, 0 *6*= 1, thus, lim*x**→**a **f*(*x*) fails to exist.

Many other examples exist.

(b) Fix any *y **∈ *R. Since *f *is not bounded from below, there exists *a **∈ *R s.t. *f*(*a*) *< y*. Since *f*

is not bounded from above, there exists *b **∈ *R s.t. *f*(*b*) *> y*. Thus, *f*(*a*) *< y < f*(*b*), and, since *f*

is monotone increasing, *a < b*. Applying the intermediate value theorem (Darboux property of

continuous functions) to *f *on [*a, b*] we conclude that *∃**x **∈ *(*a, b*) s.t. *f*(*x*) = *y*. Thus, *f *attains

all real values.

Consider an arbitrary *M > *0 since *f *is not bounded from above there exists *x*0 *∈ *R s.t.

*f*(*x*) *> M*, then *∀**x > x*0 *f*(*x*) *> f*(*x*0) *> M*. By definition it means that lim*x**→∞ **f*(*x*) = +*∞*.

The other limit lim

*x**→−∞ **f*(*x*) = *−∞ *is proven analogously.

3

Math 321 Exam 2 November 25, 2019

Problem 4 [20 points]

(a) Show that the following limit does not exist

lim

*x**→*0

sin *x *1

(b) Consider *f*(*x*) = *x*4 sin *x *1 on R, assume that *f*(0) = 0.

Show that *f *is differentiable on R. Find the explicit formula for *f**0*, explain your

answer.(You may assume that (sin *t*)*0 *= cos *t *and (1*/t*)*0 *= *−*1*/t*2.)

Solution (a) Let us use the sequential definition. Consider *x**n *= *πn *1 *→ *0, then sin *x*1 *n * = 0,

so *f*(*x**n*) *→ *0. On the other hand, *x**0**n *= 2*πn*+ 1 *π/*2 *→ *0, then sin *x*1 *0 **n * = 1, so *f*(*x**0**n*) *→ *1. Since

lim

*n**→∞ **f*(*x**n*) *6*= lim*n**→∞ **f*(*x**0**n*) while both *x**n *and *x**0**n *converge to 0, so lim*x**→*0 *f*(*x*) does not

exist.

(b) For any *x **6*= 0 we can apply the product and chain rules to obtain

*f**0*(*x*) = 4*x*3 sin *x *1 + *x*4 cos *x *1 *−**x *12 = 4*x*3 sin *x *1 *− **x*2 cos *x *1

Let us evaluate the derivative at *x *= 0:

lim

*x**→*0

*f*(*x*) *− **f*(0)

*x **− *0

= lim

*x**→*0

*x*3 sin *x *1 = 0

the last equality is implied by the squeeze theorem, since

*x*3 sin *x *1 *≤ |**x*3*| → *0

4

Math 321 Exam 2 November 25, 2019

Problem 5 [20 points]

(a) Show that any periodic function that is differentiable everywhere on R must have

infinitely many points where its derivative is equal to 0. (Hint: use Rolle’s theorem).

(b) Assume that a continuous function *f*: [*a, b*] *→ *R additionally satisfies the positivity constraint: *∀**x **∈ *[*a, b*] *f*(*x*) *> *0. Show that *f *is actually ‘bounded away from zero’:

there exists a constant *α > *0 s.t. *∀**x **∈ *[*a, b*] *f*(*x*) *≥ **α > *0. (Hint: use compactness of [*a, b*]).

Solution (a) Assume that *T > *0 is the value of the period of *f*. For any *n **∈ *N let *a**n *= *nT*,

*b*

*n *= (*n *+ 1)*T *= *a**n *+ *T*. Since *f *is periodic, *f*(*a**n*) = *f*(*b**n*). By Rolle’s theorem there exists

*x*

*n **∈ *(*a**n**, b**n*) s.t. *f **0*(*x**n*) = 0. Since the open intervals (*a**n**, b**n*) are disjoint, *x**n **6*= *x**m *when

*n **6*= *m*, therefore, we have a sequence of distinct points in which *f **0 *is zero.

(b) The shortest way to show it is to notice that a continuous function *f *attains its minimum

on the compact set [*a, b*]: *∃**x**min **∈ *[*a, b*] s.t. *f*(*x**min*) = inf*x**∈*[*a,b*] *f*(*x*). By the assumption of the

problem, *f*(*x**min*) *> *0. Thus, we may pick *α *= *f*(*x**min*).

Alternatively, we can use a compactness argument to prove it.

5

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