Math Challenge - May 2019

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1. a. Let ##(\mathfrak{su}(2,\mathbb{C}),\varphi,V)## be a finite dimensional representation of the Lie algebra ##\mathfrak{g}=\mathfrak{su}(2,\mathbb{C})##.
Calculate ##H\,^0(\mathfrak{g},\varphi)## and ##H\,^1(\mathfrak{g},\varphi)## for the Chevalley-Eilenberg complex in the cases​
(i) ##(\varphi,V)= (\operatorname{ad},\mathfrak{g})##​
(ii) ##(\varphi,V)= (0,\mathfrak{g})##​
(iii) ##(\varphi,V)= (\pi,\mathbb{C}^2)## is the natural representation on ##\mathbb{C}^2##.​

b. Consider the Heisenberg algebra ##\mathfrak{g}=\mathfrak{h}=\left\{ \left.\begin{pmatrix}0&a&c\\0&0&b\\0&0&0\end{pmatrix}\right|a,b,c\in \mathbb{R}\right\}## and calculate ##H\,^0(\mathfrak{h},\operatorname{ad})## and ##H\,^1(\mathfrak{h},\operatorname{ad})\,.##​

2. Show that the dihedral group ##D_{12}## of order twelve is the finite reflection group of the root system of type ##G_2##.

3. (solved by @Periwinkle ) Consider the set
$$
\mathcal{P}_n := \{\,\{2\},\{4\},\ldots,\{2n\}\,\} \subseteq \mathcal{P}(\mathbb{N})
$$
and determine the ##\sigma-##algebra ##\mathcal{A}_\sigma(\mathcal{P}_n) \subseteq \mathcal{P}(\mathbb{N})##, and show that ##\bigcup_{n\in \mathbb{N}}\mathcal{A}_\sigma(\mathcal{P}_n)## isn't a ##\sigma-##algebra.​

4. (solved by @Periwinkle ) Linear Operators. (Only solutions to both count!)
a. Show that eigenvectors to different eigenvalues of a self-adjoint linear operator are orthogonal and the eigenvalues real.​
b. Given a real valued, bounded, continuous function ##g \in C([0,1])## with​
$$
m = \inf_{t\in [0,1]} g(t)\; , \;M = \sup_{t\in [0,1]} g(t)
$$
and an operator ##T_g(f)(t):= g(t)f(t)## on the Hilbert space ##\mathcal{H}=L^2([0,1])\,.##​
Calculate the spectrum of ##T_g\,.##​
5. Let ##\mathbb{F}## be a field. Then for a polynomial ##f \in \mathbb{F}[X_1,\ldots,X_n]## we define ##D(f)=\{\,q\in \mathbb{A}^n(\mathbb{F})\,|\,f(q)\neq 0\,\}##.
Show that these sets build a basis of the Zariski topology on ##\mathbb{A}^n(\mathbb{F})##, and decide whether finitely many of them are sufficient to cover a given open set.​
6. Let ##R := \mathbb{Q}[x,y]/\langle x^2+y^2-1 \rangle## and ##\varphi \in \operatorname{Der}(R)## a ##\mathbb{Q}-##linear derivation such that ##\varphi(x)=y\; , \;\varphi(y)=-x\,.##
A derivation ##\varphi\, : \,R \longrightarrow R ## of an algebra ##R## is a linear function with ##\varphi(p\cdot q)=\varphi(p)\cdot q + p\cdot \varphi(q)\,.##​
a. Determine the kernel of ##\varphi\,.##​
b. Solve ##\varphi^2 + \operatorname{id} = 0\,.##​
c. Since ##x^2+y^2=1## we can apply Thales' theorem and identify ##(x,\alpha),(y,\alpha)## with the sides of a right triangle with hypotenuse (diameter) ##1## according to an angle ##\alpha\,.##​
Show that​
$$
(x,\alpha +\beta ) = (x,\alpha)(y,\beta) + (x,\beta)(y,\alpha)
$$
Thales.png


7. (solved by @Periwinkle ) Prove that for all ##a,b,c \in \mathbb{R}## holds $$a>0\wedge b>0\wedge c>0 \Longleftrightarrow a+b+c>0\wedge ab+ac+bc>0\wedge abc>0\,.$$

8. (solved by @Periwinkle ) Let ##a,b \in L^2\left( \left[ -\frac{\pi}{2},+\frac{\pi}{2} \right] \right)## given as
$$
a(x)=11\sin(x) + 8\cos(x) \; , \;b(x)=4\sin(x) + 13\cos(x)
$$
Calculate the angle ##\varphi = \sphericalangle (a,b)## between the two vectors.​

9. (solved by @Couchyam) Let ##\varepsilon_k :=\begin{cases}1&,\text{ if the decimal representation of }k\text{ has no digit }9\\0&, \text{ otherwise }\end{cases}##
Show that ##\sum_{k=1}^\infty \dfrac{\varepsilon_k}{k}## converges.​
10. (solved by @Periwinkle ) Let ##x_0\in [a,b]\subseteq \mathbb{R}## and ##f\, : \,[a,b]\longrightarrow \mathbb{R}## continuous and differentiable on ##[a,b]-\{x_0\}##.
Furthermore exists the limit ##c:=\lim_{x \to x_0}f\,'(x)\,.## Then ##f(x)## is differentiable in ##x_0## with ##f\,'(x_0)=c\,.##​
Proof: Let ##x\in [a,b]-\{x_0\}##. According to the mean value theorem for differentiable functions there is a​
$$
\xi(x) \in (\min\{x,x_0\},\max\{x,x_0\})
$$
with ##f\,'(\xi(x))=\dfrac{f(x)-f(x_0)}{x-x_0}\,.## Because ##\lim_{x \to x_0}\min\{x,x_0\}=\lim_{x \to x_0}\max\{x,x_0\}=x_0## we must have ##\lim_{x \to x_0}\xi(x)=x_0## and by assumption ##\lim_{x \to x_0}f\,'(\xi(x))=c##, hence ##\lim_{x \to x_0}\dfrac{f(x)-f(x_0)}{x-x_0}=c##.​
What has to be regarded in this proof, and is there a way to avoid this hidden assumption?​


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11. A house ##H## and a rosary ##R## are near a circular lake ##L##.​
The Gardener walks with two watering cans from the house to the lake, fills the cans and goes to the rosary. We assume ##\overline{HR}\cap L=\emptyset##.​
At which point ##S## of the shore does he have to get water, so that his path length is minimal, and why?​
12. How long is the distance on a direct flight from London to Los Angeles and where is its most northern point?​
How long will it last by an assumed average speed of ##494## knots over ground? We neglect the influence of weather, esp. wind.​
We take the values ##51°\,28'\,39''\,N,\,0°\,27'\,41''\,W## for LHR in London,​
##\,33°\, 56'\,33''\,N,\,118°\,24'\,29''\,W## for LAX in Los Angeles, and a radius of ##3,958## miles for earth.​
13. Trial before an American district court.​
The witness claims he saw a blue cab drive off after a night accident. The judge decides to test the reliability of the witness.​
Result: The witness recognizes the color correctly in the dark in ##80\%## of all cases.​
A survey also found that ##85\%## of taxis in the city are green and ##15\%## are blue.​
With which probability has the taxi actually been blue?​
14. A monk climbs a mountain.​
He starts at ##8\,##a.m. on ##1000\,##m above sea level and reaches the peak at ##8\,##p.m. at ##3000\,##m.​
After a bivouac on top of the mountain, he returns to the valley the next morning and again starts at ##8\,##a.m. and returns at ##8\,##a.m.​
a. If he wants to avoid being at the same time of day at the same place as the day before when he climbed upwards, which strategy must he use downwards, and why?​
b. Assume he climbed at a rate of height ##u(t)## proportional to the square root of time, determine his path dependent on hourly noted time ##t##.​
c. Assume he follows the same path downwards and the height of his path is given by ##d_1(t)= \dfrac{125}{9}(t-20)^2+1000## in the first three hours and ##d_2(t)=-125\,t+3500## for the rest of his way. When will he be at the same point as the day before and at which height.​
15. I'm annoyed by my two new alarm clocks. They both are powered by the grid.​
One leaps two minutes an hour and the other one runs a minute an hour too fast. Yesterday I took the effort and set them to the correct time. This morning, I assume there was a power loss, one clock showed exactly ##6\,a.m## while the other one showed ##7\,a.m.##​
When did I set the clocks and how long did they run?​
 
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  • #2
mathwonk
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For #5, if acceptable, I offer a definition of the Zariski topology: a set V is Zariski closed if and only if there is a collection of polynomials {fj} such that V consists precisely of those points p at which all polynomials fj in the collection are zero.

I also ask a clarification of the last part of the problem: does it mean
i) decide whether each open set is in fact a finite union of such sets D(f)?
or ii) decide when a finite number of D(f) suffice as the whole basis?
 
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  • #3
fresh_42
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I also ask a clarification of the last part of the problem: does it mean
i) decide whether each open set is in fact a finite union of such sets D(f)?
or ii) decide when a finite number of D(f) suffice as the whole basis?
The first option: Decide whether each open set is in fact a finite union of such sets D(f).
(added and corrected)
 
  • #4
10 -> Is ##\max\{x_0\}## a typo?
 
  • #6
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Question 7, how can ##abc>0 \implies a>0, \, b>0, \, c>0## (we can have two negatives and one positive)? Or are the three propositions one big proposition?
 
  • #7
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Question 7, how can ##abc>0 \implies a>0, \, b>0, \, c>0## (we can have two negatives and one positive)? Or are the three propositions one big proposition?
The commata have been logical ANDs. I substituted them now by ##\wedge ##.
##abc > 0 ## alone isn't sufficient.
 
  • #8
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The commata have been logical ANDs. I substituted them now by ##\wedge ##.
##abc > 0 ## alone isn't sufficient.
Hence me asking, thank you!
 
  • #9
Periwinkle
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The ##\mathcal{A}_\sigma(\mathcal{P}_n)## ##\sigma##- algebra containing the sets of ##\mathcal{P}_n := \{\,\{2\},\{4\},\ldots,\{2n\}\,\}##, consists of a finite number of elements. All the unions of the above sets, their complement, the empty set and ##\mathbb{N}##.

However, the set ##\mathcal{P}_n := \{\,\{1\},\{3\},\ldots,\{2n+1\},\ldots\,\}## is not an element for ##\mathcal{A}_\sigma(\mathcal{P}_n)##, nor is it an element of the union of all ##\mathcal{A}_\sigma(\mathcal{P}_n)## sets.

Nevertheless, ##\mathcal{P}_n := \{\,\{1\},\{3\},\ldots,\{2n+1\},\ldots\,\}## is the element of the ##\sigma##-algebra generated by the set ##\bigcup_{n\in \mathbb{N}}\mathcal{A}_\sigma(\mathcal{P}_n)##, because it is complement to the the following set: ##\{\{2\},\{4\},\ldots,\{2n\},\ldots\}##.
 
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  • #10
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I think "##\implies##" is trivial.
Suppose ##a+b+c>0 \wedge ab+ac+bc>0 \wedge abc>0##
1) Suppose ##(a<0 \wedge b>0 \wedge c>0) \vee (a<0 \wedge b<0 \wedge c<0)## then ##abc<0## therefore we can't have only one or three negative reals.
2) Suppose ##a<0 \wedge b<0 \wedge c>0## then from ##a+b+c>0## we have ##c>a+b## and from ##ab+ac+bc>0## we have ##c(a+b)>ab## but ##c(a+b)<0 \wedge ab>0##, contradiction and therefore we can't have two negative numbers.
Thus ##a>0 \wedge b>0 \wedge c>0##
 
  • #11
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"Question 3"
The ##\mathcal{A}_\sigma(\mathcal{P}_n)## ##\sigma##- algebra containing the sets of ##\mathcal{P}_n := \{\,\{2\},\{4\},\ldots,\{2n\}\,\}##, consists of a finite number of elements.
The question was: which elements, i.e. which sets belong to ##\mathcal{A}_\sigma(\mathcal{P}_n)## explicitly listed.
All the unions of the above sets, their complement, the empty set and ##\mathbb{N}##.
This is almost true. It is all countably many unions and the definition of a sigma algebra. It doesn't answer the question.
However, the set ##\mathcal{P}_n := \{\,\{1\},\{3\},\ldots,\{2n+1\},\ldots\,\}## is not an element for ##\mathcal{A}_\sigma(\mathcal{P}_n)## ...
Right, but why? Isn't it the complement of all even numbers?
... nor is it an element of the union of all ##\mathcal{A}_\sigma(\mathcal{P}_n)## sets.
This isn't requested. We have a fixed ##n##.
Nevertheless, ##\mathcal{P}_n := \{\,\{1\},\{3\},\ldots,\{2n+1\},\ldots\,\}## is the element of the ##\sigma##-algebra generated by the set ##\bigcup_{n\in \mathbb{N}}\mathcal{A}_\sigma(\mathcal{P}_n)##, because it is complement to the the following set: ##\{\{2\},\{4\},\ldots,\{2n\},\ldots\}##.
The ##\sigma##-algebra generated by the set ##\bigcup_{n\in \mathbb{N}}\mathcal{A}_\sigma(\mathcal{P}_n)## is ##\mathcal{P}(\mathbb{N})##.
 
  • #12
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"Question 7"
I think "##\implies##" is trivial.
Suppose ##a+b+c>0 \wedge ab+ac+bc>0 \wedge abc>0##
1) Suppose ##(a<0 \wedge b>0 \wedge c>0) \vee (a<0 \wedge b<0 \wedge c<0)## then ##abc<0## therefore we can't have only one or three negative reals.
2) Suppose ##a<0 \wedge b<0 \wedge c>0## then from ##a+b+c>0## we have ##c>a+b## and from ##ab+ac+bc>0## we have ##c(a+b)>ab## but ##c(a+b)<0 \wedge ab>0##, contradiction and therefore we can't have two negative numbers.
Thus ##a>0 \wedge b>0 \wedge c>0##
Where have your minuses gone? From ##a+b+c>0## we have ##c>-a-b = |a|+|b|## and we have ##0>c(a+b)>-ab##.

Hint: This question can be solved without considering any cases!
 
  • #13
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Where have your minuses gone? From ##a+b+c>0## we have ##c>-a-b = |a|+|b|## and we have ##0>c(a+b)>-ab##.

Hint: This question can be solved without considering any cases!
Well, both ##a## and ##b## are negative, and we supposed the sum is positive, so ##c## must be greater than ##a+b##.
 
  • #14
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Well, both ##a## and ##b## are negative, and we supposed the sum is positive, so ##c## must be greater than ##a+b##.
O.k., but why is ##c(a+b) > ab\,##?
 
  • #15
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O.k., but why is ##c(a+b) > ab\,##?
I have made a mistake there, my bad.
 
  • #16
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I have made a mistake there, my bad.
No problem. As a hint: if there are symmetries, and here we have a few, then it is always a good idea to consider whether they can be used.
 
  • #17
Periwinkle
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The question was: which elements, i.e. which sets belong to ##\mathcal{A}_\sigma(\mathcal{P}_n)## explicitly listed.

This is almost true. It is all countably many unions and the definition of a sigma algebra. It doesn't answer the question.

Right, but why? Isn't it the complement of all even numbers?

This isn't requested. We have a fixed ##n##.

The ##\sigma##-algebra generated by the set ##\bigcup_{n\in \mathbb{N}}\mathcal{A}_\sigma(\mathcal{P}_n)## is ##\mathcal{P}(\mathbb{N})##.

Thank you very much for your guidance. I wrote late at night yesterday, and that was the first question I tried to solve.

---------------------------------------------------------------------------------------------------------------------------------------------

It is known in the algebra of sets if we take the resolution of ##\Omega##, ie sets ##\mathcal{A}_1, \mathcal{A}_2, \dots, \mathcal{A}_n##, that ## \Omega = \mathcal{A}_1\cup \mathcal{A}_2 \cup \dots \cup \mathcal{A}_n ##, and ## \mathcal{A}_i \cap \mathcal{A}_j= \text{Ø} ## for ## i \neq j##, then the algebra generated by the resolution is finite and consists of the ## \text{Ø} ## empty set and sets of the following types: ##\mathcal{A}_{i_1}\cup \mathcal{A}_{i_2} \cup, \dots, \cup \mathcal{A}_{i_n}##.

The above algebra of sets elements can be listed by induction as follows:

First level

## \mathcal{A}_1, \mathcal{A}_2, \dots , \mathcal{A}_n##.​

If the ##k##-th level is known, then the ##k+1##-th level is produced as follows

Each ##\mathcal{A}_{i_1}\cup \mathcal{A}_{i_2} \cup \dots \cup \mathcal{A}_{i_n}## set of the ##k##-th level is supplemented by each ##\mathcal{A}_i## (but only one at a time) where ##i## is greater than ## i_1, i_2, \dots, i_n ## each.​

Applying the above general theorem to ##n + 1## to the specific question, we get the ##\mathcal{A}_\sigma(\mathcal{P}_n) ## ##\sigma ##-algebra:

## \Omega = \mathbb{N}, \mathcal{A}_1 = \{2\}, \mathcal{A}_2 =\{4\}, \dots , \mathcal{A}_n =\{2n\}, \mathcal{A}_{n+1} = \complement \{\{2\},\{4\},\ldots,\{2n\}\,\} ## complement set.

In this case, the ##\sigma##-algebra is the same as the algebra because it contains finite many elements.

The set ## \{\,\{1\},\{3\},\ldots,\{2n+1\},\ldots\,\}## is not an element for ##\mathcal{A}_\sigma(\mathcal{P}_n)##, because if an element of ##\mathcal{A}_\sigma(\mathcal{P}_n)## does not contain the above ##\mathcal{A}_{n+1}##, it consists of a finite number of natural numbers; but if it contains, it also contains all natural numbers that are greater than an ##2n##. Nor is it an element of the union of all ##\mathcal{A}_\sigma(\mathcal{P}_n)## sets.

Therefore ##\bigcup_{n\in \mathbb{N}}\mathcal{A}_\sigma(\mathcal{P}_n)## cannot be ##\sigma##-algebra because if it were, it would contain the complement of the ##\{2\}\cup\{4\}\cup\ldots \cup\{2n\},\ldots = \complement \{\{2\},\{4\},\ldots,\{2n\},\ldots\} ## set, which, however, as described above, does not contain.
 
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  • #18
Periwinkle
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## A=a+b+c ##
## B=ab+ac+bc ##
## C=abc ##

If one or three of ##a##, ##b## and ##c## are negative, or any of ##0##, then ## C\leq0##. If two are negative, let's say ##a## and ##b##. If ## A\geq0 ##, then ## |a|+ |b|\leq c ##.

But

##B= ab+(a+b)c=ab-(|a|+ |b|)c \leq ab- (|a|+ |b|)\cdot(|a|+ |b|)=##
##ab -|a|^2- |b|^2 -2|a||b|=-|a|^2- |b|^2 -|a||b|##

which is negative.
 
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  • #19
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First level

## \mathcal{A}_1, \mathcal{A}_2, \dots , \mathcal{A}_n##.​

If the ##k##-th level is known, then the ##k+1##-th level is produced as follows

Each ##\mathcal{A}_{i_1}\cup \mathcal{A}_{i_2} \cup \dots \cup \mathcal{A}_{i_n}## set of the ##k##-th level is supplemented by each ##\mathcal{A}_i## (but only one at a time) where ##i## is greater than ## i_1, i_2, \dots, i_n ## each.​
...
The question was: Determine ##\mathcal{A}_\sigma(\mathcal{P}_n)##.
This means, that ##n## is a fixed natural number and ##\mathcal{P}_n## generates the ##\sigma-##algebra over ##\Omega = \mathbb{N}##.

Therefore we have ##\mathcal{A}_\sigma(\mathcal{P}_n) = \{\,\emptyset, \mathbb{N}\,\} \cup \mathbb{P}(\mathcal{P}_n) \cup_B (\mathbb{N}-B)## where ##B\in \mathbb{P}(\mathcal{P}_n)## denotes the power set of ##\mathcal{P}_n\,.## The only difficulty is to describe the complement ##\mathbb{N}-B## in an appropriate explicit way.

An indirect proof for the second assertion is a good idea. This means we have to use that ##\bigcup_{n\in \mathbb{N}} \mathcal{A_\sigma(\mathcal{P}_n)}## is a ##\sigma-##algebra. I'm not quite sure where you used this. (My suggestion is to use the complement of the set of odd numbers. It is the set of all even numbers and as countable union of ##\{\,2k\,\}## it has to be in that union. But why can't this be?)
 
  • #20
Ben2
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In #10, suppose x < x_0. The mean value theorem in Stewart's Calculus (Section 3.2) requires that f is continuous on [x, x_0] and differentiable on (x, x_0). One could replace x_0 with x_0 - \epsilon, but to me it's not obvious that the resulting difference quotient has the same limit at x_0 as the one desired. The problem seems to ask for a proof that the derivative is continuous at x_0.
 
  • #21
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## A=a+b+c ##
## B=ab+ac+bc ##
## C=abc ##

If one or three of ##a##, ##b## and ##c## are negative, or any of ##0##, then ## C\leq0##. If two are negative, let's say ##a## and ##b##. If ## A\geq0 ##, then ## |a|+ |b|\leq c ##.

But

##B= ab+(a+b)c=ab-(|a|+ |b|)c \leq ab- (|a|+ |b|)\cdot(|a|+ |b|)=##
##ab -|a|^2- |b|^2 -2|a||b|=-|a|^2- |b|^2 -|a||b|##

which is negative.
Looks good.

Here is the solution I had in mind:

##(x-a)(x-b)(x-c) = x^3 - (a+b+c)x^2 + (ab+ac+bc)x -abc < 0## for all ##x \leq 0## so the zeroes of this polynomial, ##a,b,c## are all in ##x \in (0,\infty)##.
 
  • #22
Periwinkle
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The question was: Determine ##\mathcal{A}_\sigma(\mathcal{P}_n)##.
This means, that ##n## is a fixed natural number and ##\mathcal{P}_n## generates the ##\sigma-##algebra over ##\Omega = \mathbb{N}##.

Therefore we have ##\mathcal{A}_\sigma(\mathcal{P}_n) = \{\,\emptyset, \mathbb{N}\,\} \cup \mathbb{P}(\mathcal{P}_n) \cup_B (\mathbb{N}-B)## where ##B\in \mathbb{P}(\mathcal{P}_n)## denotes the power set of ##\mathcal{P}_n\,.## The only difficulty is to describe the complement ##\mathbb{N}-B## in an appropriate explicit way.

I meant that the elements of the finite ##\sigma##-algebra should be listed. This is not an enumeration but a general definition in other ways. My method is an algorithmic listing.

An indirect proof for the second assertion is a good idea. This means we have to use that ##\bigcup_{n\in \mathbb{N}} \mathcal{A_\sigma(\mathcal{P}_n)}## is a ##\sigma-##algebra. I'm not quite sure where you used this. (My suggestion is to use the complement of the set of odd numbers. It is the set of all even numbers and as countable union of ##\{\,2k\,\}## it has to be in that union. But why can't this be?)

I used it when I said that the set ##\{2\}\cup\{4\}\cup\ldots \cup\{2n\},\ldots## is an element of the ##\sigma##-algebra because it is the countable union of elements of finite algebras.
 
  • #23
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In #10, suppose x < x_0. The mean value theorem in Stewart's Calculus (Section 3.2) requires that f is continuous on [x, x_0] and differentiable on (x, x_0). One could replace x_0 with x_0 - \epsilon, but to me it's not obvious that the resulting difference quotient has the same limit at x_0 as the one desired. The problem seems to ask for a proof that the derivative is continuous at x_0.
Quick tips, put your math ##here## for it to be inline, ##\int_a^{b}dt## or $$here$$ for it to be like this $$\int_a^{b}dt$$
 
  • #24
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I meant that the elements of the finite ##\sigma##-algebra should be listed. This is not an enumeration but a general definition in other ways. My method is an algorithmic listing.
O.k. just for the records:
\begin{align*}
\mathcal{A}_\sigma(\mathcal{P}_n)&=\{\,\emptyset , \mathbb{N}\,\} \cup \{\,B\subseteq \mathbb{N}\,:\,B\subseteq \{2,4,,\ldots ,2n\}\,\}\\
&\phantom{{}={}} \cup \{\,B\subseteq \mathbb{N}\,:\,2k\in B \,\forall \,k>n\,\wedge\,2k-1\in B\,\forall\,k\in \mathbb{N}\,\}
\end{align*}
I used it when I said that the set ##\{2\}\cup\{4\}\cup\ldots \cup\{2n\},\ldots## is an element of the ##\sigma##-algebra because it is the countable union of elements of finite algebras.
Yes, but you said that the complement you constructed contains all even numbers greater than ##2n##. Now what if a set didn't contain any even number, why isn't its complement of all odd numbers not in the union as well? With this way of proof you end up at the point whether ##n=0##, i.e. ##\mathcal{P_n}=\emptyset## is a possibility or not, or whether the complement should be build according to this empty ##\mathcal{P}## or the natural numbers as a whole.

It is easier to say: ##B_k := \{\,2,4,\ldots,2k\,\} \in \mathcal{P}_k## and so are ##\cup_k B_k \in \cup_n\mathcal{A}_\sigma(\mathcal{P}_n)##. But as ##\cup_k B_k ## isn't an element of any ##\mathcal{A}_\sigma(\mathcal{P}_n)##, it isn't an element of the union either.
 
  • #25
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In #10, suppose x < x_0. The mean value theorem in Stewart's Calculus (Section 3.2) requires that f is continuous on [x, x_0] and differentiable on (x, x_0). One could replace x_0 with x_0 - \epsilon, but to me it's not obvious that the resulting difference quotient has the same limit at x_0 as the one desired. The problem seems to ask for a proof that the derivative is continuous at x_0.
Your on the right track and it has to do with ##x##. But the proof is technically correct. It is the same limit, since ##\xi## is chosen accordingly. It simply uses an assumption which is not explicitly mentioned, which one?
 
  • #26
Periwinkle
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Make the following assignment of ##0, 1, 2, 3 \dots ## natural numbers (the examples show the way of assignment):

## ~~~~~~~~~~~~~~~~~~ 3 ~ \to 10 \times 0.3 ##
## ~~~~~~~~~~~~~~~~~~ 21 ~ \to 10 \times 0.21 \times \frac 1 {10} ##
## ~~~~~~~~~~~~~~~~~~ 529 ~ \to 10 \times 0.529 \times \frac 1 {10^2} ##
## ~~~~~~~~~~~~~~~~~~ 4791 ~ \to 10 \times 0.4791 \times \frac 1 {10^3} ##
and so on.

This assignment assigns to all natural numbers ##k## a real number which is greater than or equal to ##\dfrac{\varepsilon_k}{k}##.

However, if in each line we add all the real numbers that are possible on the right, we get the amounts smaller than

## ~~~~~~~~~~~~~~~~~~ 10 \times 9 ##
## ~~~~~~~~~~~~~~~~~~ 10 \times 9^2 \times \frac 1 {10}##
## ~~~~~~~~~~~~~~~~~~ 10 \times 9^3 \times \frac 1 {10^2}##
## ~~~~~~~~~~~~~~~~~~ 10 \times 9^4 \times \frac 1 {10^3}##
and so on.

However, if they are added, we get

## ~~~~~~~~~~~~~~~~~~ 10 \times 9 \times (1 + \frac 9 {10} + \frac {9^2} {10^2} + \frac {9^3} {10^3}+ \dots)##

what is finite.
 
  • #27
fresh_42
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Make the following assignment of ##0, 1, 2, 3 \dots ## natural numbers (the examples show the way of assignment):

## ~~~~~~~~~~~~~~~~~~ 3 ~ \to 10 \times 0.3 ##
## ~~~~~~~~~~~~~~~~~~ 21 ~ \to 10 \times 0.21 \times \frac 1 {10} ##
## ~~~~~~~~~~~~~~~~~~ 529 ~ \to 10 \times 0.529 \times \frac 1 {10^2} ##
## ~~~~~~~~~~~~~~~~~~ 4791 ~ \to 10 \times 0.4791 \times \frac 1 {10^3} ##
and so on.

This assignment assigns to all natural numbers ##k## a real number which is greater than or equal to ##\dfrac{\varepsilon_k}{k}##.

However, if in each line we add all the real numbers that are possible on the right, we get the amounts smaller than

## ~~~~~~~~~~~~~~~~~~ 10 \times 9 ##
## ~~~~~~~~~~~~~~~~~~ 10 \times 9^2 \times \frac 1 {10}##
## ~~~~~~~~~~~~~~~~~~ 10 \times 9^3 \times \frac 1 {10^2}##
## ~~~~~~~~~~~~~~~~~~ 10 \times 9^4 \times \frac 1 {10^3}##
and so on.

However, if they are added, we get

## ~~~~~~~~~~~~~~~~~~ 10 \times 9 \times (1 + \frac 9 {10} + \frac {9^2} {10^2} + \frac {9^3} {10^3}+ \dots)##

what is finite.
Sorry, but I don't understand this. First, can you describe the assignment properly? You divided your numbers by powers of ten until their sum is finite? That doesn't prove anything, since this scaling is rarely an allowed transformation. I assume that it has to do with the quotient ##\frac{1}{k}## but you shouldn't leave the guesses to your readers.

Hint: Simply count the numbers which count of a certain length, and add over all lengths.

If you allow me to give an advice, I'd say: I think you should work on your presentations. It would be a pity if your ideas, which are good, will get lost in a mess of words. A good measure is to ask oneself: Will I be able to read and understand this a year from now?

We also have a lot of young students who read our threads even if they might not understand everything. For those it is better to write answers step by step, so one can easily follow them. This series is a good example, because it should be understandable for high schoolers as well as students of mathematics. It is not too complicated to give a formal proof.
 
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  • #28
TeethWhitener
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Looks good.

Here is the solution I had in mind:

##(x-a)(x-b)(x-c) = x^3 - (a+b+c)x^2 + (ab+ac+bc)x -abc < 0## for all ##x \leq 0## so the zeroes of this polynomial, ##a,b,c## are all in ##x \in (0,\infty)##.
How does this proof ensure that ##a, b, c## are real?
 
  • #29
fresh_42
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How does this proof ensure that ##a, b, c## are real?
They are given as real numbers, or at least integers, since I assumed implicitly an Archimedean ordering.
That the zeroes of the polynomial are real is given by the fact that ##p(a)=p(b)=p(c)=0## are the known zeroes.
 
  • #30
TeethWhitener
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They are given as real numbers, or at least integers, since I assumed implicitly an Archimedean ordering.
But for example if ##a=i, b=-i, c\in \mathbb{R}##, then $$a+b+c=c\in \mathbb{R}$$
$$ab+bc+ca = 1\in \mathbb{R}$$
$$abc=c\in \mathbb{R}$$
So the fact that those three expressions are in ##\mathbb{R}## is a necessary, but not sufficient condition (by itself) for ##a, b, c\in \mathbb{R}##.
 
  • #31
Periwinkle
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Your on the right track and it has to do with ##x##. But the proof is technically correct. It is the same limit, since ##\xi## is chosen accordingly. It simply uses an assumption which is not explicitly mentioned, which one?


I've been thinking so much about this question. If we want to be so precise, then the conditions of the theorem could be completed by that ## a \lt b##.
 
  • #32
fresh_42
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I've been thinking so much about this question. If we want to be so precise, then the conditions of the theorem could be completed by that ## a \lt b##.
No, it's more subtle than this and uses a 'tool' you wouldn't have expected in calculus.
 
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  • #33
Periwinkle
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No, it's more subtle than this and uses a 'tool' you wouldn't have expected in calculus.

Then this is just the Axiom of Choice. Then I'll explain it.
 
  • #34
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Then this is just the Axiom of Choice. Then I'll explain it.
Yes, ##\xi## is selected from an interval which depends on ##x##, and since this is done for all ##x## we have silently used a selection function ##x \longmapsto \xi(x)##.

Do you also have an idea how it can be avoided?
 
  • #35
Periwinkle
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Yes, ##\xi## is selected from an interval which depends on ##x##, and since this is done for all ##x## we have silently used a selection function ##x \longmapsto \xi(x)##.

Do you also have an idea how it can be avoided?


The question described above is written on pages 74-77 of this book. The reverse is questionable.
 

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