## 502 – Infinite well-ordered sets

October 9, 2009

I just want to record the Exercise I mentioned in class:

Suppose that $(A,<)$ is an infinite well-ordered set, and that $B\subseteq A.$ Show that there is a bijection between $A$ and the disjoint union $A\sqcup B.$

To be explicit, I want a proof that makes no use of the axiom of choice. Also, although I am not requiring this as an exercise, recall that the point is to use this result to complete the proof of the following:

Theorem. If $(A,<)$ is a well-ordered set, then there is a well-ordered set $(B,\prec)$ such that $(A,<)$ is a proper initial segment of $(B,\prec)$ and there is no injection from $B$ into $A.$

## 175 – Integrating products of secants and tangents

October 9, 2009

(In what follows, I will write ${\tan^n x}$ for ${(\tan x)^n}$ and ${\sec^m x}$ for ${(\sec x)^m.}$)

Recall that

$\displaystyle \tan x=\frac{\sin x}{\cos x}\quad\mbox{ and }\quad\sec x=\frac1{\cos x},$

that

$\displaystyle (\tan x)'=\sec^2x\quad\mbox{ and }\quad(\sec x)'=\sec x\cdot\tan x,$

and that

$\displaystyle \sec^2x=\tan^2x+1.$

The formulas below make use of these identities repeatedly.

We want a series of methods and reduction formulas that allow us to evaluate any expression of the form

$\displaystyle \int \sec^m x\cdot \tan^n x\,dx,$

for ${m}$ and ${n}$ integers, ${m,n\ge0.}$

## 502 – Ultraproducts of finite sets

October 2, 2009

I want to sketch here the proof that if ${(M_n\mid n\in{\mathbb N})}$ is a sequence of finite nonempty sets, and ${\lim_n |M_n|=\infty,}$ then ${\prod_nM_n/{\mathcal U}}$ has size ${|{\mathbb R}|}$ for any nonprincipal ultrafilter ${{\mathcal U}}$ on ${{\mathbb N}.}$

The argument I present is due to Frayne, Morel, Scott, Reduced direct products, Fundamenta Mathematica, 51 (1962), 195–228.

The topic of the size of ultraproducts is very delicate and some open questions remain. For ultraproducts of finite structures, this is continued in Keisler, Ultraproducts of finite sets, The Journal of Symbolic Logic, 32 (1967), 47–57, and finally in Shelah, On the cardinality of ultraproduct of finite sets, The Journal of Symbolic Logic, 35 (1) (Mar., 1970), 83–84. Shelah shows that if an ultraproduct of finite sets is infinite, say of size ${\kappa,}$ then ${\kappa^{\aleph_0}=\kappa.}$ His argument is a very nice application of non-standard analysis. The case that interests us is easier.

Clearly,

$\displaystyle |\prod_nM_n/{\mathcal U}|\le|\prod_n M_n|\le|{\mathbb N}^{\mathbb N}|=|{\mathbb R}|,$

so it suffices to show that ${|{\mathbb R}|\le|\prod_nM_n/{\mathcal U}|.}$

## 175 – Quiz 3

October 2, 2009

Here is quiz 3.

Problem 1 is Exercise 9.3.16 from the book. Here is a graph showing the cardioid and the circle. The easiest way to compute the required area is by subtracting the area of the circle from that of the cardioid. (Note that the circle is completely contained in the cardioid as, for each ${ \theta\in[0,\pi/2],}$ we have that ${0<\cos\theta< 1+\cos\theta.}$ This shows that in the portion of the circle in the first quadrant is within the portion of the cardiod there. The same holds in the fourth quadrant by symmetry. Of course, all of this also follows directly from the graph.)

The area of the cardioid is given by ${ \displaystyle\int_0^{2\pi}\frac12(1+\cos\theta)^2\,d\theta}$ and the area of the circle is ${ \displaystyle\int_0^\pi \frac12(\cos\theta)^2\,d\theta}$ (note that as ${ \theta}$ varies from ${ 0}$ to ${ \pi}$ the whole circle is traveled once).

By symmetry, ${ \displaystyle\int_0^{2\pi}\frac12(1+\cos\theta)^2\,d\theta= \int_0^{\pi}(1+\cos\theta)^2\,d\theta,}$ so the area we want is given by

$\displaystyle \int_0^{\pi}\left((1+\cos\theta)^2-\frac12(\cos\theta)^2\right)\,d\theta$ $\displaystyle =\int_0^\pi\left(\frac{\cos^2\theta}2+2\cos\theta+1\right)\,d\theta.$

Recall that ${ \displaystyle\cos^2\theta=\frac{1+\cos2\theta}2,}$ so the integral is ${ \displaystyle\int_0^\pi\left(\frac{1+\cos2\theta}4+2\cos\theta+1\right)\,d\theta.}$

Now note from the Cartesian graph of ${ \cos\theta}$ that ${ \displaystyle\int_0^\pi\cos2\theta\,d\theta=\int_0^\pi\cos\theta\,d\theta=0,}$ so the area we want is just ${ \displaystyle\int_0^\pi\frac14+1\,d\theta=\frac54\pi.}$

Problem 2 is Exercise 9.5.50 from the book. Recall that the standard polar equation of a line not going through the origin is given by

$\displaystyle r = \frac{r_0}{\cos(\theta-\theta_0)},$

where ${ r_0}$ is the distance from the line to the origin, and ${ \theta_0}$ is the angle of the point in the line that realizes this distance, i.e., ${ (r_0,\theta_0)}$ are the polar coordinates of the point on the line closest to the origin.

There are at least two ways we can proceed:

1) We can directly find the distance from the line to the origin. To do this, we recall that the line that goes through the origin and is perpendicular to ${ y=mx+b,}$ where ${ m\ne0,}$ is given by ${ \displaystyle y=-\frac1m x.}$ The point in ${ y=mx+b}$ closest to the origin is the intersection of these two lines. In the case that interests us, this is the intersection of ${ y={\sqrt3}x-1}$ and ${ \displaystyle y=-\frac1{\sqrt3} x,}$ so ${ \displaystyle{\sqrt3} x -1=-\frac1{\sqrt3} x}$ or ${ \displaystyle\left({\sqrt3}+\frac1{\sqrt3}\right)x=1,}$ or ${ \displaystyle x=\frac{\sqrt3}4,}$ so ${ \displaystyle y=-\frac1{\sqrt3} x=-\frac14.}$ We have found that the point ${ \displaystyle\left(\frac{\sqrt3}4,-\frac{1}4\right)}$ is closest in ${ y=\sqrt3-1}$ to the origin. Its distance is ${ r_0=\sqrt{x^2+y^2}=2/4=1/2.}$ Its angle ${ \theta_0}$ is on the fourth quadrant and satisfies ${ \tan\theta_0=1/\sqrt3,}$ i.e., ${ \theta_0=-\pi/6.}$

Putting this together, ${ \displaystyle r=\frac{1/2}{\cos(\theta+\pi/6)}}$ is the desired equation.

2) The other method we saw in lecture is based in the fact that

$\displaystyle \cos(\theta-\theta_0)=\cos\theta_0\cos\theta+\sin\theta_0\sin\theta,$

and consists in first writing the Cartesian equation of the line directly in polar coordinates, and then using the resulting expression to find ${ r_0}$ and ${ \theta_0:}$ We have ${ y={\sqrt 3} x-1,}$ so ${ r\sin\theta=\sqrt 3 r\cos\theta-1,}$ or ${ r(\sqrt 3\cos\theta-\sin\theta)=1.}$ This gives us

$\displaystyle r=\frac 1{\sqrt3 \cos\theta-\sin\theta}.$

We find a constant ${ k}$ such that ${ k\sqrt3}$ and ${ -k}$ are the cosine and the sine of some angle, precisely ${ \theta_0.}$ We then have ${ k\sqrt3\cos\theta-k\sin\theta=\cos\theta_0\cos\theta+\sin\theta_0\sin\theta=\cos(\theta-\theta_0).}$

To find this ${ k,}$ we use that ${ \cos^2\alpha+\sin^2\alpha=1}$ for any ${ \alpha,}$ so ${ k}$ must satisfy ${ {k^2}3+k^2=1,}$ or ${ k^2=1/4,}$ so ${ k=1/2.}$ The value of ${ \theta_0}$ such that ${ \cos\theta_0={\sqrt3}/2}$ and ${ \sin\theta_0=-1/2}$ is ${ \theta_0=-\pi/6.}$ This means that the equation we are looking for is

$\displaystyle r=\frac k{k\sqrt3 \cos\theta-k\sin\theta}=\frac{1/2}{\cos(\theta+\pi/6)},$

just as with the previous method.

Typeset using LaTeX2WP.

## 502 – Notes on compactness

October 1, 2009

The goal of this note is to present a proof of the following fundamental result. A theory ${T}$ is said to be satisfiable iff there is a model of ${T.}$

Theorem 1 (Compactness) Let ${T}$ be a first order theory. Suppose that any finite subtheory ${T_0\subseteq T}$ is satisfiable. Then ${T}$ itself is satisfiable.

As indicated on the set of notes on the completeness theorem, compactness is an immediate consequence of completeness. Here I want to explain a purely semantic proof, that does not rely on the notion of proof.

The argument I present uses the notion of ultraproducts. Although their origin is in model theory, ultraproducts have become an essential tool in modern set theory, so it seems a good idea to present them here. We will require the axiom of choice, in the form of Zorn’s lemma.

The notion of ultraproduct is a bit difficult to absorb the first time one encounters it. I recommend working out through some examples in order to understand it well. Here I confine myself to the minimum necessary to make sense of the argument.

## Why dose nobody ask the students

October 1, 2009

From Tampa Bay Times, October 1, 2009, page T10.

A student’s letter to tbt*

Editor’s note: On Monday, tbt* published a story describing President Obama’s desire to increase the number of days U.S. children go to school each year. A junior at a Pasco high school e-mailed the following response — with the subject line “why dose nobody ask the students’’ — to tbt*. It is printed here, unedited:

Dear “tbt” editor of which it may concern I’m a student of pasco coun ty a junior to be precise and pleas do not mistake this for a Dear Abby seg ment I am not a 40 year old women concerned about her felines. I am just appalled as a member of the student body that in ever y in other words the only article I could find of president Obahmas plan to extend the school year, school days, and e ven rumored around school of mak ing us work weekends that they have not a single students opinion on the subject.

more so that they believe that the school boards “research” should stand as good evidence of the im provement in our “higher achieve ment”. a very intelligent man once said if you obser ve you have therefor changed the out come (now notice I didn’t abbreviate that phrase as I did the other stated phrases that’s because it’s not exact not as an in sult to any of the readers but some would use this against the students to further there pointless war against other country’s grade average).

at any rate not only do I belie ve it to be unreliable research but I also believe it dose not properly address to how our mind will adjust to the added stress. now personally I don’t ge t what in heavens name these peo ple they put in these articles have to do with the students spending more time in school it’s not like there stuck in the classroom ding the e xtra work the teachers prepare to compensate for the e xtra time. are they siting first row in this teenage bomb they’re building.

the plan reducing summer well first why don’t they come clean that they all ready reduced it we’re only on vacation for two months give or take a few days and for years they lied to our faces saying we have three months vaca.

to sum it all up I just wish he pub lic the pollutions the school board the reporters would address us on things that affect US not them now granted there may have been a seg ment on the news some were that addressed students about it but if there was did you even pay attention. and let me end this by saying the subduction of certain rights in order to maintain the learning enviorment making your usage of the 1 st amend ment right lesser then you would have in normal society is the same as removing them at the front step!

(I would like to stay anonymous)