real number system 11

In the preceding posts, we have seen some properties of real numbers.

(1) A cut of a real number line satisfies the condition of Dedekind cut in which there exists $\max S$  and does not $\min T$ ,  or there exists $\min T$  and does not $\max S$ .

(2) Bounded monotone real number sequences will converge.

(3) If a set of real numbers is bounded above or below, then it has a supremum or a infimum.

(4) If $a$ and $b$ are positive real numbers, then there is an $n\in \mathbb{N}$ such that $na>b$. (It is called the Archimedean property. ) In an addition, nested intervals have a limit real number.

By the strict proofs, we will get that these four properties are equivalent. Therefore, we are able to adopt which properties as an axiom of completeness.

When four properties are apposed, you will feel a similar work, won't you?

This is a last post of 2013. Best wishes for a happy new year  2014 ! !

one-sided continuity of a funtion

The definition of the continuity of a function is as follow.

The function $f$  is continuous, if for any $\epsilon>0$  there is a $\delta>0$  such that
\[ |f(x)-f(c)|<\epsilon  \]
for every points $x$  for which $|x-c|<\delta$ .

It means that $\lim_{x\rightarrow -c}f(x)=f(c)$  and $\lim_{x\rightarrow +c}f(x)=f(c)$ .
By the preceding definition of one-sided limits, we are able to expand the continuity or discontinuity of a function.

A function $f(x)$  is continuous from the right at point $c$  if $\lim_{x\rightarrow +c}f(x)=f(c)$ .
Similarly, A function $f(x)$  is continuous from the left at point $c$  if $\lim_{x\rightarrow -c}f(x)=f(c)$ .

There are three kinds of discontinuity at point $c$ .

(1)Removable discontinuity :  $\lim_{x\rightarrow c}f(x)$  exists. But $\lim_{x\rightarrow c}f(x)\ne f(c)$ . If we can redefine the function $f$ except a point $c$ , the discontinuity will be removed.
\[  (\mbox{example})\quad f(x)=\frac{x^2+x-2}{x-1} \]

(2)Jump or Step discontinuity : One-sided limits exists.
\[ (\mbox{example})\quad f(x)=\left\{ \begin{array}{cccc}
 -x^2 & (x<0)& & \\
 x^2+1 & (x\geq 0)& &
\end{array}
\right. \]
This example means $f(x)$ is continuous from the right at point 0, but not from the left.

(3)Infinite or essential discontinuity : One or both of the one-sided limits do not exist or infinite.
\[  (\mbox{example})\quad f(x)=\frac{1}{x-1} \]

limits of a function (one-sided limits)

If a function $f(x)$  approaches $c$  as its argument $x$  approaches a point $x_0$ , the function $f$  is said to approach the limit $c$ . Now we shall address the following function.
\[ f(x)=\left\{ \begin{array}{cccc}
 -x^2 & (x<0)& & \\
 1& (x=0)& & \\
 x^2 & (x>0)& &
\end{array}
\right. \]
As this function is not continuous, it is not easy for us to understand the limit of the function.

If $x$  approaches $0$  from the left-side hand, $f(x)\rightarrow 0$ . This result is also same in the case of $x$  approaching from the right-side hand. However, as $f(0)=1$  by its own definition, $\lim_{x\rightarrow 0} f(x)\ne f(0)$ .

We shall define two kinds of limit of a function.  The limit of $f(x)$  in the case which $x\rightarrow x_0$  and $x<x_0 $   is said to the left hand limit and is written by
\[ \lim_{x\rightarrow -x_0}f(x) \]
If the right hand limit, then
\[ \lim_{x\rightarrow +x_0}f(x) \]
These are called one-sided limits. "left hand" is equivalent to "below" and "right hand" is equivalent to "above".

If for any $\epsilon>0$ there is a $\delta>0$ such that $|f(x)-c|<\epsilon$  whenever $0<x_0-x <\delta$ , the limit of $f(x)$  is $c$ as $x$ approaches $x_0$ from below. If the case is from above or right hand, the condition becomes $0<x-x_0<\delta$ .

In the above example, both the right hand and left hand limit of $f(x)$  at the point $0$  are not $f(0)$ unfortunately.

real number system 10 (Cauchy sequence)

In nested intervals, let $a_n$  be the increasing sequence from the left-hand side and $b_n$  be the decreasing sequence from the right-hand side. In the definition of the convergence the crucial condition was $|a_n-b_n|\rightarrow 0$  as $n\rightarrow \infty$ .

Given a sequence $c_n$  and consider a sequence $n_i$  of positive integers such that  $n_1<n_2<\cdots $ . Then the sequence $c_{n_i}$  is called a sub sequence of $c_n$ .
It is clear that $c_n$  converges to $c$  if and only if every sub sequence of $c_n$  converges to $c$ .

A sequence $c_n$  is said to be a Cauchy sequence if for any $\epsilon>0$ there is a $N$  such that $|c_n-c_m|<\epsilon$  if $n,m\geq N$ . (It means $|c_n-c_m|\rightarrow 0$ but $[c_n,c_m]$ are not always nested. )

We need to know the theorem in which every convergent sequence is a Cauchy sequence. However in the theorem the limit is not explicitly involved.

real number system 9 (bounded monotone sequences)

In the method of nested intervals, we supposed the following proposition.

If a real number sequence increasing or decreasing monotonically is bounded,  it will converge.

We shall prove it simply. A monotonical sequence is either

$(increasing)\quad a_1\leq a_2\leq\cdots\leq a_n\leq\cdots$ , or
$(decreasing)\quad a_1\geq a_2\geq\cdots\geq a_n \geq\cdots$ .

We will adderess the case of "a increasing sequence", as both cases are same.

If a increasing sequence is bounded above, it has $\sup a_n=A$ (by the real number system property). Therefore for any $\epsilon>0$,
\[  A-\epsilon < a_m\leq a_{m+1}\leq \cdots \leq A  \]
Namely for all $n$ such that $n>m$ , $|a_n-A|<\epsilon $  because $A-\epsilon<a_n\leq A$ .
As it means $\lim_{n\rightarrow\infty}a_n=A$ , we get the desired result.

coffee break 6

I finished "Batman Arkham city" on PS3. I have already played "Batman Arkham asylum" in last year and now I am playing the second time on PC. 

One of the thing which I was charmed is cool motions in the battles of Batman.

Of course the player has to accurately push the buttons which enable the motion. However, the cool motions or combos will succeed by pushing buttons gently in these games.

The timing at which I (feeling impatiently) find myself pushing the buttons is slightly overquicker than the timing at which nice players recognize. From the result, the number of times pushing the buttons are too many.

Cool motions or combos are needed appropriate pushing operation times and accurate timings. Action games are quite difficult, but much exciting for me.

real number system 8 (nested intervals)

We shall prove that a contracting closed interval becomes a point. It is definitely no doubt. But it is only enabled by infinite operations and is in the base of the real number system.

For any bounded closed interval sequences $I_n=[a_n,b_n], (-\infty<a_n\leq b_n<\infty)$ , if $I_n\supset I_{n+1}, (n\in\mathbb{N})$ and $|a_n-b_n|\rightarrow 0$ as $n\rightarrow \infty$, there exists a real number $c$ such that
\[ \bigcap_{n=1}^{\infty}I_n = c \]
In other words, if the real number sequence $a_1,a_2,\cdots $ is monotonic increasing bounded above, the bounded real number sequence $b_1,b_2,\cdots $ is monotonic decreasing bounded below and $a_n\leq b_n$ for any $n$ , there is a real number $c$ such that $a_n\leq c\leq b_n$ . Furthermore, if $|a_n-b_n|\rightarrow 0$ as $n\rightarrow \infty$, $c$ is a point. That is to say, $\lim_{n\rightarrow\infty}a_n=\lim_{n\rightarrow\infty}b_n=c$ .

In real number system, as a bounded monotone real number sequence has a limit number, we put $\lim_{n\rightarrow\infty}a_n=a,\quad \lim_{n\rightarrow\infty}b_n=b$ . Hence, for any $n$ ,
\[ a_n\leq a\leq b\leq b_n \]
and
\[ I_n\supset [a,b]  \]
It means $\bigcap_{n=1}^{\infty}I_n\ne \phi$ . There is a real number $c\in [a,b]$ .

As $a_n\leq c\leq b_n$ for any $n$ , $|a_n-c|\leq b_n-a_n$ . Therefore $a=c$, because $|a_n-b_n|\rightarrow 0$ . Similarly $b=c$ .

It is called the method of nested intervals. By the method we are able to get a root of some kinds of equations computationally.

intervals and segments

An interval or a segment in $\mathbb{R}$ means the set of every points $x$ between $a\in \mathbb{R}$ and $b\in \mathbb{R}$ , ($a<b$).

We shall define intervals and segments in $\mathbb{R}$ accuratelly.

If the set does not include both endpoints, it is called the open interval or the segment. We write the open interval or the segment $(a,b)$ .

The interval or the closed interval which includes both endpoints is expressed $[a,b]$ .Furthermore, it is easy for you to understand the left half open interval $(a,b]$ and the right half open interval $[a,b)$ .

As $\pm\infty$  are not real numbers, the following intervals are impossible.
\[ [-\infty,a),\quad [-\infty,a],\quad (a,\infty],\quad [a,\infty],\quad [-\infty,\infty] \]
The intervals $(-\infty,a), (a,\infty), (-\infty,\infty)$  are regarded as open and $(-\infty,a], [a,\infty)$ are regarded as half open.

epsilon-delta proofs 8

In a metric space, we are able to estimate the distance between elements with the distance function. Therefore, without a real numerical sequence explicitly we can define the "limit".

We write $\lim_{x\rightarrow c}f(x)=y$  when $f(x)$  approaches $y$ as $x$ approaches $c$. Suppose that the set $A$ including $x$ and the set $B$ onto which $f(x)$ maps are both metric spaces. Hence, there are the distance function $d_A(p,q)$  of $A$  and $d_B(p,q)$  of $B$.

$\lim_{x\rightarrow c}f(x)=y$  means that for any $\epsilon>0$ , there exists a $\delta>0$  such that if
\[ d_A(x,c)<\delta \] 
then 
\[ d_B(f(x),y)<\epsilon \]

Preceding definition was as follow.

$a_n$ approches $a$, when for any $\epsilon >0$, there is a $N>0$ such that if $n\geq N$,
\[ |a_n-a|< \epsilon \]

Let us equate $d_A(x,c)$  with $n\geq N$  and $d_B(f(x),y)$  with $|a_n-a|$ . Only $n\geq N$ is not a distance function. Please understand carefully that $d_B(f(x),y)<\epsilon$ is true for "every" $x$ which satisfies $d_A(x,c)<\delta$ .

metric spaces

In an arbitrary set, we are able to set some relations between the elements.

The order relationship in real number system is one of them.

[31] For each pair of real numbers $a$ and $b$, exactly one of the following is true:
       $a=b$,  $a<b$,  or  $b<a$
[32] If $a<b$ and $b<c$, then $a<c$
[33] If $a<b$, then $a+c<b+c$
[34] If $a<b$, then $ac<bc$, whenever $0<c$

Now we will introduce a new relation. For any elements $a, b$ in an arbitrary set $A$, there is a function $d(x,y)\in \mathbb{R}$ which satisfies following conditions.

(1)$d(a,b)\geq 0, \quad (d(a,b)=0 \quad \mbox{if and only if}  a=b)$
(2)$d(a,b)=d(b,a)$
(3)$d(a,b)+d(b,c)\geq d(a,c), \quad (\mbox{for any}   c\in A)$

Then, we call the set $A$ a metric space. (3) is the triangle inequality. The function $d$ is namely a distance function. The metric space is a set which has a quantifiable distance between elements.

For example, $d(a,b)=|b-a|, (a,b\in\mathbb{R})$ is a distance function. Therefore, $\mathbb{R}$ becomes a metric space. Please note that there are various kinds of sets, and every set may adopt multiple distance functions to determine the corresponding distance.

coffee break 5

I read "Winter Frost".

In the past, I have read "Frost at Christmas",
"Night Frost",
"A touch of Frost",
"Hard Frost",
 and short story "Early Morning Frost".
All works is very funny and interesting. This is what workaholic is.

It is a very sad fact that R. D. Wingfield has gone.
We can not be lost in the world of Frost and merry companions. 
Only "Killing Frost" is left for me.

weak law of large numbers

Let random variables$X_1,X_2,\cdots,X_n$ be $i.i.d.$ and any $E(X_i)=\mu<\infty$. If we put
\[ Y_n=\frac{1}{n}\sum_{i=1}^nX_i \]
then,
\[ P(\lim_{n\rightarrow\infty}Y_n=\mu)=1 \]
It is called strong law of large numbers.

On the other hand, weak law of large numbers has a weaker assertion in which $Y_n$ approches $\mu$ than strong law of large numbers. It says, for any $\epsilon>0$,
\[ P(\lim_{n\rightarrow\infty}|Y_n-\mu|>\epsilon )=0\quad  \]
Weak law of large numbers does not say that $Y_n$ approaches $\mu$ with probability 1.

On mathematical terms, strong law of large numbers is a almost everywhere convergence. Weak law of large numbers is a convergence in probability. Therefore, in the random variables, if strong law of large numbers is true, then weak law of large numbers is true. However the converse is not true.

epsilon-delta proofs 7

Let random variables$X_1,X_2,\cdots,X_n$ be i.i.d. and any $E(X_i)=\mu<\infty$. If we put
\[ Y_n=\frac{1}{n}\sum_{i=1}^nX_i \]
then law of large numbers says,
\[ P(\lim_{n\rightarrow\infty}Y_n=\mu)=1 \]
It means, for any $\epsilon>0$ there is a $N$ such that if $n>N$, $|Y_n-\mu|\leq \epsilon$ and it's probability becomes 1.

In the probability theory, as we are used to write elements $w$ of the sample space explicitly,
\[ P (w |\lim_{n\rightarrow\infty}Y_n(w)=\mu )=1. \]
Therefore, when $1/j,(j\in \mathbb{N})$  is used in place of $\epsilon$, if a set $A$ is defined by
\[ A=\left\{w |\forall j, \exists N, n\geq N, |Y_n-\mu|<1/j \right\} \]
, then $P(A)=1$  .  Furthermore, using symbols of the set theory, if a set $A$ is
\[ A=\bigcap_{j=1}^{\infty}\bigcup_{N=1}^{\infty}\bigcap_{n=N}^{\infty}  \left\{w  |  |Y_n-\mu|<1/j \right\} \]
, then $P(A)=1$  .

A epsilon-delta technique can be written in the form like this.

(strong) law of large numbers

Once more we will begin with the epsilon-delta technique.

In the probability theory there is a theorem called "(strong) law of large numbers". It is one of most famous theorems in the probability or statistics.

Random variables $X_1,X_2,\cdots ,X_n$ are independent identically distributed(i. i. d.). $E(X_i)=\mu<\infty$. Then,
\[ Y_n=\frac{1}{n}\sum_{i=1}^nX_i\rightarrow\mu\quad (n\rightarrow\infty)\quad a.s. \]
In other words,
\[  P\left\{ \lim_{n\rightarrow\infty}Y_n=\mu \right\}=1 \]

Law of large numbers means the average of sample data will converge to the expectation value of each random variable (which all will become same),  when sample data increase infinitely.
In general, law of large numbers indicates the strong law.

In next post, I will explain how applied the epsilon-delta technique is.  

During a long intermission

From June to September, I worked for a special lecture of the probability theory for finance. The themes of the lecture are as follow.

1. Introduction of the probability theory

2. Lebesgue integral theory

3. Normal distribution and Brownian motion

4. Conditional expectation and Martingales

5. Ito Calculus

The work was much overburdened for me. At 28/Sep it was completed.
So now I am easy, I am able to restart my blog. Hereafter, in this blog, I will induct the contents of the lecture in a different form.

Around the same time I read books, saw movies, and play games.
However, now I can not remember all of them. I will present them at every
opprtunity, too.

coffee break 4

Almost in April/2013, I have enjoyed following entertainments.

[Translated Novel]
(1)☆☆Cogan's Trade(George V.Higgins)

[Japanese Novel]
(2)☆☆Kokumin no Kotoba(Genichiro Takahashi)
(3)☆☆Gakuhi den 4(Kenzou Kitakata)
(4)☆☆☆☆Namonaki sekai no endrole(Kaoru Kouzei)

[Movie]
(5)☆☆☆The tourist(Johnny Depp, Angelina Jolie)
(6)☆☆☆Hanamizuki（Yui Aragaki, Touma Ikuta）

In this month, I was very busy in business. I could not address Mathematical issues,
and I am afraid I have so little time to see books and movies.

Although the ending is very painful, (4)"Namonaki" makes us sympathetic.
(2)"Genichiro Takahashi" has nice expressions. (1)"G.V.Higgins" has interesting conversations.

coffee break 3

Almost in March/2013, I have enjoyed following entertainments.

[Translated Novel]
(1)☆☆☆The Cobra(Frederick Forsyth)
(2)☆☆☆Avenger(Frederick Forsyth)

[Japanese Novel]
(3)☆☆☆☆Momiji machi eki mae Jisatsu center(Masaki Mitsumoto)
(4)☆☆Taitei no ken 5(Baku Yumemakura) - completed
(5)☆☆Henshin shashin kan(Tomoko Mano)
(6)☆☆☆"Ougon no Bantam" wo yabutta otoko(Naoki Hyakuta)

[Movie]
(7)☆☆☆☆Frozen River(Charlie McDermott, Melissa Leo)
(8)☆☆Nankyoku Ryori nin(Masato Sakai)
(9)☆☆☆Alice in Wonderland（Mia Wasikowska, Johnny Depp）

[Games]
(10)☆☆☆☆Borderlands(ps3)

In this month some works were fine with me. Among them, (3)"Momiji machi" is good. This work describes a extreme grief of losing a child. In (7)"Frozen River", heroine loses her husband and has a small income from a dangerous job and arrests. However, she carries on with her life overcoming the sadness. It is strong and beautiful. (10)"Borderlands" is much fun.

real number system 7

We address how to handle $\pm \infty $ in real number system. It is said to be a extended real number system.

In general, $x\in \mathbb{R}$ means $-\infty<x<+\infty$. However, if $x\in \mathbb{R}$ is not bounded above, we understand $x=+\infty$. It is convenient to make the rule of the fictitious number $\infty$.

["$\infty$"] for any $x\in \mathbb{R}, (x\ne 0)$ , that is, $-\infty<x<+\infty, (x\ne 0)$,
$x+\infty=\infty$, $x-\infty=-\infty$, $\infty+\infty=\infty$,
$(\infty)\cdot (\infty)=\infty$, $(-\infty)\cdot (-\infty)=\infty$, $(-\infty)\cdot (\infty)=-\infty$,
If $x>0$, then $x\cdot (+\infty)=\infty$, $x\cdot (-\infty)=-\infty$,
If $x<0$, then $x\cdot (+\infty)=-\infty$, $x\cdot (-\infty)=\infty$,
$\frac{x}{+\infty}=0$, $\frac{x}{-\infty}=0$,

Unfortunately we are not able to define the following forms.
$\infty-\infty$, $\frac{\pm\infty}{\pm\infty}$, $0\cdot\infty$

Therefore, the extended real number system is ordered, but it is not a field. Do not take any notice of these definitions.

real number system 6

We summarize the properties of real numbers based upon preceding posts.

(1)Real numbers are the set of  all decimals. That is, every real numbers has the form as follow.
\[ C.c_1c_2c_3\cdots, \quad \mbox{or} \quad -C.c_1c_2c_3\cdots \]
where $C$ is any nonnegative integer and $c_i$ is a single digit figure between $0$ and $9$ inclusive.

(In this definition, a finite decimal number always has two forms. For example one is $1.000\cdots $ and $0.999\cdots $. After this, in these cases we shall promise that if $x$ is $0.999\cdots$, we make $x$ be $1$. The same applies to another numbers.)

(2)Real numbers consist of rational numbers and irrational numbers.

(3)Real numbers are a field. On a field we can calculate as prescribed.

(4)Real numbers are ordered by a natural magnitude relationship.

(5)Real numbers are complete.

From (1), (3) and (4), we are not able to get (5). Therefore, we must accept Axiom for completeness or Dedekind cut is satisfied in real numbers. If one is accepted, the other can be proved.

(6)Real numbers are dense. Rational numbers have been already dense in real numbers.

(7)Real numbers are uncountable. Although both $\mathbb{R}$ and $\mathbb{N}$ are infinite sets, $\aleph$ is greater than $\aleph_0$.

(8)There is a one to one correspondence between points of the real number line and real numbers. The one to one function from real numbers to points in the square having a side length $1$ also exists.

Essentially we need (1), (3), (4) and (5). However, do you think all things are very interesting?

cardinal numbers 3

We will see a representative relationship lying between the cardinal number $\aleph_0$ of a countable set and $\aleph$ of an uncountable set.

Suppose a set $B$ be $\left\{0,1 \right\}$. For example, the cardinal number of the direct product set $B\times B\times B$ is card$(B\times B\times B)=2^3=8$, i.e.,
\[ 000, 001, 010, 011, 100, 101, 110, 111 \]
Hence, the cardinal number of the finite direct product set $B_1\times \cdots \times B_n$ is $2^n$.

We put a set $C$ as the infinite direct product set of $B$.
\[ C=\left\{ B\times B\times B\times \cdots   \right\} \]
card$C$ is expressed by $2^{\aleph_0}$ if $n\rightarrow \infty$. The set $C$ has all sequences of a infinite combinations of $0$ and $1$. In other words, $C$ contains all results of having thrown a coin with no end. We can show that $C$ is uncountable.

If $C$ is countable, all elements of $C$ can be listed in any order. However, after all elements were listed, by Cantor's diagonal argument, we can find a new sequence which should be in $C$. It is a contradiction. Therefore, $C$ must be uncountable. (refer to "countable sets 2")

This conclusion will be accepted by the simple fact that the binary number system is equivalent to the decimal in a scale of infinite digits.

As $C$ is uncountable, the cardinal number of $C$ becomes $\aleph$. However, since card$C$ is $2^{\aleph_0}$,
\[  2^{\aleph_0}=\aleph\]
It also means that
\[  \aleph_0 < 2^{\aleph_0} \]
Will you be convinced that it is true?

cardinal numbers 2

The cardinal number of an infinite set is very interesting. We are able to find some examples.

(1)Suppose $A_1=\left\{1,3,5,\cdots \right\}$ and $A_2=\left\{2,4,6,\cdots \right\}$. Then, obviously, card$A_1$=card$A_2$=$\aleph_0$. But, since $A_1\cup A_2=\mathbb{N}$, card$(A_1\cup A_2)=\aleph_0$. In the generality, for the direct sum of two infinite countable sets,
\[ \aleph_0+\aleph_0=\aleph_0 \]
(2)Similarly, in the cardinality of the continuum like the cardinal number of real numbers,
\[ \aleph +\aleph =\aleph \]
It can be understood by following facts. Let two sets be interval $A_1=(0,1]$ and $A_2=(1,2)$. The cardinal number card$A_1$=card$A_2$=$\aleph$. But, card$(A_1\cup A_2)=$card$(0,2)=\aleph$.

(3)We have already seen the open interval $(0,1)$ is one to one corresponding to $\mathbb{R}.  $Let a set $S$ be constructed by all points in a rectangle of the length $a>0$ and width $b>0$. Hence, $S$ is written by
\[ S=\left\{ (x,y)|x\in [0,a], y\in [0,b], a,b\in \mathbb{R}, a,b>0  \right\} \]
Do not confuse the direct product $(x,y)$ with an open interval.

We are able to arrange a one to one function $(x,y)\in S\rightarrow z\in\mathbb{R}$. Supposing $a=b=1$ and $z\in [0,1]$, we shall simplify the problem to understand. Let $c_i$ be a single digit figure from $0$ to $9$. If we define $x,y,z$ as follow,

$x=0.c_1c_3c_5\cdots $
$y=0.c_2c_4c_6\cdots $
$z=0.c_1c_2c_3c_4c_5c_6\cdots $

this function satisfies the conditions. That is, the function $(x,y)\rightarrow z$ is a one to one correspondence. Therefore, the cardinal number of $S$ is $\aleph$. It means in a direct product set that the following equation is true.
\[ \aleph^2=\aleph \]
Furthermore, in n-dimensions,
\[ \aleph^n=\aleph \]
You should note that a deciding factor of the cardinal number of a set is the existence of a one to one function. Will your intuition say yes?

coffee break 2

Almost in February/2013, I have enjoyed following entertainments.

[Translated Novels]
(1)☆☆Stark's War, Stark's Command, Stark's Crusade (John G. Hemry)
(2)☆☆Double play (Robert B. Parker)

[Japanese Novels]
(3)☆☆☆☆Sekai de ichiban shiawase na Okujyo, Bolero (On Yoshida)
(4)☆☆☆☆Uyoku to iu shokugyo (Hiroshi Take)
(5)☆☆Taitei no ken 4 (Baku Yumemakura) - to be continued
(6)☆☆☆Genso Yubin kyoku (Asako Horikawa)
(7)☆☆Neko machi (Sakutaro Hagiwara, Etsuko Kanai)

[Movies]
(8)☆☆Snatch (Jason Statham, Brad Pitt)
(9)☆☆Kokuriko zaka kara (animation, Ghibli)
(10)☆☆Metro ni notte (Shinichi Tutumi, Aya Okamoto)

In this month, Japanese novels are much fine. (3) is the sequel to the book "Yoru ni neko ga mi wo hisomeru tokoro, Think ". The latest is more excellent. The story is written well-carefully and the text is very stylish. We are surprised that the author On Yoshida is a teenager in addendum. However, she is actually a fictional author.

(4) is written by a former right-winger cadre. The story said to be a non-fiction shows much extremities by which I have been taken back. Those are very terrible and scare me.

cardinal numbers

The number of elements of a set is said to be the cardinal number. The cardinal number of a set $A$ is written by $\mbox{card}A$. If a set is finite, the cardinal number of the set becomes the number of elements. That is, if $A=\left\{-1,2,\sqrt{3},4,15 \right\}$, then $\mbox{card}A=5$.

Of course, $\mbox{card}\left\{0,1 \right\}=2$. We are able to define the cardinal number of the direct product of sets, too. If $A=\left\{0,1 \right\}\times \left\{0,1 \right\}\times \left\{0,1 \right\}$, or $A=\left\{(x_1, x_2, x_3)|x_i\in \left\{0, 1\right\}, i=1, 2, 3 \right\}$, then $\mbox{card}A=2^3=8$. If a set is finite, it is very simple.

In the case of an infinite set, you may not to be able to define the cardinal number, or may think that there is just only one cardinal number which is defined. However, we have understood an infinite set could be countable or uncountable. Hence, there are two types of infinite sets.

We define $\aleph_0$ as the cardinal number of a countably infinite set. For example,
\[ \mbox{card}\mathbb{Q}=\aleph_0  \]
Next, we define $\aleph $ as the cardinal number of real numbers.
\[ \mbox{card}\mathbb{R}=\aleph  \]
As real numbers are rational numbers plus irrational numbers, the number of elements of real numbers is more than those of rational numbers. Therefore,
\[ \aleph_0<\aleph \]
It was an astonishing fact discovered by G. Cantor that the number of elements of an infinite set also had a magnitude relationship.

countable sets 2

We accepted that although a set (for example rational numbers) was infinite, it could be countable. But all infinite sets are not countable. We should show that real numbers are uncountable.

At first, we show that $\mathbb{R}=(-\infty,\infty)$ is corresponding to open interval $(0,1)$. It is very simple, because we have only to arrange the function
\[ f(x)=\tan \left(x-\frac{1}{2}\right)\pi, \quad x\in (0,1) \]
It may be kind of weird. But the function must have been familiar to you.

Hence, we decide to show that open interval $(0,1)$ is uncountable. The proof is by contradiction. Let the open interval $(0,1)$ be the set $I$. If $I$ is countable, then all elements $x$ of $I$ can be listed in any order.

$x_1=0.c_{11}c_{12}c_{13}\cdots$
$x_2=0.c_{21}c_{22}c_{23}\cdots$
$x_3=0.c_{31}c_{32}c_{33}\cdots$
$\quad \cdots \cdots \cdots$
$x_n=0.c_{n1}c_{n2}c_{n3}\cdots$
$\quad \cdots \cdots \cdots$

In here, $c_{ij}$ is a single digit figure from $0$ to $9$ and suppose that for a $i$, all $c_{ij} (j=0,1,\cdots \infty)$ of $x_i$ are not nine and are not zero, because $0.999\cdots =1$ and $0.000\cdots =0$ are out of interval.

This list is with no limit. However, after listing, we are able to make a new number $y$ by the following procedure.

Let $y$ be $0.d_1d_2d_3\cdots$ where $d_i (i=0,1,\cdots \infty)$ is also a single digit figure. Then, we pick a number different from $c_{11}$ as $d_1$. Next, we pick a number different from $c_{22}$ as $d_2$, and also do a number different from $c_{33}$ as $d_3$. Furthermore we keep carrying on this choice with no limit. As a result, the number $y$ becomes a new number different from every $x_i$. Although all $x_i$ is all elements of $I$, we have gotten a new number $y$. It is a contradiction. Therefore, open interval $I$ is not countable and real numbers become uncountable.

It is the famous Cantor's diagonal argument. How do you feel about two procedures with no end?

countable sets

A one-to-one correspondence of two sets means that there is a injective or surjective function between them. That is, two sets are one-to-one correspondent if and only if there is a into or onto mapping.

If one of two sets is the domain $\mathbb{N}$ and the other is the range $A=\left\{ a_n | n\in\mathbb{N}, a_n\in \mathbb{R} \right\}$, the function expresses a infinite real number sequence as follow.
\[ f : n\in\mathbb{N}\rightarrow a_n\in A \]
You may not suspect that the number of $a_n$ is countable with no limits.

[countable or uncountable]
A set is said to be countable, if it is finite or countably infinite. Otherwise, a set is uncountable.

In other words, if there is a one-to-one correspondence between $\mathbb{N}$ and a set $A$, then $A$ is countable. If a range set $A$ is finite, it bothers nobody. Even if the range set is infinite, it is possible for the set to be countable as a sequence noted above.

In this point of view, rational numbers are countably infinite. We will be able to accept it by making the sequence as follow.
\[  \frac{1}{1}, \frac{1}{2}, \frac{2}{2}, \frac{1}{3}, \frac{2}{3}, \frac{3}{3}, \frac{1}{4}, \frac{2}{4}, \frac{3}{4}, \frac{4}{4}, \cdots \]
This sequence is clearly countable and becomes rational numbers in addition to a plus or minus sign and zero. It may be most simple than any other proofs.

coffee break

Recently, almost in Jan/2013, I enjoyed following entertainments. All is made in Japan or is translated into Japanese.

[Translated Novels]
☆☆The Grayman (Mark Greaney)
☆☆Mission M.I.A. (J.C.Pollock)
☆☆Dead Shot (Jack Coughlin with Donald A.Davis)
☆☆Soft Target (Stephen Hunter)

[Japanese Novels]
☆☆Diner (Yumeaki Hirayama)
☆☆Shukusin 4 (Baku Yumemakura) -completed
☆☆Sora no Kobushi (Mitsuyo Kakuta)
☆☆Jyugan (Arimasa Oosawa)

[Comics]
☆☆☆☆Kyo no necomura-san 4,5 (Yoriko Hoshi) -to be continued

[Movies]
☆☆☆☆Mahoro eki mae Tada Benriken(Eita, Ryuhei Matsuda)
☆☆Shodou girls(Youko Narumi, Rio Yamashita, Nanami Sakuraba)

[Games]
☆☆☆☆Ninja Gaiden sigma(ps3)
☆☆☆Ninja Gaiden sigma2(ps3)
☆☆Ninja Gaiden 3(ps3)

As there are many holidays from end of December to January, I enjoyed many books. However, I was not able to find exciting books which I want to push.

Movie "Mahoro eki mae Tada Benriken" was nice. It is better than the original(Novel awarded the prize of Naoki-sho, Shion Miura). I think it owed good performances of actors Eita and Ryuhei Matsuda. The sequel is broadcast on the TV program by same casts.

In series of  games "Ninja Gaiden", "sigma " which is the first work is best. As the number of the work goes on, the contents become more joyless. No cigar. Originals are made in Xbox360 and I do not have those. As those are said to be different from ps3 works, I regret it.

functions

In the preceding post (real number system 5), we assumed any point on a real number line was corresponding to a real number with no fail. Nobody will not believe it. If two sets (in the case, a real number line and real numbers) have some kind of relationship, there will be a function. Let us define a function.

[function, domain, and range]
A relation from a set $X$ to a set $Y$ which is many to one or one to one is called a function from $X$ to $Y$. Then, a set $X$ is said to be the domain of the function and a set $Y$ is said to be the range of the function. These relations are written by
\[ f : X\rightarrow Y \]
or, in familiar form,
\[ f(x)=y,\quad (x\in X, y\in Y)  \]
Here, $x$ is said to be a argument of the function, and $y$ is said to be a value of the function.

The expressions, which you have known enough, could be extended in general. For instance, the following forms are available.
\[ f : X\times X\rightarrow Y, \quad or\quad f(x_1,x_2)=y,\quad (x_1,x_2\in X, y\in Y) \]
If every element of the range of the function is related to some elements of the domain of the function, the function is surjective, or is said to be the mapping from $X$ onto $Y$. That is, for all $y\in Y$, there are some $x$ such that $f(x)=y$. Do not confuse, it is not always true that $x$ is one. The term of 'mapping ' is equivalent to 'function '.

If one element of the range of the function is related to only one element of the domain of the function, the function is one to one, or injective, or is said to be the mapping from $X$ into $Y$. That is, if $f(x_1)=f(x_2)$, then $x_1=x_2$. The contrapositive of the statement is that if $x_1\ne x_2$, then $f(x_1)\ne f(x_2)$.

Especially, two sets $X$ and $Y$ are said to be in one-to-one correspondence if there exists a one to one surjective function with domain $X$ and range $Y$. A One-to-one correspondence means the mapping from $X$ onto and into $Y$. Hence, two sets in one-to-one correspondence have the same number of elements.

real number system 5 (Dedekind cut)

Given a real number line $(-\infty, \infty)$, for example a coordinate axis, we are able to cut the line in two at an arbitrary point. The point will correspond one to one with a real number. Let the left side of the point on the line be $S$, and the right side be $T$. $S$ and $T$ are nonempty sets of real numbers such that
\[ S\cup T=\mathbb{R}, \quad  S\cap T=\phi .  \]
That is, all real numbers are mutually exclusive and collectively exhaustive in $S$ or $T$. Furthermore, if $s\in S$ and $t\in T$, then $s<t$. It is said 'Dedekind cut ' expressed by $(S,T)$.

In general, a cut of one ordered set into two sets $S,T$ logically has one of following opportunities.

(1) There exist both $\max S$ and $\min T$.
(2) There exists $\max S$, and does not $\min T$.
(3) There exists $\min T$, and does not $\max S$.
(4) There do not exist both $\max S$ and $\min T$.

You should note that we have to use $\max, \min$ in substitution for $\sup, \inf$. The reason is because conditions $\max S\in S$ and $\min T\in T$ should be met.

Any Dedekind cut of the real number line can not bring the result (1) in the dense field. Because, if we put $a=\max S$, and $b=\min T$,  then $a<b$. However, as real numbers are dense, there is a real number $x$ such that $a<x<b$. it is the contrary of $S\cup T=\mathbb{R} $. Hence, the result (1) is cleared out.

Suppose that $c$ is a point of 'Dedekind cut '. If $c$ is in $S$, the result (2) occurs, or if $c$ is in $T$, (3) does. As $c$ could be any real number, an irrational number $c=\sqrt{2}$  is available.

In the past, R.Dedekind explained real numbers were a continuum, since (2) or (3) was true by assuming that irrational numbers would exist in addition to rational numbers on real number
system.Then, although we intuitively feel it true, as it is impossible to reject the result (4) from all Dedekind cut in a dense ordered field, especially real numbers, we have to decide not to accept (4).

[41''] Dedekind cut at every point of a real number line brings just only the result (2) or (3).

It's assertion is equivalent to the axiom for completeness of real numbers [41] and  means the continuity of real numbers.

How do you feel about all of this?

real number system 4

 We will return to discuss the axiom for completeness [41]. In general, given an arbitrary ordered field $A$, and a nonempty subset $M\in A$, if every $M$ which is bounded above always has a least upper bound $\sup M$, then it is said  "the field $A$ is complete."

This is the reason why the term of "completeness" is used in [41]. As we have accepted the assertion [41] as an axiom, real numbers become complete. But rational numbers are incomplete.

The completeness axiom [41] will be also written as below.

[41'] If a nonempty set $M$ of real numbers is bounded above, then there is a unique real number $\sup M $ such that
(1) $m\leq \sup M $ for all $m\in M$
(2) if $\epsilon>0$ (no matter how small), there is a $m\in M$ such that $m>\sup M - \epsilon$.

As real numbers are dense and a ordered field, $\sup M$ is unique (why?). (1) is the definition of $\sup M$. If $\sup M \in M$, (2) is also no doubt. Moreover (2) is true even though $\sup M\notin M$. We already have used (2) in the proof of Archimedean property.  (2) indicates that we should examine the cut in between  $M$ and $\sup M$.

rational numbers

It is clearly true that rational numbers are dense, because for rational numbers $a$ and $b$ with $a<b$, there is a rational number $(a+b)/2$. Furthermore this dense property of rational numbers is satisfied in real numbers.

If $a$ and $b$ are real numbers with $a<b$, then there is a rational number $p/q$ such that $a<p/q<b$.

It is said that rational numbers are dense in real numbers. That is, between any different two real numbers, there is a rational number.

[wrong answer]
For any rational number $q>0$, $qa<qb$. As real numbers are dense, there is a $p$ such that
\[ qa<p<qb \]
Therefore,
\[ a<\frac{p}{q}<b \]
But, we are not able to state assertively that $p$ and $p/q$ are rational numbers.

[almost sure]
On Archimedean property " $na>b$ ", let $a$ be $b-a$, $b$ be $1$, and $n$ be $q$. Then,
\[ q(b-a)>1 \]
In this form $a$ and $b$ are real numbers and $q$ is a integer number. Next, if we choose $p$ be a smallest integer number such that $p>qa$, then $p-1\leq qa$. Hence, as
\[ qa<p\leq qa+1<qa+q(b-a)=qb, \]
if we divide above all sides by $q$, then we get the equation which would be hoped. Obviously, $p/q$ is a rational number.

real number system 3 (Archimedean property)

For understanding the completeness axiom [41], we should find out the density of real numbers and Archimedean property.

[density of real numbers]
If $a$ and $b$ are real numbers with $a<b$, then there is a real number $x$ such that $a<x<b$.

The proof of this proposition is very simple. If we put $x=(a+b)/2$, the proposition is always satisfied. Moreover, you will immediately find that there are so many real numbers in open interval $(a,b)$. For example, let $x$ be $a+(b-a)/n, (n>1, n\in \mathbb{N})$. This property is called the density of real numbers.

 The equation $0.999\cdots=1$ which we have addressed at first post is consistent with the dense property. Because if $0.999\cdots\ne 1$, as $0.999\cdots<1$ is a natural relationship on the ordered field, then there is an $x$ such that $0.999\cdots<x<1$. However, we can never accept it.

[Archimedean property of real numbers]
If $a$ and $b$ are positive real numbers, then there is an $n\in \mathbb{N}$ such that $na>b$.

The proof is by contradiction. For any $n\in \mathbb{N}$, if $na\leq b$, the set $A=\left\{ na | n\in \mathbb{N} \right\}$ has an upper bound $b$. By axiom for completeness [41], there exists $\sup A=s $.

For a $n'\in \mathbb{N}$, if $s=n'a\in A$, $s<(n'+1)a$. It is impossible, since $s$ is an upper bound of $A$. Therefore for any $n\in \mathbb{N}$, $na<s$. That is, $s\notin A$.

However by the density of real numbers, for any $\epsilon>0$(no matter how small), as we can put $s-\epsilon<s$, $s-\epsilon$ is not an upper bound of $A$. Therefore, we can choose an $n$ such that $s -\epsilon<na$.

At this time, if $\epsilon=a$, $s - a<na$, and $s<(n+1)a$. As $s$ is the least upper bound, it is a contradiction. Thus the proposition is correct.

Real number system became a dense ordered field satisfied completeness.