We shall give the proof by which following (1) is eqivalent to (2) .
Given two topological spaces $S,T$ and the function $f:S\rightarrow T$ .
If $f$ is continuous,
(1) for an arbitrary open set $D_1\in T$ , $f^{-1}(D_1)$ becomes open set in $S$ .
(2) for an $x\in S$ , let $f(x)=y$ be. For the arbitrary neighbourhood $B_{\epsilon}(y)\in T$
$f^{-1}(B_{\epsilon}(y))$ becomes the neighbourhood of $x$ in $S$ .
$(1)\rightarrow (2)$
If $B'(y)$ is a neighbourhood of $y$ , there is an $\epsilon$ such that
$y\in B_{\epsilon}(y)\subset B'(y)\in T$ .
Here, as we have accepted (1), $x\in f^{-1}(B_{\epsilon}(y))\subset f^{-1}(B'(y))$ ,
and both open sets are neighbourhoods.
$(2)\rightarrow (1)$
Let $Y\in T$ and $f^{-1}(Y)=X\subset S$ be. If $x\in X$ and $f(x)=y$ ,
as $y\in Y$ , there is a neighbourhood $B_{\epsilon}(y)$ which is in $Y$ .
Therefore, $f^{-1}(B_{\epsilon}(y))$ is a neighbourhood of $x$ and in $S$ .
As the neighbourhood is a open set, (1) is gotten.
[proposition]
A set $O$ is open if and only if, for an arbitrary element $x$ in $O$ ,
$O$ becomes a neighbourhood of $x$ .
2014/12/07
2014/11/23
topological spaces 5
Given two distance spaces $D,E$ . If the function
$f:x\in D\rightarrow f(x)=y\in E$
is continuous, for an $x_n\rightarrow x\quad (n=1,2,\cdots)$
$f(x_n)\rightarrow f(x)\quad (n\rightarrow \infty)$ .
On the other hand, on topological spaces $S,T$,
if the function $f(S\rightarrow T)$ is continuous,
for an arbitrary open set $C_1\in T$ , $f^{-1}(C_1)$ becomes the open set
in $S$ .
This definition is equivalent to next two definitions.
For an arbitrary closed set $D_1\in T$ , if the function is continuous,
$f^{-1}(D_1)$ becomes the closed set in $S$ .
For $x\in S$ , let $f(x)=y\in T$ be. For the arbitrary neighbourhood $B_{\epsilon}(y)\in T$ ,
$f^{-1}(B_{\epsilon}(y))$ becomes the neighbourhood of $x$ in $S$ .
As no one will have the questions for the equality, the important thing is a proof.
$f:x\in D\rightarrow f(x)=y\in E$
is continuous, for an $x_n\rightarrow x\quad (n=1,2,\cdots)$
$f(x_n)\rightarrow f(x)\quad (n\rightarrow \infty)$ .
On the other hand, on topological spaces $S,T$,
if the function $f(S\rightarrow T)$ is continuous,
for an arbitrary open set $C_1\in T$ , $f^{-1}(C_1)$ becomes the open set
in $S$ .
This definition is equivalent to next two definitions.
For an arbitrary closed set $D_1\in T$ , if the function is continuous,
$f^{-1}(D_1)$ becomes the closed set in $S$ .
For $x\in S$ , let $f(x)=y\in T$ be. For the arbitrary neighbourhood $B_{\epsilon}(y)\in T$ ,
$f^{-1}(B_{\epsilon}(y))$ becomes the neighbourhood of $x$ in $S$ .
As no one will have the questions for the equality, the important thing is a proof.
2014/11/12
topological spaces 4
A closed set is a complementary set of the open set.
Therefore, we are able to define the topology by closed sets.
As with open sets, a collection of closed sets $\mathcal{F}_c\in\Omega$
has following properties.
(1) If $C_1,C_2\in\mathcal{F}_c$ , then $(C_1\cup C_2)\in \mathcal{F}_c$ .
(2) If $\left\{C_i(i=1,2,\cdots)\right\}$ is a collection of the elements of $\mathcal{F}_c$ ,
then $\cap_{i=1}^{\infty}C_i\in\mathcal{F}_c$ .
These are gotten by properties of open sets. The topology by closed sets
will be defined by adding next requirements.
(3) $\Omega,\phi\in \mathcal{F}_c$ .
The proof of (2) is based on the famous De Morgan's laws.
$(\cap C_i)^c=\cup C_i^c$
Please try the challenge.
Therefore, we are able to define the topology by closed sets.
As with open sets, a collection of closed sets $\mathcal{F}_c\in\Omega$
has following properties.
(1) If $C_1,C_2\in\mathcal{F}_c$ , then $(C_1\cup C_2)\in \mathcal{F}_c$ .
(2) If $\left\{C_i(i=1,2,\cdots)\right\}$ is a collection of the elements of $\mathcal{F}_c$ ,
then $\cap_{i=1}^{\infty}C_i\in\mathcal{F}_c$ .
These are gotten by properties of open sets. The topology by closed sets
will be defined by adding next requirements.
(3) $\Omega,\phi\in \mathcal{F}_c$ .
The proof of (2) is based on the famous De Morgan's laws.
$(\cap C_i)^c=\cup C_i^c$
Please try the challenge.
2014/10/25
topological spaces 3
If $A_1$ and $A_2$ are open sets in $\Omega$, then $(A_1\cap A_2)$ is a open set. We have already proved it in "open sets 2(the intersection is open)" generally.
If $(A_1\cap A_2)$ has no elements, or $A_1$ is equal to $A_2$ ,
then the proof would even be clear.
In the case which $(A_1\cup A_2\cup\cdots )$ is open, the proof would not also be required.
Therefore, the definition of a topological space essentially gives the collection of open sets in $\Omega$ .
Precisely the collection of open sets is called the topology,
and $\Omega$ is called the support or the set theoretic support.
For example, as $(\Omega, \phi)$ is the collection of open sets,formally $(\Omega, \phi)$ gives a topology.
Let $\left\{ (-\infty,0)\cup (0,\infty)\right\}$ be $\Omega$ . $(-\infty,0)$ and $(0,\infty)$ give a topology, too.
That is to say, on a $\Omega$ , various kinds of a toplogical space can be defined.
If $(A_1\cap A_2)$ has no elements, or $A_1$ is equal to $A_2$ ,
then the proof would even be clear.
In the case which $(A_1\cup A_2\cup\cdots )$ is open, the proof would not also be required.
Therefore, the definition of a topological space essentially gives the collection of open sets in $\Omega$ .
Precisely the collection of open sets is called the topology,
and $\Omega$ is called the support or the set theoretic support.
For example, as $(\Omega, \phi)$ is the collection of open sets,formally $(\Omega, \phi)$ gives a topology.
Let $\left\{ (-\infty,0)\cup (0,\infty)\right\}$ be $\Omega$ . $(-\infty,0)$ and $(0,\infty)$ give a topology, too.
That is to say, on a $\Omega$ , various kinds of a toplogical space can be defined.
2014/10/18
topological spaces 2
The standard definition of a topological space in the preceding post might have surprised you. Because there are no distance functions, no definitions of open sets, and so on.
However it is enough in this. We shall understand the reason gradually.
The story will begin in $\mathbb{R}^n$ . In $\mathbb{R}^n$ we may define
some kinds of a distance function $d(x,y)$.
Let $a,b,c$ be elements in $\mathbb{R}^n$ .
A distance function $d(a,b)$ has to satisfy following conditions;
(1) $d(a,a)=0$ and if $d(a,b)=0$ , then $a=b$ .
(2) $d(a,b)=d(b,a)$ .
(3) $d(a,b)+d(b,c)\geq d(a,c)$ .
Most popular form of a distance function in $\mathbb{R}^n$ is
\[ d(a,b)=\sqrt{(b_1-a_1)^2+(b_2-a_2)^2+\cdots +(b_n-a_n)^2} . \]
We will usually write $d(a,b)=|b-a|$ or $\lVert b-a \rVert$ .
Using a distance function, a neighborhood $B$ of a point $a$ in $\mathbb{R}^n$ is defined.
\[ B_{\epsilon}(a)=\left\{x| d(a,x)<\epsilon \right\} \quad (\epsilon>0) \]
Let $A$ be a subset of $\mathbb{R}^n$ . For any point $x\in A$ ,
if there is a $\epsilon >0$ such that $B_{\epsilon}(x)\subset A$ , then $A$ is a open set of $\mathbb{R}^n$ .
A closed set is a complementary set of the open set in $\mathbb{R}^n$.
As you may know well, it is a standard definition of a open set.
However it is enough in this. We shall understand the reason gradually.
The story will begin in $\mathbb{R}^n$ . In $\mathbb{R}^n$ we may define
some kinds of a distance function $d(x,y)$.
Let $a,b,c$ be elements in $\mathbb{R}^n$ .
A distance function $d(a,b)$ has to satisfy following conditions;
(1) $d(a,a)=0$ and if $d(a,b)=0$ , then $a=b$ .
(2) $d(a,b)=d(b,a)$ .
(3) $d(a,b)+d(b,c)\geq d(a,c)$ .
Most popular form of a distance function in $\mathbb{R}^n$ is
\[ d(a,b)=\sqrt{(b_1-a_1)^2+(b_2-a_2)^2+\cdots +(b_n-a_n)^2} . \]
We will usually write $d(a,b)=|b-a|$ or $\lVert b-a \rVert$ .
Using a distance function, a neighborhood $B$ of a point $a$ in $\mathbb{R}^n$ is defined.
\[ B_{\epsilon}(a)=\left\{x| d(a,x)<\epsilon \right\} \quad (\epsilon>0) \]
Let $A$ be a subset of $\mathbb{R}^n$ . For any point $x\in A$ ,
if there is a $\epsilon >0$ such that $B_{\epsilon}(x)\subset A$ , then $A$ is a open set of $\mathbb{R}^n$ .
A closed set is a complementary set of the open set in $\mathbb{R}^n$.
As you may know well, it is a standard definition of a open set.
2014/10/11
topological spaces
We shall give the standard definition of a topological space.
For a set $\Omega$ , given $\mathcal{F}$ which is the family of subsets of $\Omega$
(it is a element of the power set of $\Omega$) .
$\Omega$ is a topological space, if following conditions are satisfied;
(1)$\Omega\in \mathcal{F}$ , and $\phi\in \mathcal{F}$ .
(2)If $A_1\in \mathcal{F}$ and $A_2\in \mathcal{F}$, then $A_1\cap A_2\in \mathcal{F}$ .
(3)If $\left\{A_i(i=1,2,\cdots )\right\}$ is a family of the elements of $\mathcal{F}$
(namely, all $A_i$ is the element of $\mathcal{F}$), then $\cup_{i}A_i\in \mathcal{F}$ .
We also say $A_i$ a open set of $\Omega$ .
It is equivalent to the preceding definition(coffee break 8-2).
However, the definition of a open set is given additionally.
For a set $\Omega$ , given $\mathcal{F}$ which is the family of subsets of $\Omega$
(it is a element of the power set of $\Omega$) .
$\Omega$ is a topological space, if following conditions are satisfied;
(1)$\Omega\in \mathcal{F}$ , and $\phi\in \mathcal{F}$ .
(2)If $A_1\in \mathcal{F}$ and $A_2\in \mathcal{F}$, then $A_1\cap A_2\in \mathcal{F}$ .
(3)If $\left\{A_i(i=1,2,\cdots )\right\}$ is a family of the elements of $\mathcal{F}$
(namely, all $A_i$ is the element of $\mathcal{F}$), then $\cup_{i}A_i\in \mathcal{F}$ .
We also say $A_i$ a open set of $\Omega$ .
It is equivalent to the preceding definition(coffee break 8-2).
However, the definition of a open set is given additionally.
2014/09/30
coffee break 9
I have finished the open seminar.
As I hoped that participants would understand the contents more easily,
I had to have some improvements or Kaizens in the previous materials and explanations.
Such works, （as you also know well, ) are very difficult.
Because, as usual, the improvements is likely to be increasing the contents.
However, the time of the seminar may not be changed.
Therefore It was neccessary for me to cut down previous contents,
and to shorten the description.
The result will have possibilities to make the seminar easy or complicated.
Now, I hope that my improvements were good for the participants.
As I hoped that participants would understand the contents more easily,
I had to have some improvements or Kaizens in the previous materials and explanations.
Such works, （as you also know well, ) are very difficult.
Because, as usual, the improvements is likely to be increasing the contents.
However, the time of the seminar may not be changed.
Therefore It was neccessary for me to cut down previous contents,
and to shorten the description.
The result will have possibilities to make the seminar easy or complicated.
Now, I hope that my improvements were good for the participants.
2014/07/27
Hausdorff spaces
In the preceding post (coffee break 8-2), a topological space
is defined by open subsets.
Putting
simply, a topological space is a collection of open sets.
However, as this
definition is too general, some problems will occur.
First of all, when
$a_n\rightarrow x$ and $a_n\rightarrow y$ , $x=y$ may not be
proved.
$a_n\rightarrow
x$ means that, for an $N<\infty$ , if $N<n$ , all $a_n$ is included
in the
neighborhood of $x$ , where the
neighborhood of $x$ is an open subset in the topological space which includes $x$ .
In the definition of a topological space, we may not say the neighborhood of $x$ and $y$ is
same or different.
Therefore, we
prepare the topological space such that, if two elements $x$ and $y$
is different,
each neighborhood
of elements is pairwise disjoint. Such a space is called Hausdorff space, a separable space, or $T2$ space.
2014/07/19
coffee break 8-2 (topological spaces)
Axiomatic structure is not just only in probability theory.
All fields in mathematics is constructed by axiomatic structure.
It is no problems to say that Mathematics itself has axiomatic structure.
Except Axioms which are accepted by no proofs, as definitions, propositions,
theorems, and corollaries are all needed a proof of validity or consistency,
at this point, all is same.
Axioms are assertions in which no one has questions.
However, we are not always able to address mathematical issues from using axioms,
because the explanations are very long.
Then, we usually begin with basic definitions which have been consistently gotten by axioms.
For example, a topological space is defined as follows.
A topological space $T$ is a set which has open subsets and satisfies the following conditions.
(1) $T$ itself (and empty set ) is open.
(2) the intersection of two open subsets in $T$ is open.
(3) the union of any open subsets in $T$ is open.
As you know, a topological space is a generalized distance space.
However, there are not distance functions in the definition.
It is necessary for us to understand that open sets can take a role of distance functions.
I think that such a beginning makes the gate of mathematics narrow and narrow.
2014/07/06
coffee break 8 (probability measures)
It is well known
that one of the main ideas underlying in Black-Scholes’ formula is the risk
neutral probability.
As the risk neutral probability might be gotten by solving given equations, it is usually explained in Finance that the risk neutral probability is artificial.
However, in mathematics, every probability measures $P$ is artificial. Because those may just satisfy the only following definitions.
(1)$P(\Omega)=1,\quad
P(\phi)=0$
(2)$0\leq P(A)
\leq 1\quad (A\subset \Omega)$ (3)$P(\cup A_i)=\sum P(A_i)$
Such the axiomatic
probability theory was founded by Kolmogorov in the 1930s.
Many mathematicians accepted gladly the new paradigm.
Then, the axiomatic
structure based upon Lebesgue theory gave birth to big progress in the
probability theory.
However, on the other hand, as Kiyoshi Ito has once noted, the structure has made an enormous gap between those who are familiar with mathematics and those who are not.
2014/06/09
coffee break 7
I have a plan of the open lecture in which I explain the probability theory for Finance. The contents are similar to the one which has been held in last summer.
1.Basic concepts of probabilities
2.Elementary Lebesgue theory
3.Normal distributions and Brownian motions
4.Conditional expectations
5.Ito's calculus
The lecture will be given over a period of 12 hours. Because the time of the lecture is extended one hour, I will give some changes in the explanations.
Especially, in introducing two methods for solving Black-Scholes formula, I will explain how to use definitions and propositions of the probability theory.
If you are interested in, please join the seminar (← link) in Aug-Sept.
1.Basic concepts of probabilities
2.Elementary Lebesgue theory
3.Normal distributions and Brownian motions
4.Conditional expectations
5.Ito's calculus
The lecture will be given over a period of 12 hours. Because the time of the lecture is extended one hour, I will give some changes in the explanations.
Especially, in introducing two methods for solving Black-Scholes formula, I will explain how to use definitions and propositions of the probability theory.
If you are interested in, please join the seminar (← link) in Aug-Sept.
2014/06/01
functions 2
In preceding post, we defined a function as follows.
A relation $f$ from a member of 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, we write
$f:X\rightarrow Y$
There are usually two metric spaces $S,T$ , and $X\subset S$ and $Y\subset T$ are assumed.
$Y=f(X)=\left\{ f(x) | x\in X \right\}$ is the range or the image of the function.
Conversely, the inverse image of the function is
$f^{-1}(Y)=\left\{ x | f(x)\in Y \right\}$
As it is the domain or a part of the domain of the function, $f^{-1}(Y)\subset X$
By using the distance function $d$ on $S$ and $d'$ on$T$ , we shall define again continuous functions by epsilon-delta proofs.
A function $f$ is continuous at a point c if, for any $\epsilon>0$, there is a $\delta>0$
such that if $d(x,c)<\delta$ , $d'(f(x),f(c))<\epsilon$
Next definition is derived from above. But no distance fuctions are used.
A function is continuous if and only if the inverse image of a open set is open.
Similarly, the definition by a closed set is available.
A function is continuous if and only if the inverse image of a closed set is closed.
Two definitions are equivalent each other. However these are not true on the range or the image.
A relation $f$ from a member of 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, we write
$f:X\rightarrow Y$
There are usually two metric spaces $S,T$ , and $X\subset S$ and $Y\subset T$ are assumed.
$Y=f(X)=\left\{ f(x) | x\in X \right\}$ is the range or the image of the function.
Conversely, the inverse image of the function is
$f^{-1}(Y)=\left\{ x | f(x)\in Y \right\}$
As it is the domain or a part of the domain of the function, $f^{-1}(Y)\subset X$
By using the distance function $d$ on $S$ and $d'$ on$T$ , we shall define again continuous functions by epsilon-delta proofs.
A function $f$ is continuous at a point c if, for any $\epsilon>0$, there is a $\delta>0$
such that if $d(x,c)<\delta$ , $d'(f(x),f(c))<\epsilon$
Next definition is derived from above. But no distance fuctions are used.
A function is continuous if and only if the inverse image of a open set is open.
Similarly, the definition by a closed set is available.
A function is continuous if and only if the inverse image of a closed set is closed.
Two definitions are equivalent each other. However these are not true on the range or the image.
2014/05/25
compact sets 2
We shall summarize the definitions of compact sets.
At first glance, you may not think these are equivalent.
However, these present a same concept and show the properties of compact sets.
(1)A set is compact if and only if any open cover of the set has a finite open sub cover.
It depends on the theorem of Heine-Borel. Please note "finite".
(2)A set in $\mathbb{R}$ is compact if and only if the set is complete and bounded.
Hence, a closed interval in $\mathbb{R}$ is compact. It is understood easily.
(3)A metric space is compact if and only if any infinite sequences has a convergent sub sequence.
It is called the theorem of Bolzano-Weierstrass.
Note that these theorems would not be true if the existing conditions were changed even if only slightly.
At first glance, you may not think these are equivalent.
However, these present a same concept and show the properties of compact sets.
(1)A set is compact if and only if any open cover of the set has a finite open sub cover.
It depends on the theorem of Heine-Borel. Please note "finite".
(2)A set in $\mathbb{R}$ is compact if and only if the set is complete and bounded.
Hence, a closed interval in $\mathbb{R}$ is compact. It is understood easily.
(3)A metric space is compact if and only if any infinite sequences has a convergent sub sequence.
It is called the theorem of Bolzano-Weierstrass.
Note that these theorems would not be true if the existing conditions were changed even if only slightly.
2014/05/18
compact sets
We will again give the definition of the compactness.
If a open cover of $X$ always contains a finite sub cover, the set $X$ is compact.
The reasons why the compactness is more stronger than the completeness depend
on two propositions as follow.
(1)A compact space becomes complete.
(2)A compact set becomes bounded.
We leave out the proofs. However, as a open cover of $X$ always contains a finite sub cover,
no one will have a feeling of strangeness.
Therefore, a set becomes compact if and only if a set is complete and has a finite open cover.
Then we reach the theorem of Heine Borel because a complete subset in a complete space is a closed set. (a finite open cover means bounded.)
2014/05/10
some definitions related to open sets 5 (compact)
In preceding posts, a definition of the completeness of real numbers was given.
Here we shall give the similar definition of the completeness of a metric space.
In a metric space if any Cauchy sequences is always convergent, the metric space is complete.
Therefore, closed intervals on the real number line is complete. That is to say,
a closed subset in a complete metric space is complete.
The compactness is more stronger than the completeness. However the definition is most
difficult and abstract. A open cover must be prepared at first.
Given a set $X\subset S$ , $S=\cup A_i$ and all $A_i$ is open, $S$ is called a open cover of $X$ .
$[0,1)\subset \cup_{i=1}^{\infty}\left( -\frac{1}{i},1-\frac{1}{i} \right)$
Hence, $\cup_{i=1}^{\infty}\left( -\frac{1}{i},1-\frac{1}{i} \right)$ is a open cover of $[0,1)$
In general, there are many open covers of $X$ .
If $S'$ which is made by joining some selected $A_i$ also becomes a open cover of $X$ ,
$S'$ is called a sub cover. If $i$ is finite, $S$ is called a finite open cover.
If a open cover of $X$ always contains a finite open sub cover, the set $X$ is compact.
The famous Heine Borel theorem is as follows.
A set $X\subset \mathbb{R}$ is compact if and only if $X$ is closed and bounded.
It is not easy to understand the essence which the theorem insists.
Here we shall give the similar definition of the completeness of a metric space.
In a metric space if any Cauchy sequences is always convergent, the metric space is complete.
Therefore, closed intervals on the real number line is complete. That is to say,
a closed subset in a complete metric space is complete.
The compactness is more stronger than the completeness. However the definition is most
difficult and abstract. A open cover must be prepared at first.
Given a set $X\subset S$ , $S=\cup A_i$ and all $A_i$ is open, $S$ is called a open cover of $X$ .
$[0,1)\subset \cup_{i=1}^{\infty}\left( -\frac{1}{i},1-\frac{1}{i} \right)$
Hence, $\cup_{i=1}^{\infty}\left( -\frac{1}{i},1-\frac{1}{i} \right)$ is a open cover of $[0,1)$
In general, there are many open covers of $X$ .
If $S'$ which is made by joining some selected $A_i$ also becomes a open cover of $X$ ,
$S'$ is called a sub cover. If $i$ is finite, $S$ is called a finite open cover.
If a open cover of $X$ always contains a finite open sub cover, the set $X$ is compact.
The famous Heine Borel theorem is as follows.
A set $X\subset \mathbb{R}$ is compact if and only if $X$ is closed and bounded.
It is not easy to understand the essence which the theorem insists.
2014/05/03
some definitions related to open sets 4 (a interval is connected)
We reached in the preceding post to a proposition which insisted that a connected set on the real number line was a interval. So, we shall give a brief proof of the proposition.
Suppose that $I$ is a connected set on the real number line
and $I=A\cup B$ , where $A,B\subset \mathbb{R}$ , $A\cap B=\phi$
and both $A$ and $B$ are closed and not empty.
We are able to choose two points $a_1$ from $A$ and $b_1$ from $B$ , and put $a_1<b_1$ .
Dividing the new interval $(a_1,b_1)$ in half, the one must contain points of $A$ and $B$ .
Let the one be a small interval $(a_2,b_2)$ and we repeat the same operation.
Then, we get the sequence of the interval $(a_n,b_n)$ where $a_n\in A$ and $b_n\in B$
and $(a_n,b_n)\supset (a_{n+1},b_{n+1})$ .
If the operation is repeated infinitely, as the new interval becomes smaller and smaller, the real number sequence $a_n$ and $b_n$ have a same limit point $c$ .
As $A$ and $B$ are closed, $c$ is a intersection point of two sets. However, it is in contradiction to the precondition.
Suppose that $I$ is a connected set on the real number line
and $I=A\cup B$ , where $A,B\subset \mathbb{R}$ , $A\cap B=\phi$
and both $A$ and $B$ are closed and not empty.
We are able to choose two points $a_1$ from $A$ and $b_1$ from $B$ , and put $a_1<b_1$ .
Dividing the new interval $(a_1,b_1)$ in half, the one must contain points of $A$ and $B$ .
Let the one be a small interval $(a_2,b_2)$ and we repeat the same operation.
Then, we get the sequence of the interval $(a_n,b_n)$ where $a_n\in A$ and $b_n\in B$
and $(a_n,b_n)\supset (a_{n+1},b_{n+1})$ .
If the operation is repeated infinitely, as the new interval becomes smaller and smaller, the real number sequence $a_n$ and $b_n$ have a same limit point $c$ .
As $A$ and $B$ are closed, $c$ is a intersection point of two sets. However, it is in contradiction to the precondition.
2014/04/26
some definitions related to open sets 3 (connected spaces)
Open sets are a most important and basic concept. We will also understand it by following explanations.
Giving a metric space $\Omega$ and subsets $A,B\subset \Omega$ .
If $A,B\ne \phi$ , $A\cap B=\phi$ and both $A$ and $B$ are open sets,
$S=A\cup B$ is not a connected space (or set).
$S$ is called a disconnected space (or set).
A connected set means we can not divide it two open sets which are not empty
and of which the intersection is empty. If $S$ is connected, $A$ or $B$ should be empty.
A trivial example of a connected space is a set which consists of a point and the empty set.
Namely, a point is the connected space.
As it is a obviously definite thing that the line of real numbers is connected,
we get to a following basic proposition.
A connected set on the real number line is a interval. (a set is not empty and the interval may be open or closed. )
Giving a metric space $\Omega$ and subsets $A,B\subset \Omega$ .
If $A,B\ne \phi$ , $A\cap B=\phi$ and both $A$ and $B$ are open sets,
$S=A\cup B$ is not a connected space (or set).
$S$ is called a disconnected space (or set).
A connected set means we can not divide it two open sets which are not empty
and of which the intersection is empty. If $S$ is connected, $A$ or $B$ should be empty.
A trivial example of a connected space is a set which consists of a point and the empty set.
Namely, a point is the connected space.
As it is a obviously definite thing that the line of real numbers is connected,
we get to a following basic proposition.
A connected set on the real number line is a interval. (a set is not empty and the interval may be open or closed. )
2014/04/19
some definitions related to open sets 2 (isolated and limit points)
Given a metric space $\Omega$ , a open set $A\subset \Omega$ and a member $a\in A$ .
We say a point $a$ is isolated, if, for any $\delta>0$ , a intersection of the open ball
$B(a,\delta )$ and $A$ is $a$ . It means there is not any point of $A$ except $a$
in a neighborhood of $a$ .
If any neighborhoods of $a$ has infinite points of $A$ , $a$ is called a accumulation point
or a limit point.
These will be understood easily. However, these are important definitions in the topology.
Please try to assume relations of various kinds of a set.
We say a point $a$ is isolated, if, for any $\delta>0$ , a intersection of the open ball
$B(a,\delta )$ and $A$ is $a$ . It means there is not any point of $A$ except $a$
in a neighborhood of $a$ .
If any neighborhoods of $a$ has infinite points of $A$ , $a$ is called a accumulation point
or a limit point.
These will be understood easily. However, these are important definitions in the topology.
Please try to assume relations of various kinds of a set.
2014/04/12
some definitions related to open sets
We shall introduce some definitions related to open sets. Here is a metric space $\Omega$ and $A,B\subset\Omega$ .
At first, the closed set means the complementary set of a open set. If $A$ is a open set,
$X=A^c$ is a closed set. Hence, $\mathbb{R}$ and the empty set $\phi$ are both closed sets.
Then $\mathbb{R}$ and the empty set $\phi$ become open sets and closed sets, too.
In a general metric space we have to accept sets which have these two properties at a same time.
There are sets which are not open and closed. You may remember half open intervals.
The interior $X$ of a set $A$ is the maximum open set of the set $A$ . Of course, if $A$ is a open set, $X=A$ . In other words, $X$ is the open set including all open sets which belong to $A$ . We often write just like $X=\cup\left\{ Y\subset A | Y \mbox{ is open} \right\}$ .
The closure $X$ of a set $A$ is the minimum closed set of the set $A$ . If $A$ is a closed set, $X=A$ . Correctly, $X$ is the minimum closed set which includes the set $A$ . We also often write $X=\cap\left\{ Y\supset A | Y \mbox{ is closed} \right\}$ .
The boundary $X$ of a set $A$ is the set whose elements are the closure minus the interior.
Namely, the intersections of the neighborhoods of any elements in the boundary $X$ and $A$ is not empty, and the intersections of the neighborhood and the complementary of $A$ is not empty, too.
These must be the most familiar definitions to you. However, you have to note that these are only based upon the definition of open sets.
At first, the closed set means the complementary set of a open set. If $A$ is a open set,
$X=A^c$ is a closed set. Hence, $\mathbb{R}$ and the empty set $\phi$ are both closed sets.
Then $\mathbb{R}$ and the empty set $\phi$ become open sets and closed sets, too.
In a general metric space we have to accept sets which have these two properties at a same time.
There are sets which are not open and closed. You may remember half open intervals.
The interior $X$ of a set $A$ is the maximum open set of the set $A$ . Of course, if $A$ is a open set, $X=A$ . In other words, $X$ is the open set including all open sets which belong to $A$ . We often write just like $X=\cup\left\{ Y\subset A | Y \mbox{ is open} \right\}$ .
The closure $X$ of a set $A$ is the minimum closed set of the set $A$ . If $A$ is a closed set, $X=A$ . Correctly, $X$ is the minimum closed set which includes the set $A$ . We also often write $X=\cap\left\{ Y\supset A | Y \mbox{ is closed} \right\}$ .
The boundary $X$ of a set $A$ is the set whose elements are the closure minus the interior.
Namely, the intersections of the neighborhoods of any elements in the boundary $X$ and $A$ is not empty, and the intersections of the neighborhood and the complementary of $A$ is not empty, too.
These must be the most familiar definitions to you. However, you have to note that these are only based upon the definition of open sets.
2014/04/05
open sets 2 (the intesection is open)
We will give a general proof by which, given open sets $O_1,O_2$ in a metric space $\Omega$ ,
the intersection of $O_1$ and $O_2$ is open.
A intersection of a finite number of open sets is open. Namely,
If $O_1,O_2,\cdots , O_k\in\Omega$ , $O_1\cap O_2,\cap \cdots \cap O_k\in\Omega$ is open.
If a element $e$ is in $\cap_{i=1}^k O_i$ , as $e\in O_i$ for all $i=1,\cdots ,k$ ,
there is a $\delta_i>0$ such that open ball $B(e, \delta_i)$ is in $O_i$ .
We choose $\delta=\min(\delta_1,\cdots\delta_k)$ .
Then, since $B(e, \delta)\subset B(e, \delta_i)$ for all $i=1,\cdots ,k$ ,
$B(e, \delta)\subset \cap_{i=1}^k O_i$ .
Thus, we obtain the result which we want. We should remember that a set $O$ is open
if and only if, for any element $e\in O$ , there is a $\delta>0$ such that the open ball $B(e,\delta)\subset O$ .
the intersection of $O_1$ and $O_2$ is open.
A intersection of a finite number of open sets is open. Namely,
If $O_1,O_2,\cdots , O_k\in\Omega$ , $O_1\cap O_2,\cap \cdots \cap O_k\in\Omega$ is open.
If a element $e$ is in $\cap_{i=1}^k O_i$ , as $e\in O_i$ for all $i=1,\cdots ,k$ ,
there is a $\delta_i>0$ such that open ball $B(e, \delta_i)$ is in $O_i$ .
We choose $\delta=\min(\delta_1,\cdots\delta_k)$ .
Then, since $B(e, \delta)\subset B(e, \delta_i)$ for all $i=1,\cdots ,k$ ,
$B(e, \delta)\subset \cap_{i=1}^k O_i$ .
Thus, we obtain the result which we want. We should remember that a set $O$ is open
if and only if, for any element $e\in O$ , there is a $\delta>0$ such that the open ball $B(e,\delta)\subset O$ .
2014/03/29
open sets
In a metric space $\Omega$ we shall define open sets. Given sets $A,B\subset \Omega$ and
the distance function $d(x,y)$ .
We call a set $B(x,\delta)\in\Omega$ a open ball centered at a point $x$ which has the radius $\delta$ if, for any element $x\in\Omega$ and a real number $\delta>0$ , $B$ is a set $\left\{ y | d(x,y)<\delta \right\}$ .
Using a open ball, a set $A$ is called a neighbourhood of $x$ if $A$ has a subset $B(x,\delta)$ .
Open sets in $\Omega$ is the set in which all elements has a neighbourhood of the element
and which contains the neighbourhood.
Namely, a set $O$ is called open if, for any point $x\in O$ . there is a real number $\delta$
such that a set $\left\{ y | d(x,y)<\delta \right\}$ belongs to $O$ .
$R^n$ is open and the empty set $\phi$ is too. (After this, you will find that these two sets are also closed sets. )
Open sets are the base of topological space.
the distance function $d(x,y)$ .
We call a set $B(x,\delta)\in\Omega$ a open ball centered at a point $x$ which has the radius $\delta$ if, for any element $x\in\Omega$ and a real number $\delta>0$ , $B$ is a set $\left\{ y | d(x,y)<\delta \right\}$ .
Using a open ball, a set $A$ is called a neighbourhood of $x$ if $A$ has a subset $B(x,\delta)$ .
Open sets in $\Omega$ is the set in which all elements has a neighbourhood of the element
and which contains the neighbourhood.
Namely, a set $O$ is called open if, for any point $x\in O$ . there is a real number $\delta$
such that a set $\left\{ y | d(x,y)<\delta \right\}$ belongs to $O$ .
$R^n$ is open and the empty set $\phi$ is too. (After this, you will find that these two sets are also closed sets. )
Open sets are the base of topological space.
2014/03/08
direct product
If two sets $A$ and $B$ are given, we are able to make ordered combinations of each element of the set. Namely if $a\in A$ and $b\in B$ , one of ordered combinations is $(a,b)$ . These are called direct product. We will show all combinations $A\times B$ . Hence,
$A\times B=\left\{ (a,b) | a\in A, b\in B \right\}$
You have to note that direct product is ordered. Therefore, $(a,b)$ is not equal to $(b,a)$ . Because two sets are arbitrary, $A=B$ is allowed. If $A=B=\mathbb{R}$ , $(a,b)$ means the point of Cartesian coordinates of $\mathbb{R}\times \mathbb{R}=\mathbb{R}^2$ . Then, elements $a$ and $b$ become coordinate axes.
Direct product is expanded over two sets. Given n sets, it means n-dimensional space. You must have known $\mathbb{R}^3$ very well. Having already proved, you may still remember that, for $C=\left\{ 0,1 \right\}$ ,
$C^{\infty}=\left\{ 0,1 \right\}\times \left\{ 0,1 \right\}\times \cdots$
is uncountable.
$A\times B=\left\{ (a,b) | a\in A, b\in B \right\}$
You have to note that direct product is ordered. Therefore, $(a,b)$ is not equal to $(b,a)$ . Because two sets are arbitrary, $A=B$ is allowed. If $A=B=\mathbb{R}$ , $(a,b)$ means the point of Cartesian coordinates of $\mathbb{R}\times \mathbb{R}=\mathbb{R}^2$ . Then, elements $a$ and $b$ become coordinate axes.
Direct product is expanded over two sets. Given n sets, it means n-dimensional space. You must have known $\mathbb{R}^3$ very well. Having already proved, you may still remember that, for $C=\left\{ 0,1 \right\}$ ,
$C^{\infty}=\left\{ 0,1 \right\}\times \left\{ 0,1 \right\}\times \cdots$
is uncountable.
2014/02/23
symmetric difference
Symmetric difference is one of forms which are infrequently used in elementary level. Given a set $A$ and $B$, it is denoted by $A\vartriangle B$ .
By using preceding terms it means that
$A\vartriangle B=(A-B)\cup (B-A)=(A\cap B^c)\cup (A^c\cap B)$ or
$A\vartriangle B=\left\{ x | (x\in A, x\notin B)\cup (x\notin A, x\in B) \right\}$ .
For example, if $A=\left\{ 1,2,3,4 \right\}$ and $B=\left\{ 3,4,5,6 \right\}$ ,
$A\vartriangle B=\left\{ 1,2,5,6 \right\}$ .
Therefore, these below are true.
$A\vartriangle B=B\vartriangle A$ (Commutative law is satisfied)
$(A\vartriangle B)\vartriangle C=A\vartriangle (B\vartriangle C)$ (Associative law is satisfied)
$A\cap (B\vartriangle C)=(A\cap B)\vartriangle (A\cap C)$ (Distributive law in intersection is satisfied)
$A\vartriangle \phi=A$
$A\vartriangle A=\phi$
By using preceding terms it means that
$A\vartriangle B=(A-B)\cup (B-A)=(A\cap B^c)\cup (A^c\cap B)$ or
$A\vartriangle B=\left\{ x | (x\in A, x\notin B)\cup (x\notin A, x\in B) \right\}$ .
For example, if $A=\left\{ 1,2,3,4 \right\}$ and $B=\left\{ 3,4,5,6 \right\}$ ,
$A\vartriangle B=\left\{ 1,2,5,6 \right\}$ .
Therefore, these below are true.
$A\vartriangle B=B\vartriangle A$ (Commutative law is satisfied)
$(A\vartriangle B)\vartriangle C=A\vartriangle (B\vartriangle C)$ (Associative law is satisfied)
$A\cap (B\vartriangle C)=(A\cap B)\vartriangle (A\cap C)$ (Distributive law in intersection is satisfied)
$A\vartriangle \phi=A$
$A\vartriangle A=\phi$
2014/02/09
power sets
If the element of a set is also a set, the set whose elements are a set is called a power set.
Given a set $X=\left\{1, 2, 3 \right\}$ , the number of all subsets of $X$ is eight,
$\phi, \left\{1 \right\}, \left\{2 \right\}, \left\{3 \right\}, $
$ \left\{1,2 \right\}, \left\{2,3 \right\}, \left\{1,3 \right\}, \left\{1,2,3 \right\}$ .
Therefore, the power set of $X$ is the set whose elements are all above.
$\left\{\phi, \left\{1 \right\}, \left\{2 \right\}, \left\{3 \right\}, \left\{1,2 \right\}, \left\{2,3 \right\}, \left\{1,3 \right\}, \left\{1,2,3 \right\} \right\}$
In general, if the number of the elements of a finite set is $n$ , the number of the elements of the power set becomes $2^n$ . It will be understood easily.
As this sample is very easy, you should visualize the power set of $[0,1]$
(the interval of real numbers). You will find it is impossible to imagine or write the result.
But it exists definitely.
If the set $X$ is the empty set $\phi$ , the power set of $X$ is $\left\{\phi \right\}$ .
Strictly, this power set of $X$ is not the empty set. That is to say,
$\left\{\phi \right\}\ne \phi$ .
Because the power set of $X$ has one element, but $X$ has no element.
Given a set $X=\left\{1, 2, 3 \right\}$ , the number of all subsets of $X$ is eight,
$\phi, \left\{1 \right\}, \left\{2 \right\}, \left\{3 \right\}, $
$ \left\{1,2 \right\}, \left\{2,3 \right\}, \left\{1,3 \right\}, \left\{1,2,3 \right\}$ .
Therefore, the power set of $X$ is the set whose elements are all above.
$\left\{\phi, \left\{1 \right\}, \left\{2 \right\}, \left\{3 \right\}, \left\{1,2 \right\}, \left\{2,3 \right\}, \left\{1,3 \right\}, \left\{1,2,3 \right\} \right\}$
In general, if the number of the elements of a finite set is $n$ , the number of the elements of the power set becomes $2^n$ . It will be understood easily.
As this sample is very easy, you should visualize the power set of $[0,1]$
(the interval of real numbers). You will find it is impossible to imagine or write the result.
But it exists definitely.
If the set $X$ is the empty set $\phi$ , the power set of $X$ is $\left\{\phi \right\}$ .
Strictly, this power set of $X$ is not the empty set. That is to say,
$\left\{\phi \right\}\ne \phi$ .
Because the power set of $X$ has one element, but $X$ has no element.
2014/02/02
complementary sets
If every sets which we are addressing in a problem is a subset of a big set, the big set which contains all sets is called a universal set. Given the universal set $X$ and a object set $A$ in the problem, a difference set $X-A$ is able to be defined. We usually express it by $A^c$ .
$A^c$ is also called a complementary set of $A$ . By connotative form,
$A^c=\left\{x | x\notin A, x\in X \right\}$
In general as $x\in X$ has been understood and omitted, $A^c$ becomes $\left\{x | x\notin A\right\}$ .
When the universal set is implicit, it is very convenient. Of course,
$(A^c)^c=A$
$\phi^c=X$ and $X^c=\phi$ ($\phi$ is the empty set)
The expressions of complementary sets is very famous in De Morgan's law.
$(A\cup B)^c=A^c\cap B^c$
$(A\cap B)^c=A^c\cup B^c$
If one universal set for $A$ and $B$ exists, these are true. Please try proofs. However, if the universal set of $A$ is not equal the universal set of $B$ , these are not true.
$A^c$ is also called a complementary set of $A$ . By connotative form,
$A^c=\left\{x | x\notin A, x\in X \right\}$
In general as $x\in X$ has been understood and omitted, $A^c$ becomes $\left\{x | x\notin A\right\}$ .
When the universal set is implicit, it is very convenient. Of course,
$(A^c)^c=A$
$\phi^c=X$ and $X^c=\phi$ ($\phi$ is the empty set)
The expressions of complementary sets is very famous in De Morgan's law.
$(A\cup B)^c=A^c\cap B^c$
$(A\cap B)^c=A^c\cup B^c$
If one universal set for $A$ and $B$ exists, these are true. Please try proofs. However, if the universal set of $A$ is not equal the universal set of $B$ , these are not true.
2014/01/26
difference sets
There are two sets whose elements are overlapped partially. For example,
$A=\left\{ 1,2,3,4,5,6 \right\}$
$B=\left\{ 3,4,5,6,7,8,9 \right\}$
In preceding post as we have studied,
$A\cup B=\left\{ 1,2,3,4,5,6,7,8,9 \right\}$ .
In this case overlapped elements $\left\{ 3,4,5,6 \right\}$ become one element.
If we want to clean the overlapped elements, we use the difference set.
$A-B=A\backslash B=\left\{ 1,2 \right\}$
In this case the elements of $B$ $\left\{ 7,8,9 \right\}$ have no contributions to the calculation.
Note that $A-B\ne \left\{ 1,2,-7,-8,-9 \right\}$
Of course, $B-A=\left\{ 7,8,9 \right\}$ . We will ignore the elements $\left\{ 1,2 \right\}$ .
Hence $A\cup B=(A-B)\cup B=A\cup (B-A)$ .
These are the differences of the calculations of sets with those of real numbers.
$A=\left\{ 1,2,3,4,5,6 \right\}$
$B=\left\{ 3,4,5,6,7,8,9 \right\}$
In preceding post as we have studied,
$A\cup B=\left\{ 1,2,3,4,5,6,7,8,9 \right\}$ .
In this case overlapped elements $\left\{ 3,4,5,6 \right\}$ become one element.
If we want to clean the overlapped elements, we use the difference set.
$A-B=A\backslash B=\left\{ 1,2 \right\}$
In this case the elements of $B$ $\left\{ 7,8,9 \right\}$ have no contributions to the calculation.
Note that $A-B\ne \left\{ 1,2,-7,-8,-9 \right\}$
Of course, $B-A=\left\{ 7,8,9 \right\}$ . We will ignore the elements $\left\{ 1,2 \right\}$ .
Hence $A\cup B=(A-B)\cup B=A\cup (B-A)$ .
These are the differences of the calculations of sets with those of real numbers.
2014/01/19
equal sets
A set is decided by the own elements or members. Therefore, if every elements of two sets is according, two sets is equivalent. We shall some examples. The sets are as follows.
$A=\left\{ 1,2,3 \right\}$
$B=\left\{ 4,5,6 \right\}$
$C=\left\{ 1,2,3,4,5,6 \right\}$
$D=\left\{ 1,3,5 \right\}$
$E=\left\{ 2,4,6 \right\}$
$F=\left\{ 1,3 \right\}$
$G=\left\{ 2,4 \right\}$
$H=\left\{ 3,5 \right\}$
$I=\left\{ 6 \right\}$
Then, all below is true.
$A\cup B=C$ , $D\cup E=C$ , $F\cup G\cup H\cup I=C$ ,
$A\cup C=C$ , $B\cup C=C$ , $D\cup C=C$ , $E\cup C=C$ ,
$F\cup C=C$ , $G\cup C=C$ , $H\cup C=C$ , $I\cup C=C$ ,
$A\cap D=F$ , $B\cap E=G$ , $A\cap E=A\cap G$ , $B\cap D=B\cap H$ ,
$A\cup G=F\cup G$ , $E\cap H=G\cup H\cup I$ ,
Please note that, in elementary, overlapped elements should be deleted. i.e,
$A\cup F=\left\{ 1,1,2,3,3 \right\}=\left\{ 1,2,3 \right\}$
$A=\left\{ 1,2,3 \right\}$
$B=\left\{ 4,5,6 \right\}$
$C=\left\{ 1,2,3,4,5,6 \right\}$
$D=\left\{ 1,3,5 \right\}$
$E=\left\{ 2,4,6 \right\}$
$F=\left\{ 1,3 \right\}$
$G=\left\{ 2,4 \right\}$
$H=\left\{ 3,5 \right\}$
$I=\left\{ 6 \right\}$
Then, all below is true.
$A\cup B=C$ , $D\cup E=C$ , $F\cup G\cup H\cup I=C$ ,
$A\cup C=C$ , $B\cup C=C$ , $D\cup C=C$ , $E\cup C=C$ ,
$F\cup C=C$ , $G\cup C=C$ , $H\cup C=C$ , $I\cup C=C$ ,
$A\cap D=F$ , $B\cap E=G$ , $A\cap E=A\cap G$ , $B\cap D=B\cap H$ ,
$A\cup G=F\cup G$ , $E\cap H=G\cup H\cup I$ ,
Please note that, in elementary, overlapped elements should be deleted. i.e,
$A\cup F=\left\{ 1,1,2,3,3 \right\}=\left\{ 1,2,3 \right\}$
2014/01/13
notations of sets
A set is a collection of objects of any kind. The most important thing is that elements (i.e. objects themselves ) of a set have been defined mathematically with no doubt.
(a) a set $A$ of integer numbers, whose elements are very great numbers.
(b) a set $A$ of integer numbers, whose elements are greater than $10$ .
You must have understood (a) is a insufficient definition of a set. Because with (a) we are not able to decide whether the number $1,000,000$ should be in $A$ . (please recall Archimedean properties)
A set $A$ with the definition (b) is expressed by two ways, as are well known.
(1) $\left\{ 11,12,13,\cdots \right\}$
(2) $\left\{ x | x>10, x\in\mathbb{Z} \right\}$
(1) is said to be a extensional definition and (2) is connotative. For a finite set, if we adopt a extensional definition, we have to write all elements in principle. However, for a set whose elements are much more or infinite, as it is impossible to do, we can use "$\cdots$ ". In that case, we have to leave no ambiguities for "$ \cdots $ " .
In a connotative definition $\left\{x | C(x) \right\}$ , $x$ is a variable number satisfied the conditions $C(x)$ . Of course, $\left\{x | C(y) \right\}$ is unacceptable.
(a) a set $A$ of integer numbers, whose elements are very great numbers.
(b) a set $A$ of integer numbers, whose elements are greater than $10$ .
You must have understood (a) is a insufficient definition of a set. Because with (a) we are not able to decide whether the number $1,000,000$ should be in $A$ . (please recall Archimedean properties)
A set $A$ with the definition (b) is expressed by two ways, as are well known.
(1) $\left\{ 11,12,13,\cdots \right\}$
(2) $\left\{ x | x>10, x\in\mathbb{Z} \right\}$
(1) is said to be a extensional definition and (2) is connotative. For a finite set, if we adopt a extensional definition, we have to write all elements in principle. However, for a set whose elements are much more or infinite, as it is impossible to do, we can use "$\cdots$ ". In that case, we have to leave no ambiguities for "$ \cdots $ " .
In a connotative definition $\left\{x | C(x) \right\}$ , $x$ is a variable number satisfied the conditions $C(x)$ . Of course, $\left\{x | C(y) \right\}$ is unacceptable.