Athens Math Circle

Posted by: Christian | 27/05/2011

Infinitude of Primes essay

A few days ago, I decided to rewrite in tex format a previous post about Fustenberg’s proof on the infinitude of prime numbers. This pdf after a number of corrections is ready and you can find it here Primes.

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Posted by: Christian | 27/05/2011

A Mathematical Gift, I

Looking for something interesting, I fell on a book by mathematicians from China. From a quick look, it made me great impression for the reason that these guys presented to high-schools of China advanced topics of mathematics like Euler Characteristic, the Poincare – Hopf theorem, surfaces in 3 and 4 dimensions. So, this post (and maybe a few next) will be some personal notes about these topics.

The Euler Characteristic

Let K be a figure of the plane. Using triangles of arbitrary size, someone can triangulate this figure according the next rule

the triangles must fit together along the edges
the triangles should not overlap each other
the vertex of a triangle cannot touch the edge of another triangle

Without loss of generality, we suppose that such a figure is a plane square and such triangulation (which obey the above rule) is

Triangulated Square

Before, we define the main notion of this post, let’s first make some considerations. After triangulate the figure, we have to compute three very important numbers, which numbers will be the key points for defining -in full generality- the Euler Characteristic. These numbers are

the number of the triangular faces, f
the number of edges, e
the number of vertices, v

Therefore, we define the Euler Characteristic of the figure K, denoted by $\chi (K)$ be the formula

$\chi (K) = f-e+v.$

After a few work, and using mathematical induction, one can prove that the Euler Characteristic of a polygon on the plane is equal to 1 (and hence of the square). At this point, we have to point out, that the Euler Characteristic, doesn’t depend on the triangulation of the figure. In other words, for a figure there exist a number of triangulations, and a number of different partitions of the triangles. But in every case, the Euler Characteristic doesn’t change.

Figures in space

To close this post, we pass to figures in the space. We consider again, without loss of generality, that such figures are the Sphere and the Torus ( in common, a doughnut) whose schemes are given below.

Sphere

Torus

Again, the triangulation of these figures follow the same rule as mentioned above, the Euler Characteristic is described by the same formula ( $\chi (K) = f-e+v$ ), and we have

$\chi (sphere) = 2$ and $\chi (torus) = 0.$

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Posted in Topology/Geometry

Posted by: Christian | 20/05/2011

Scrambled thoughts.

I know, that passed a lot of time from the last post. And unfortunately, my obligations didn’t let me to work over a new post… So, it’s time for some “scrambled” (what a rough pronunciation) thoughts…

First of all, after a small search for new blogs and adding some other to blogroll, I noticed that graphs is a mode around the world… Unfortunately, I hadn’t had the chance to attend a course about graphs at UoA and I think that the next topic I will try to study is sth about them…

Next, I consider that modern geometry stands around algebraic geometry… A large amount of blogs, examine topics around algebraic varietes, shemes, projective planes, algebraic notions like modules, ideals, topics very closed to this kind of geometry, not so familiar to me but very very interesting…

Last but not least, I was excited when I saw how a combinatorial theorem like Sauer-Shelah Lemma with a lot of set-theoreric notions, is used to approach a problem of Geometry of Banach spaces.. If the question, how we can do geometry on abstract spaces like Banach spaces has already arisen and to you, then the only (and for now acceptable by me) answer is the following, given by a professor of mine. In general words, a group of prestigious mathematicians who were working primary on the inequalities which arise on Banach spaces, tried to separate them from mathematicians who were working (and work until now) on linear programming, applications of convex sets, Hilbert spaces etc. For this reason, remained the name Geometry of Banach spaces and not just Banach spaces or sth else, although there isn’t any notion of classical geometry on Banach spaces…

Until next post, enjoy life and mathematics….

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Posted in Uncategorized

Posted by: Christian | 10/03/2011

The Infinitude of Primes via Topology

One of the most beautiful theorems in number theory, is the Infinitude of Primes, with the well known proof given by Euclid. Through the passage of the centuries, different proofs were presented, sometimes by famous mathematicians, like P.Erdos. At this post, we will try to present Furstenberg’s proof for this theorem, using topological methods. Before we are ready to prove our theorem, we have to do a few work. In particular, to define a topology on $\mathbb{Z}$ . Something neither so obvious nor so usual.

Topology on $\mathbb{Z}$

Let $\mathbb{Z}$ be the set of all integers and $\mathbb{P}$ the set of all primes. We call the set $\mathcal{O} \subset \mathbb{Z}$ open if $\mathcal{O}$ is the empty set, or if $\mathcal{O}$ contains all the integers of the form $a+b \mathbb{Z} = \{a+bn|n \in \mathbb{Z}\}$ for integers $a,b$ and $b>0$ .

The first step for our proof is done. We define in a set the open sets. Next, we will consider the family of all these open sets. By definition, a topology on a space X, denoted as $\mathcal{T}$ , is the family of all open sets of the space. Of course, there exist some properties of this family. In next lines, we will present the properties of topology, and in the same time, we will check that the open sets $\mathcal{O}$ which defined above define by themselves a topology on $\mathbb{Z}$ . So,

The empty set $\emptyset$ and $\mathbb{Z}$ belongs in $\mathcal{T}$ .
If $\mathcal{O}_1$ , $\mathcal{O}_2$ are open sets in $\mathbb{Z}$ , then $\mathcal{O}_1 \cap \mathcal{O}_2$ belongs in $\mathcal{T}$ .
If $\mathcal{O}_1$ , $\mathcal{O}_2$ are open sets $\mathbb{Z}$ , then $\mathcal{O}_1 \cup \mathcal{O}_2$ belongs in $\mathcal{T}$ .

As a result, the open sets $\mathcal{O}$ satisfy the properties for a topology and as a result they define a topology on $\mathbb{Z}$ . In addition, we have to underline two very important facts:

If $\mathcal{O}$ is open and non-empty, then it has order infinity ( $|\mathcal{O}|=\infty$ )
If $a,b \in \mathbb{Z}$ and $b>0$ , then the set $\{ a+b\mathbb{Z}\}$ is clopen (i.e. both open and close). To prove this, we note first of all that the union of open sets in a topology is also open and $a+b\mathbb{Z}=\mathbb{Z} \backslash (\cup^{b}_{i=1}(a+i)+b\mathbb{Z})$ . From this, and having in mind at the same time that the complement of an open set is close, we have the result.

Number-theoretic properties of $\mathbb{Z}$ .

Until now, we have define open sets on $\mathbb{Z}$ and a topology of this sets on $\mathbb{Z}$ again. Let’s return to the algebraic properties of $\mathbb{Z}$ . For every $n \in \mathbb{Z}\backslash \{-1,+1\}$ , $n$ is prime or has some prime divisor $p$ . Therefore, we can write the set of integers without $\pm 1$ as the union of $p\mathbb{Z}$ . In other words we have,

$\mathbb{Z}\backslash \{-1,+1\}=\cup_{p \in \mathbb{P}}0+p\mathbb{Z}$ .

Infinitude of Primes

Theorem. There are infinitely many prime numbers.

Proof. We suppose that $\mathbb{P}$ is finite. This means, that the set $\cup_{p \in \mathbb{P}}0+p\mathbb{Z}$ is also closed and as a result the set $\{-1,+1 \}$ is open, with order 2. But this contradicts with the first fact (that any open and non-empty set has order infinity). As a result, the set $\mathbb{P}$ is infinity.

q.e.d.

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Posted in Number Theory

Posted by: Christian | 16/01/2011

A naive introduction to Measure Theory

It was not in my plan such a post at the moment, but the decisions of the moment are sometimes very interesting. Well, the topic of this post is a so naive and so descriptive view to very first things of measure theory. Contributors of this theory was Paul Halmos, Kolmogorov, Caratheodory, Egorov, the Polish School of Mathematics, all great mathematicians. Let’s begin here our journey with objects which called Algebras. An algebra, is a family $\mathcal{A}$ of subsets of a set X with the following properties:

$X\in\mathcal{A}$
If $A\in\mathcal{A}$ , then the complement of $A$ , $X-A \in \mathcal{A}$
If $A_1, A_2,...,A_n \in \mathcal{A}, n\in \mathbb{N}$ , then $\cap^{n}_{i=1}A_{n}\in \mathcal{A}$

After that, we define a more general notion of algebra, this one of $\sigma -algebra$ . A $\sigma -algebra$ has the same properties like an algebra with a vey important difference. In the third property above we substitute the finite union by an infinity. So, a family $\mathcal{A}$ is called $\sigma-algebra$ if satisfy the following properties

$X\in\mathcal{A}$

If $A\in\mathcal{A}$ , then the complement of $A$ , $X-A \in \mathcal{A}$

If $A_1, A_2,...,A_n \in \mathcal{A}, n\in \mathbb{N}$ , then $\cap^{\infty}_{i=1}A_{n}\in \mathcal{A}$

It is obvious that a $\sigma-algebra$ is an algebra but the inverse doesn’t hold (Why?). So, until now we define two (not so different) families of sets. Next step, will be to define set-functions on this families. As a result, for a set X and a $\sigma-algebra$ $\mathcal{A}$ a function $\mu : \mathcal{A} \rightarrow [0,+\infty]$ is called measure, if it satisfy the following:

$\mu (\emptyset)=0$
$\mu(\cup ^{\infty}_{i=1}A_n)=\sum^{\infty}_{i=1}\mu(A_n)$ where ${A_{n}}$ is a sequence of disjoint sets

So, we define sets, set-functions on these sets, and after a lot of moments of thought Caratheodory, defined a class of special measures, the well-known, outer measures, which denoted by $\mu^{*}$ . What special have these set-functions? First of all, they defined on the powerset $\mathcal{P}(x)$ rather than $\mathcal{A}$ . The important thing is that the powerset is a rich, from a topological viewpoint, set. Secondly, outer measures are monotones. This means that, if $A \subset B \subset X$ , then $\mu ^{*}(A) \le \mu^{*} (B)$ . The other important thing we have to underline here for the outer measure is the countable subadditive. That means, that if ${A_n}$ is an infinity sequence of subsets of X, then $\mu^{*}(\cup_{n}A_{n}) \le \sum_{n}\mu^{*}(A_{n})$ . Of course, it is obvious that hold $\mu^{*}(\emptyset)=0$ .

To conclude, we have to underline that a measure, after a number of operations, it can be outer measure. From the other hand, an outer measure fails to be a measure, because of the fact that is subadditive.

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Posted in Analysis

Posted by: johngreek90 | 24/12/2010

A common mistake.

Let $f:A \mapsto \mathbb{R}$ a function and $\epsilon>0$ .

We say that $f$ is uniformly continuous if $\forall x,y \in A$ $\exists \delta(\epsilon)>0$ so that if $|x-y|<\delta$ , then $|f(x)-f(y)|<\epsilon$ .

We say that $f$ is Lipschitz continuous, if there exists $M>0$ so that $\forall x,y\in A$ $|f(x)-f(y)|\leqslant M|x-y|$ .

Theorem 1: Every function $f:A \mapsto \mathbb{R}$ which is Lipschitz continuous is uniformly continuous.

Proof: Since $f$ is Lipschitz continuous, there exists $M>0$ so that $\forall x,y\in A$ $|f(x)-f(y)|\leqslant M|x-y|$ .

We want to prove that $f$ is uniformly continuous: Let $\epsilon>0$ . We’ll find a $\delta>0$ so that if $|x-y|<\delta$ , then $|f(x)-f(y)|<\epsilon$ .

For $\delta=\epsilon/M$ , we have $|f(x)-f(y)|\leqslant M|x-y|<M\delta={M\epsilon}/M=\epsilon$ .

Theorem 2: Let $f:A \mapsto \mathbb{R}$ be a differentiable function. If $f'$ is bounded $\forall x\in A$ , then $f$ is Lipschitz continuous.

Proof: Since $f'$ is bounded, there exists a $M>0$ so that $|f'(x)|\leqslant M$ $\forall x \in A$ .

Now we will use the mean value theorem: if $x<y$ , $x,y\in A$ $\exists {\xi}_x \in (x,y)$ so that $f'(\xi)=\frac{f(y)-f(x)}{y-x}$ .

Hence we have $|f(x)-f(y)|=|f'(\xi)||x-y|\leqslant M|x-y|$ , which implies that $f$ is Lipschitz continuous.

Theorem 3: Let $f:A \mapsto \mathbb{R}$ be a differentiable function. If $f$ is Lipschitz continuous, $f'$ is bounded $\forall x\in A$ .

Proof: Since $f$ is Lipschitz continuous, there exists $M>0$ so that $\forall x,y\in A$ $|f(x)-f(y)|\leqslant M|x-y|$ . Then, $\frac{|f(x)-f(y)|}{|x-y|}\leqslant M$ .

For $y=x_0$ , we have $|f'(x_0)|=\lim _{x \to x_0}\frac{|f(x)-f(x_0)|}{|x-x_0|} \leqslant M$ .

But $x_0$ was random. Therefore $|f'(x)|\leqslant M$ $\forall x\in A$ .

Time for practice!

Problem: Suppose $f,g:I \mapsto{\mathbb{R}}$ functions which are bounded $\forall x\in I$ and uniformly continuous. Prove that $fg$ is also uniformly continuous.

Right solution: Let $\epsilon>0$ .

$f$ is bounded, so $\exists M>0$ so that $|f(x)|\leqslant M$ $\forall x\in I$ .

Since f is uniformly continuous, $\exists \delta_1 >0$ so that if $x,y\in I$ and $|x-y|<\delta_1$ then $|f(x)-f(y)|<\frac{\epsilon}{M+N}$ .

$g$ is bounded, so $\exists M>0$ so that $|g(x)|\leqslant M$ $\forall x\in I$ .

Since g is uniformly continuous, $\exists \delta_2 >0$ so that if $x,y\in I$ and $|x-y|<\delta_2$ then $|g(x)-g(y)|<\frac{\epsilon}{M+N}$ .

For $\delta$ =min{ $\delta_1 ,\delta_2$ }, we have that if $x,y\in I$ and $|x-y|<\delta$ ,

$|f(x)g(x)-f(y)g(y)|=|f(x)g(x)-f(x)g(y)+f(x)g(y)-f(y)g(y)|$

$\leqslant|f(x)g(x)-f(x)g(y)|+|f(x)g(y)-f(y)g(y)|$

$\leqslant |f(x)||g(x)-g(y)|+|g(y)||f(x)-f(y)|$

$\leqslant M\frac{\epsilon}{M+N}+N\frac{\epsilon}{M+N}=\frac{(M+N)\epsilon}{M+N}=\epsilon$ .

Therefore $fg$ is uniformly continuous.

Wrong solution: We’ll use the 3 theorems!

We have that $|f(x)g(x)|'=|f'(x)g(x)+f(x)g'(x)|\leqslant N|f'(x)|+M|g'(x)|$ .

But $f,g$ uniformly continuous, so from Theorem 3 $\exists M'>0$ so that $|f'(x)|\leqslant M'$ and $\exists N'>0$ so that $|g'(x)|\leqslant N'$ $\forall x\in I$ .

So it follows that $|f(x)g(x)|'\leqslant NN'+MM'$ .

Hence $(fg)'$ is bounded $\forall x\in I$ , and from theorem 2, we have that $(fg)$ is Lipschitz continuous.

Now, from theorem 1, we finally have that $fg$ is uniformly continuous.

Where is the mistake???

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Posted in Analysis

Posted by: Christian | 09/10/2010

Something beautiful.

Looking again some concepts of Calculus, I realized how beautiful is the following property of sequences. We consider a sequence $a_{n}$ and let a subsequence of $a_{k_{n}}$ . Then, if we also suppose that our sequence is strictly increasing, then holds the following

$k_{n} \ge n$ for any $n \in \mathbb{N}$

In general, this means that the $a_5$ is always greater than 5. For me is wonderful. And another one reason for this is that we can prove it using mathematical induction. WOW!!!!

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Posted in Analysis

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Athens Math Circle

Infinitude of Primes essay

A Mathematical Gift, I

The Euler Characteristic

Scrambled thoughts.

The Infinitude of Primes via Topology

Topology on $\mathbb{Z}$

Number-theoretic properties of $\mathbb{Z}$ .

A naive introduction to Measure Theory

A common mistake.

Something beautiful.

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