Academic Maths http://www.academicmaths.com Original Academic Maths Web Site Sun, 12 Dec 2010 18:33:29 +0000 en hourly 1 http://wordpress.org/?v=3.0.1 Construction of The Real Numbers http://www.academicmaths.com/analysis/construction-of-the-real-numbers.html http://www.academicmaths.com/analysis/construction-of-the-real-numbers.html#comments Sun, 12 Dec 2010 18:32:24 +0000 ufukkaya http://www.academicmaths.com/?p=101

DEFINITION1: The set \mathbb{R} having at least two distinct elements and satisfying the following five axioms is called the set of real numbers and each element of \mathbb{R} is called a real number:

I. AXIOMS OF ADDITION:

The function +:\mathbb{R}\times \mathbb{R}\to \mathbb{R} defined as \left( x,y \right)\to x+y\in \mathbb{R} for each \left( x,y \right) in \mathbb{R}\times \mathbb{R} satisfies the following properties:

I{{}_{1}}.\,\forall a,b\in \mathbb{R},a+b=b+a,

I{{}_{2}}.\,\forall a,b,c\in \mathbb{R},a+(b+c)=(a+b)+c,

I{{}_{3}}.\,\exists 0\in \mathbb{R}:\forall a\in \mathbb{R},a+0=a (0 is called the additive identity element),

I{{}_{4}}.\,\forall a\in \mathbb{R},\exists b\in \mathbb{R}:a+b=0 (b is called the additive inverse of a).

A pair of \left( X,+ \right) satisfying the properties {{I}_{1}},{{I}_{2}},{{I}_{3}},{{I}_{4}} is called a commutative additive group (or called an Abelian group). According to this, \mathbb{R} is a commutative additive group.

II. AXIOMS OF MULTIPLICATION:

The function \cdot:\mathbb{R}\times \mathbb{R}\to \mathbb{R} defined as \left( x,y \right)\to x\cdot{y}\in \mathbb{R} for each \left( x,y \right) in \mathbb{R}\times \mathbb{R} satisfies the following properties:

II{{}_{1}}.\,\forall a,b\in \mathbb{R},a\cdot b=b\cdot a,

II{{}_{2}}.\,\forall a,b,c\in \mathbb{R},a\cdot (b\cdot c)=(a\cdot b)\cdot c,

II{{}_{3}}.\,\exists 1\in \mathbb{R}:\forall a\in \mathbb{R},a\cdot 1=a (1 is called the multiplicative identity element),

II{{}_{4}}.\,\forall a\in \mathbb{R}\backslash \left\{ 0 \right\},\exists b\in \mathbb{R},a\cdot b=1 (b is called the multiplicative inverse of a).

The multiplication of a and b is generally denoted by ab instead of a\cdot{b}.

III. DISTRIBUTIVITY OF MULTIPLICATION OVER ADDITION:

For each a,b,c in \mathbb{R}, (a+b)\cdot c=ac+bc.

A trilogy of (X,+,\cdot) satisfying the axioms I, II, III is called a field. According to this, \mathbb{R} is a field.

IV. ORDER AXIOMS:

The relation “<” is defined over \mathbb{R}. For each distinct a,b in \mathbb{R}, one and only one of the  propositions a<b and b<a is true. By the help of the relation “<“, the “at most” relation “\le” is defined as a\le b \Leftrightarrow a<b or a=b. Furthermore, the relation “<” satisfies the following properties:

IV{{}_{1}}.\,a<b ve b<c\Rightarrow a<c,

IV{{}_{2}}.\,a<b\Rightarrow \,\forall c\in \mathbb{R},\,a+c<b+c,

IV{{}_{3}}.\,a<b ve 0<c\Rightarrow \,ac<bc.

\mathbb{R} is a totally ordered set with the relation “\le“.

V. COMPLETENESS AXIOM:

If it holds \forall a\in A, \forall b\in B, a\le b for any two nonempty subsets A and B of \mathbb{R}, then there exists a real number c such as \forall a\in A, \forall b\in B , a\le c\le b.

All the properties of the real numbers except for I, II, III, IV, V can be easily proved by using the axioms I, II, III, IV, V. Now, we give some of them as a theorem:

THEOREM1:

1. In \mathbb{R}, the additive identity element 0 is unique.

2. The additive inverse of each real number is unique. (The additive inverse of a real number a is denoted by -a. By the help of this, the substruction over \mathbb{R} is defined by a-b:= a+\left( -b \right))

3. For each a,b in \mathbb{R}, the unique solution of the equation x+a=b is x=b+(-a)=b-a.

4. In \mathbb{R}, the multiplicative identity element 1 is unique.

5. The multiplicative inverse of each nonzero real number is unique. (The multiplicative inverse of a nonzero real number a is denoted by a^{-1} or \displaystyle{\frac{1}{a}}. By the help of this, the fraction over \mathbb{R} is defined by \displaystyle{\frac{b}{a}}=b\cdot\frac{1}{a}:=b{{a}^{-1}}. Where a is nonzero)

6. For each a,b in \mathbb{R} (a\ne{0}), the unique solution of the equation ax=b is \displaystyle{x=b\cdot {{a}^{-1}}=b\cdot \frac{1}{a}=\frac{b}{a}}.

7. \forall{a}\in \mathbb{R}, a\cdot 0=0.

8. a\cdot b=0\,\Rightarrow \,a=0\lor{b=0}.

9. \forall{a}\in \mathbb{R}, (-1)\cdot a=-a.

10. \forall{a}\in \mathbb{R}, (-1)\cdot (-a)=a.

11. \forall{a}\in \mathbb{R}, (-a)\cdot (-a)=a\cdot a={{a}^{2}}.

12. For any a,b in \mathbb{R},

a<b\wedge b\le c\Rightarrow a<c,

a\le b\wedge b<c\Rightarrow a<c.

13. For any a,b,c,d in \mathbb{R},

0<a\Rightarrow -a<0,

a\le b\wedge c\le d\Rightarrow a+c\le b+d,

a\le b\wedge c<d\Rightarrow a+c<b+d.

14. For any a,b,c in \mathbb{R},

0<a\wedge 0<b\Rightarrow 0<ab,

a<0\wedge 0<b\Rightarrow ab<0,

a<0\wedge b<0\Rightarrow 0<ab,

a<b\wedge c<0\Rightarrow bc<ac.

15. 0<a\Rightarrow 0<{{a}^{-1}}.

16. 0<a\wedge a<b\Rightarrow 0<{{b}^{-1}}\wedge {{b}^{-1}}<{{a}^{-1}}.

PROOF: (will be added)

The set of the real numbers satisfying the inequality 0<a (or a>0) is called the positive real numbers and the set of  the real numbers satisfying the inequality a<0 is called the negative real numbers. They are denoted by {{\mathbb{R}}^{+}}=\left\{ x\in \mathbb{R}\:|\:x>0 \right\} and {{\mathbb{R}}^{-}}=\left\{ x\in \mathbb{R}\:|\:x<0 \right\}, respectively.

DEFINITION2: Let X be a nonempty subset of \mathbb{R}.

(i) If there exists a b in \mathbb{R} such that \forall{x}\in{X}, x\le{b}, then b is called an upper bound of X.

(ii) If there exists an a in \mathbb{R} such that \forall{x}\in{X}, a\le{x}, then a is called an lower bound of X.

(iii) If X has an upper and a lower bound i.e., \exists{a,b}\in{\mathbb{R}}: a\le{x}\le{b}, then X is called a bounded set.

(iv) If there exists an M in X such that \forall{x}\in{X}, x\le{M}, then M is called the maximum element of X and denoted by

M=\underset{x\in X}{\mathop{\max }}\,\left\{ x \right\} or M=\max \left\{ x|x\in X \right\}

(v) If there exists an m in X such that \forall{x}\in{X}, m\le{x}, then m is called the minimum element of X and denoted by

m=\underset{x\in X}{\mathop{\min }}\,\left\{ x \right\} or m=\min \left\{ x|\,x\in X \right\}

For example, the set X=\left\{ x\in \mathbb{R}|\,-1<x<1 \right\} has no minimum and maximum elements. However, the set Y=\left\{ x\in \mathbb{R}|\,-1\le x\le 1 \right\} has minimum and maximum elements. They are -1 and 1, respectively.

When the set X\subset \mathbb{R} has an upper bound, the set

B=\left\{ b\in \mathbb{R}|\,\forall{x}\in{X}, x\le{b} \right\}

is nonempty. Similarly, when the set X\subset \mathbb{R} has a lower bound, the set

A=\left\{ a\in \mathbb{R}|\,\forall{x}\in{X}, a\le{x} \right\}

is nonempty.

DEFINITION3: Let X be a nonempty subset of \mathbb{R}.

(i) if X has an upper bound, then the minimum of the set B=\left\{ b\in \mathbb{R}|\,\forall{x}\in{X}, x\le{b} \right\} is called the supremum of X and denoted by \sup{X}.

(ii) if X has a lower bound, then the maximum of the set A=\left\{ a\in \mathbb{R}|\,\forall{x}\in{X}, a\le{x} \right\} is called the infimum of X and denoted by \inf{X}.

According to this definition, it holds,

\sup{X}=\min \left\{ b\in \mathbb{R}|\,\forall{x}\in{X}, x\le{b} \right\},

\inf{X}=\max \left\{ a\in \mathbb{R}|\,\forall{x}\in{X}, a\le{x} \right\}.

There may not be the minimum and the maximum of any nonempty subset of \mathbb{R}.  Is the property valid for the infimum and supremum? The following theorem is the answer of this question:

THEOREM2 (Least Upper Bound Property): There exists the supremum of any nonempty subset of \mathbb{R} having an upper bound.

PROOF: (will be added)

Similarly, the following theorem can be proved:

THEOREM3 (Greatest Lower Bound Property): There exists the infimum of any nonempty subset of \mathbb{R} having a lower bound.

THEOREM4: Let \varnothing\ne{X}\subset{\mathbb{R}} and l,L\in{\mathbb{R}}. Then,

(i) \sup{X}=L \iff

(a) \forall{x}\in{A}, x\le{L},

(b) \forall{\varepsilon}>0, \exists{x_{\varepsilon}}\in{X}: L-\varepsilon<x_{\varepsilon}.

(ii) \inf{X}=l \iff

(a) \forall{x}\in{X}, l\le{x},

(b) \forall{\varepsilon}>0, \exists{x_{\varepsilon}}\in{X}: x_{\varepsilon}<l+\varepsilon.

PROOF: (will be added)

DEFINITION4: Let a and b be two real numbers and a<b. The set \left\{ x\in \mathbb{R}\,|\,a<x<b \right\} is called an open interval with the endpoint a and b and denoted by \left( a,b \right) (or \left] a,b \right[). Similarly, the set \left\{ x\in \mathbb{R}\,|\,a\le x\le b \right\} is called a closed interval with the endpoint a and b and denoted by \left[ a,b \right]. A left-open right-closed interval and a left-closed right-open interval are defined by the follows:

\left( a,b \right]=\left] a,b \right]=\left\{ x\in \mathbb{R}\,|\,a<x\le b \right\} left-open right-closed interval

\left[ a,b \right)=\left[ a,b \right[=\left\{ x\in \mathbb{R}\,|\,a\le x<b \right\} left-closed right open interval

For a and b in \mathbb{R}, the unbounded intervals (a,+\infty), [a,+\infty), (-\infty,b), (-\infty,b] and (-\infty,+\infty) are defined by the follows:

(a,+\infty)=\left\{ x\in \mathbb{R}\,|\,x>a \right\},

[a,+\infty)=\left\{ x\in \mathbb{R}\,|\,x\ge a \right\},

(-\infty,b)=\left\{ x\in \mathbb{R}\,|\,x<b \right\},

(-\infty,b]=\left\{ x\in \mathbb{R}\,|\,x\le b \right\},

(-\infty,+\infty)=\mathbb{R}.

The set obtained by adding two elements -\infty and +\infty read as “negatif infinity” and “positive infinity” to \mathbb{R} is called the extended real number line or the extended real number system and denoted by \overline{\mathbb{R}}=\mathbb{R}\cup \{-\infty,+\infty\}. Henceforth, we assume that the extended real number system satisfies the following properties:

1. \forall x\in \mathbb{R},

a) -\infty <x<+\infty,

b) x-(+\infty )=x-\infty =-\infty,

c) x+(+\infty )=x+\infty =+\infty,

d) x-(-\infty )=x+\infty =+\infty,

2.

a) +\infty +(+\infty )=+\infty,

b) -\infty +(-\infty )=-\infty,

3. \forall x\in {{\mathbb{R}}_{+}},

a) x(+\infty )=+\infty,

b) x(-\infty )=-\infty,

4. \forall x\in {{\mathbb{R}}_{-}},

a) x(+\infty )=-\infty,

b) x(-\infty )=+\infty,

5.

a) (+\infty )(+\infty )=+\infty,

b) (-\infty )(-\infty )=+\infty,

c) (+\infty )(-\infty )=-\infty,

6. \forall x\in \mathbb{R},

a) \displaystyle{\frac{x}{+\infty }=0},

b) \displaystyle{\frac{x}{-\infty }=0}.

Let X be a nonempty subset of \overline{\mathbb{R}}. If X has no a lower bound, then the infimum is defined as\inf X=-\infty and if X has no an upper bound, then the supremum is defined as \sup X=+\infty. According to this, each nonempty subset of \overline{\mathbb{R}} has both the infimum and the supremum.

DEFINITION5: The absolute value or modulus of a real number x is defined as follows:

|x|=\bigg\{ x,\:\:\:\:\:\text{if}\:x\ge{0}\\-x,\,\text{if}\: x<0

THEOREM5:

a) \forall{x}\in{\mathbb{R}}, |-x|=|x|\ge{0},

b) |x|=0\Leftrightarrow{x=0},

c) \forall{x}\in{\mathbb{R}}, -x\le{|x|} and x\le{|x|},

d) \forall{a,b}\in{\mathbb{R}}, |ab|=|a||b| and \displaystyle{\left| \frac{a}{b} \right| =\frac{|a|}{|b|}} (b\ne{0}),

e) \forall{a,b}\in{\mathbb{R}}, |a\pm{b}|\le{|a|+|b|},

f) \forall{a,b}\in{\mathbb{R}}, \big| |a|-|b| \big| \le{|a-b|},

g) |x|<r\Leftrightarrow{-r<x<r},

h) |x|\le{r}\Leftrightarrow{-r\le{x}\le{r}},

i) |x|>r\Leftrightarrow{x<-r\lor{x>r}},

j) |x|\ge{r}\Leftrightarrow{x\le{-r}\lor{x\ge{r}}}.

PROOF (will be added)

For two real numbers a,b, the number \left| a-b \right|=\left| b-a \right| is called the distance between a and b and denoted by d(a,b).

Let a and b be two real numbers and a<b. The number b-a>0 is called the length, width, measure or size of the intervals (a,b), [a,b), (a,b] and [a,b].

]]> http://www.academicmaths.com/analysis/construction-of-the-real-numbers.html/feed 0 Function http://www.academicmaths.com/analysis/function.html http://www.academicmaths.com/analysis/function.html#comments Sat, 18 Sep 2010 20:13:59 +0000 ufukkaya http://www.academicmaths.com/?p=37 DEFINITION1: Let X and Y be two sets and f\subset{X\times{Y}} be a relation. If the following two conditions are provided, then the relation f is called a “function” with domain X and codomain Y and denoted by f:X\to{Y} or X\stackrel{f}{\rightarrow}{Y}.

1. \forall{x}\in{X}, \exists{y}\in{Y}: (x,y)\in{f},

2. (x,y),(x,y')\in{f}\Rightarrow{y=y'}.

Henceforth, when we write f:X\to{Y}, we will consider that “f is a function from X to Y”.

As is seen from the definition, a function is a relation mapping each element in the domain to a unique element in the codomain. So, the notation y=f(x) is generally used instead of the notations xfy and (x,y)\in{f} and read “x maps to y” or “x maps to f(x)”. The notation f(x) is read “f of x”. Each element of the domain is called an “argument” and for each x in the domain, the corresponding unique element y in the codomain is called “the function value at x”, “output f for an element x” or “the image x” under the function f. The set defined as \{f(x)\:|\:x\in{X}\}\subset{Y} is called the “image” or the “range” of f. Sometimes a function is called a “map” or a “mapping”.

EXAMPLE1 (Identity function): Let X be a set. The function I_{X} over X defined as \forall{x}\in{X}, I_{X}(x)=x is called the “identity function” of X.

EXAMPLE2 (Constant function): Let X and Y be two non-empty set and c be a constant in Y. The function f from X to Y defined as \forall{x}\in{X}, f(x)=c is called a “constant function”.

EXAMPLE3 (Inclusion function): Let X be a set and A\subset{X}. The function i_{A} from A to X defined as \forall{a}\in{A}, i_{A}(a)=a is called an “inclusion function”.

EXAMPLE4 (Restriction and extension): Let f:X\to{Y} and A\subset{X}. The function f|_{A} from A to Y defined as \forall{a}\in{A}, f|_{A}(a)=f(a) is called a “restriction” of f. Besides, f is called an “extension” of f|_{A}.

EXAMPLE5 (Characteristic function): Let X be a set and A\subset{X}. The function \chi_{A} from X to \{0,1\} defined as

\chi_{A}(x)=\bigg\{ 1,\text{ }x\in{A}\\0,\text{ }x\in{X\setminus{A}}

is called the “characteristic function” of A.

DEFINITION2: Let f,g:X\to{Y}. If \forall{x}\in{X}, f(x)=g(x), then the functions f and g are called “equal functions”.

DEFINITION3: Let f:X\to{Y}. If it holds f(x_{1})=f(x_{2})\Rightarrow{x_{1}=x_{2}} for x_{1},x_{2}\in{X}, then f is called an “injective function”, an “injection” or an “one to one function”. Since the condition of being an injection is equivalent to the condition x_{1}\ne{x_{2}}\Rightarrow{f(x_{1})\ne{f(x_{2})}}, the last condition can be used as the definition of injection. As is seen from the definition, an injection is a function mapping two distinct arguments to two distinct images.

DEFINITION4: Let f:X\to{Y}. If it holds \forall{y}\in{Y}, \exists{x}\in{X} : f(x)=y, then f is called a “surjective function”, a “surjection” or an “onto function”. A function is a surjection if and only if its image equal to its codomain.

DEFINITION5: A function is called a “bijective function”, a “bijection” or a “one to one correspondence” if it is both injective and surjective. The identity function of a set in Example1 can be given as an example.

EXAMPLE6: Let’s examine the following three functions f, g and h:

f is injective because it separately maps the elements of X to the elements of Y. However, for e in Y, there is no an element x in X satisfying f(x)=e.

g is surjective because each element of B is the image of at least one element in A (g(2)=a, g(5)=b, g(1)=c, g(4)=d). However, g is not injective because g(3)=g(4) but 3\ne{4}.

As is seen from the above figure, the function h is both injective and surjective i.e., h is a bijection.

The above three examples show us a function is injective if and only if any two arrows leaving from the domain don’t arrive the same images in the codomain, and a function is surjective if and only if for each element in the codomain, there exists at least one arrow leaving from the domain such that the arrow arrives this element. Unfortunately, these necessary and sufficient conditions cannot be used for a function whose domain or codomain has infinitely many elements because an infinite set cannot be displayed by the Venn diagram. There are many different ways for testing whether a function whose domain or codomain has infinitely many elements is injective (or surjective) or not. Let’s give some examples concerning the infinite sets:

EXAMPLE7: Let a,b\in{\mathbb{R}}, a\ne{0}, f:\mathbb{R}\to{\mathbb{R}}, \forall{x}\in{\mathbb{R}}, f(x)=ax+b. f is injective because x_{1},x_{2}\in{\mathbb{R}}, f(x_{1})=f(x_{2})\Rightarrow{ax_{1}+b=ax_{2}+b}\Rightarrow{ax_{1}=ax_{2}\land{a\ne{0}}} \Rightarrow{x_{1}=x_{2}}.

Choose \displaystyle{x=\frac{y-b}{a}} for any y in \mathbb{R}. Then, \displaystyle{f(x)=f(\frac{y-b}{a})=a\frac{y-b}{a}+b=y-b+b=y}. Hence f is surjective. Since f is both injective and surjective, it is a bijection.

EXAMPLE8: Let f:\mathbb{R}\to{\mathbb{R}}, \forall{x}\in{\mathbb{R}}, f(x)=x^{2}.

f is not injective because f(-1)=(-1)^{2}=1=1^{2}=f(1) but -1\ne{1}.

f is not surjective because there is no an element in \mathbb{R} such that f(x)=x^{2}=-1 for y=-1 in \mathbb{R}.

EXAMPLE9: Consider the above function f(x)=x^{2} as \mathbb{R}\to{[0,+\infty)}. Actually, the function is the same as the above function. Its codomain is only changed. Under the given conditions, we will investigate whether this function is injective (surjective) or not:

Let y\in{[0+\infty)}. Since y\ge{0}, it has the square root \sqrt{y}. Choose x=\sqrt{y}\in{\mathbb{R}}. Since f(x)=f\left( \sqrt{y} \right)=\left( \sqrt{y} \right)^{2}=|y|=y, then f is surjective. As above, f is not injective because f(-1)=f(1)=1 but -1\ne{1} for -1,1 in the domain.

EXAMPLE10: This time, we consider the function f(x)=x^{2} as [0,+\infty)\to{[0,+\infty)}. Then, similar to example9, one can prove that f is surjective. Now, we will investigate f is whether injective or not: Take {{x}_{1}},{{x}_{2}} in [0,+\infty).

f\left( {{x}_{1}} \right)=f\left( {{x}_{2}} \right)\Rightarrow x_{1}^{2}=x_{2}^{2}\Rightarrow \sqrt{x_{1}^{2}}=\sqrt{x_{2}^{2}}\Rightarrow \left| {{x}_{1}} \right|=\left| {{x}_{2}} \right|\Rightarrow {{x}_{1}}={{x}_{2}}.

So, f is injective.

Depending on the domain and the codomain of a function, the injectivity or the surjectivity can be changed. When the injectivity or the surjectivity of the function f(x)=x^{2} is asked, one should ask this question: What are the domain and the codomain of this function.

PROPOSITION1: Let X and Y be two sets with n elements and f:X\to{Y}. Then,

f is injective if and only if it is surjective.

PROOF:

The above proposition is available for the finite sets. Over the infinite sets, an injective function doesn’t have to be surjective and a surjective function doesn’t have to be injective. The function f(x)=e^{x} from \mathbb{R} to \mathbb{R} is injective. However, it is not surjective because it is positive-valued.

DEFINITION6: Let X and Y be two sets, f:X\to{Y}, A\subset{X} and B\subset{Y}. The sets defined as

f(A)=\{f(x)\text{ }\vert\text{ }x\in{A}\},

f^{-1}(B)=\{x\in{X}\text{ }\vert\text{ }f(x)\in{B}\}

are called the image of A under f and the preimage or the inverse image of B under f, respectively. As is seen from the definition, f(A) is formed the images of the elements of A and f^{-1}(B) is formed the elements in X whose images are in B. Particularly, if it is chosen by A=X, then f(X) is range and denoted by \text{Im}f. It is clear that f(A)\subset{Y}, f^{-1}(B)\subset{X}, f(\varnothing)=\varnothing and f^{-1}(\varnothing)=\varnothing.

EXAMPLE11: X=\{1,2,3,4\}, Y=\{a,b,c,d,e\}, f(1)=a, f(2)=e, f(3)=e, f(4)=d, A_{1}=\{1,4\}, A_{2}=\{2,3\}, A_{3}=\{1,2,4\}, A_{4}=X, B_{1}=\{e\}, B_{2}=\{b,c\}, B_{3}=\{a,b,c,d\} and B_{4}=Y. Then,

f(A_{1})=\{a,d\},

f(A_{2})=\{e\},

f(A_{3})=\{a,d,e\},

f(A_{4})=f(X)=\{a,d,e\},

f^{-1}(B_{1})=\{2,3\},

f^{-1}(B_{2})=\varnothing,

f^{-1}(B_{3})=\{1,4\},

f^{-1}(B_{4})=f^{-1}(Y)=X.

PROPOSITION2: Let X and Y be two sets, f:X\to{Y}, A,B\subset{X} and C,D\subset{Y}. Then,

a) A\subset{B}\Rightarrow{f(A)\subset{f(B)}},

b) C\subset{D}\Rightarrow{f^{-1}(C)\subset{f^{-1}(D)}}.

PROOF:

PROPOSITION3: Let X and Y be two sets and f:X\to{Y}. If I and \Lambda are two index sets and \{A_{i}\}_{i\in{I}}\subset{\mathbf{P}(X)}, \{B_{\lambda}\}_{\lambda\in{\Lambda}}\subset{\mathbf{P}(Y)} are two families, then

a) \displaystyle{f\Big{(}\bigcup_{i\in{I}}{A_{i}}\Big{)}=\bigcup_{i\in{I}}{f(A_{i})}},

b) \displaystyle{f\Big{(}\bigcap_{i\in{I}}{A_{i}}\Big{)}\subset{\bigcap_{i\in{I}}{f(A_{i})}}},

c) \displaystyle{f^{-1}\Big{(}\bigcup_{\lambda\in{\Lambda}}{B_{\lambda}}\Big{)}=\bigcup_{\lambda\in{\Lambda}}{f^{-1}(B_{\lambda})}},

d) \displaystyle{f^{-1}\Big{(}\bigcap_{\lambda\in{\Lambda}}{B_{\lambda}}\Big{)}=\bigcap_{\lambda\in{\Lambda}}{f^{-1}(B_{\lambda})}}.

PROOF:

COROLLARY1: Let X and Y be two sets, f:X\to{Y}, A,B\subset{X} and C,D\subset{Y}. Then,

a) f(A\cup{B})=f(A)\cup{f(B)},

b) f(A\cap{B})\subset{f(A)\cap{f(B)}},

c) f^{-1}(C\cup{D})=f^{-1}(C)\cup{f^{-1}(D)},

d) f^{-1}(C\cap{D})=f^{-1}(C)\cap{f^{-1}(D)}.

The property “b” in the proposition3 and its corollary attracts our attention.  Although there are the equalities in the properties “a”, “c” and “d”, we can only say that f(A\cap{B}) is a subset of f(A)\cap{f(B)}. It is interesting. Since f(A\cap{B}) is a subset of f(A)\cap{f(B)}, it may be equal. According to proposition1, there are two probabilities: f(A\cap{B})\subsetneqq{f(A)\cap{f(B)}} and f(A\cap{B})=f(A)\cap{f(B)}. The identity function of a set is an example for the second probability. Is there any example for the first probability? The following example is the answer of this question:

EXAMPLE12: Choose X=\{1,2\}, Y=\{a\}, A=\{1\}, B=\{2\} and define  f:X\to{Y} as f(1)=f(2)=a. It is clear that f is a function. Since A\cap{B}=\varnothing, then f(A\cap{B})=\varnothing. On the other hand, since f(A)=f(B)=\{a\}, then f(A)\cap{f(B)}=\{a\}. As is seen, f(A\cap{B})\subsetneqq{f(A)\cap{f(B)}}.

There are many examples for the two probabilities. Is there any criterion determining whether the left side is equal to the right side or not? The following proposition is the answer this question:

PROPOSITION4: Let f:X\to{Y}. Then,

\forall{A,B}\subset{X}, f(A\cap{B})=f(A)\cap{f(B)}\Leftrightarrow{f} is injective.

PROOF:

PROPOSITION5: Let f:X\to{Y}. Then,

a) \forall{A}\subset{X}, A\subset{f^{-1}(f(A))},

b) \forall{B}\subset{Y}, f(f^{-1}(B))\subset{B}.

PROOF:

PROPOSITION6: Let f:X\to{Y}. Then,

a) f is injective \iff \forall{A}\subset{X}, f^{-1}(f(A))=A,

b) f is surjective \iff \forall{B}\subset{Y}, f(f^{-1}(B))=B,

c) f is bijective \iff \forall{A}\subset{X}, f(A^{C})=f(A)^{C}.

PROOF:

DEFINITION7: Let f:X\to Y and g:Y\to Z. The function g\circ f:X\to Z defined as \forall x\in X,\left( g\circ f \right)\left( x \right)=g\left( f\left( x \right) \right) is called the composition of the functions f and g. (Note that the codomain of f is equal to the domain of g)

EXAMPLE13: Choose X=Y=Z=\mathbb{R}, f\left( x \right)=\sin x and g\left( x \right)={{e}^{x}}. Then, \left( g\circ f \right)\left( x \right)=g\left( f\left( x \right) \right)=g\left( \sin x \right)={{e}^{\sin x}}.

Let f and g be two functions with suitably chosen domains and codomains. Is g\circ{f} equal to f\circ{g}? We will answer this question by the help of an example:

EXAMPLE14: Choose X=Y=Z=\mathbb{R}, f\left( x \right)={{x}^{2}} and g\left( x \right)=x+1. Since

\left( g\circ f \right)\left( x \right)=g\left( f\left( x \right) \right)=g\left( {{x}^{2}} \right)={{x}^{2}}+1

and

\left( f\circ g \right)\left( x \right)=f\left( g\left( x \right) \right)=f\left( x+1 \right)={{\left( x+1 \right)}^{2}},

then g\circ f\ne f\circ g. We have just obtained an example about g\circ f\ne f\circ g. Is the proposition g\circ f\ne f\circ g always true? The answer of this question is “no”. For example, if we choose f\left( x \right)=2x and g\left( x \right)=3x, then

\left( g\circ f \right)\left( x \right)=g\left( f\left( x \right) \right)=g\left( 2x \right)=3\left( 2x \right)=6x

and

\left( f\circ g \right)\left( x \right)=f\left( g\left( x \right) \right)=f\left( 3x \right)=2\left( 3x \right)=6x.

I.e., When we choose f\left( x \right)=2x and g\left( x \right)=3x, it holds g\circ f=f\circ g. Consequently, if f and g are two functions with suitably chosen domains and codomains, then there are two probabilities: g\circ f=f\circ g and g\circ f\ne f\circ g.

PROPOSITION7: Let f:X\to{Y}. Then, f\circ {{I}_{X}}=f and {{I}_{Y}}\circ f=f.

PROOF:

PROPOSITION8: Let f:X\to Y, g:Y\to Z and h:Z\to T. Then, h\circ \left( g\circ f \right)=\left( h\circ g \right)\circ f.

PROOF:

PROPOSITION9: Let f:X\to Y and g:Y\to Z. Then,

a) If f and g are injective, then g\circ f is also injective.

b) If f and g are surjective, then g\circ f is also surjective.

c) If g\circ f is injective, then f is also injective.

d) If g\circ f is surjective, then g is also surjective.

PROOF:

DEFINITION8: Let f:X\to Y.

i) If there exists at least one function R:Y\to X such that f\circ R={{I}_{Y}}, then the function R is called a right inverse of f.

ii) If there exists at least one function L:Y\to X such that L\circ f={{I}_{X}}, then the function L is called a left inverse of f.

The definitions (i) and (ii) can be given by the follows:

i*) If there exists at least one function R:Y\to X such that \forall y\in Y,\left( f\circ R \right)\left( y \right)=y, then the function R is called a right inverse of f.

ii*) If there exists at least one function L:Y\to X such that \forall x\in X,\left( L\circ f \right)\left( x \right)=x, then the function L is called a left inverse of f.

PROPOSITION10: Let f:X\to{Y}. Then,

a) There exists at least one right inverse of f \Leftrightarrow f is surjective.

b) There exists at least one left inverse of f \Leftrightarrow f is injective.

PROOF:

EXAMPLE15: Choose f:\mathbb{R}\to \mathbb{R} by f\left( x \right)={{e}^{x}}. If we define L:\mathbb{R}\to{\mathbb{R}} as

L(y)=\bigg\{ \ln{y},\: y>0\\0,\:\:\:\:\: y\le{0}

then it holds \left( L\circ f \right)\left( x \right)=L\left( f\left( x \right) \right)=L\left( {{e}^{x}} \right)=\ln {{e}^{x}}=x \ln{e}=x because \forall x\in \mathbb{R}, {{e}^{x}}>0. So, f has a left inverse.

Since the function f given in example15 is injective, then it has a left inverse. However, according to proposition10, it has no a right inverse because it is not surjective. Example15 shows that a function can have a left inverse although it has no a right inverse.

EXAMPLE16: Choose f:\mathbb{R} \to [0,+\infty) by f\left( x \right)={{x}^{2}}. Since it is surjective, then it has a right inverse. Define R:[0,+\infty)\to \mathbb{R} as R\left( y \right)=\sqrt{y}. Since \forall y\in [0,+\infty), \left( f\circ R \right)\left( y \right)=f\left( R\left( y \right) \right)=f\left( \sqrt{y} \right)={{\left( \sqrt{y} \right)}^{2}}=|y|=y, then R is a right inverse of f. According to proposition10, f has no a left inverse because it is not injective.

PROPOSITION11: Let f:X\to{Y}, {{f}_{1}},{{f}_{2}}:T\to X and {{g}_{1}},{{g}_{2}}:Y\to Z. Then,

a) If f is injective and f\circ {{f}_{1}}=f\circ {{f}_{2}}, then {{f}_{1}}={{f}_{2}} (an injective function has the left cancellation property)

b) If f is surjective and {{g}_{1}}\circ f={{g}_{2}}\circ f, then {{g}_{1}}={{g}_{2}} (a surjective function has the right cancellation property)

PROOF:

Now, let’s introduce the inverse function:

Let f:X\to{Y}. Assume that L and R are left and right inverses of f, respectively. Then,

L=L\circ {{I}_{Y}}=L\circ \left( f\circ R \right)=\left( L\circ f \right)\circ R={{I}_{X}}\circ R=R. I.e., if a function has both left and right inverses, then these are equal to each other. A function being both left inverse and right inverse of f is called an inverse of f. Now, let’s give the definition of “inverse”, more properly:

DEFINITION9: Let f:X\to{Y}. If there exists a function g such as g\circ f={{I}_{X}} and f\circ g={{I}_{Y}}, then the function g is called an inverse of f. Besides, f is called invertible.

PROPOSITION12: Let f:X\to{Y}. Then, f has an inverse if and only if f is bijective.

PROOF: The proof of this proposition can be directly obtained from Proposition10.

EXAMPLE17: If we choose f:\mathbb{R}\to {{\mathbb{R}}^{+}} by f\left( x \right)={{e}^{x}}, then the function g:{{\mathbb{R}}^{+}}\to \mathbb{R} defined by g\left( y \right)=\ln y is an inverse of f because

\forall x\in \mathbb{R},\left( g\circ f \right)\left( x \right)=g\left( f\left( x \right) \right)=g\left( {{e}^{x}} \right)=\ln {{e}^{x}}=x

and

\forall y\in {{\mathbb{R}}^{+}},\left( f\circ g \right)\left( y \right)=f\left( g\left( y \right) \right)=f\left( \ln y \right)={{e}^{\ln y}}=y.

EXAMPLE18: If we choose f:[0,+\infty)\to [0,+\infty) by f\left( x \right)={{x}^{2}}, then the function g:[0,+\infty)\to [0,+\infty) defined by g\left( y \right)=\sqrt{y} is an inverse of f because

\forall x\in [0,+\infty),\left( g\circ f \right)\left( x \right)=g\left( f\left( x \right) \right)=g\left( {{x}^{2}} \right)=\sqrt{{{x}^{2}}}=\left| x \right|=x

and

\forall y\in [0,+\infty),\left( f\circ g \right)\left( y \right)=f\left( g\left( y \right) \right)=f\left( \sqrt{y} \right)={{\left( \sqrt{y} \right)}^{2}}=y.

EXAMPLE19: If we choose \displaystyle{f:\left( -\frac{\pi }{2},\frac{\pi }{2} \right)\to \mathbb{R}} by f\left( x \right)=\tan x then the function \displaystyle{g:\mathbb{R}\to \left( -\frac{\pi }{2},\frac{\pi }{2} \right)} defined by g\left( y \right)=\arctan y is an inverse of f because

\displaystyle{\forall x\in \left( -\frac{\pi }{2},\frac{\pi }{2} \right),\left( g\circ f \right)\left( x \right)=g\left( f\left( x \right) \right)=g\left( \tan x \right)=\arctan \left( \tan x \right)=x}

and

\forall y\in \mathbb{R},\left( f\circ g \right)\left( y \right)=f\left( g\left( y \right) \right)=f\left( \arctan y \right)=\tan \left( \arctan y \right)=y.

Let the functions {{g}_{1}},{{g}_{2}}:Y\to X be two inverses of the function f:X\to Y. Then {{g}_{1}} is a left inverse of f and {{g}_{2}} is a right inverse of f. We have shown that a left inverse and a right inverse of a function are equal to each other. So, {{g}_{1}}={{g}_{2}}. Consequently, If there exists an inverse of a function, then it is unique. This unique inverse is denoted by f^{-1}.

PROPOSITION13: Let f:X\to Y, g:Y\to Z. Then,

a) If f is invertible, then {{f}^{-1}} is also invertible and {{\left( {{f}^{-1}} \right)}^{-1}}=f. Besides, {{f}^{-1}} is bijective.

b) If f and g are invertible, then g\circ f is invertible and {{\left( g\circ f \right)}^{-1}}={{f}^{-1}}\circ {{g}^{-1}}.

PROOF:

THEOREM1: Let X be a non-empty set and G=\{f\;|\;f:X\to{X}\:\:\text{is bijective}\}. Then (G,\circ) is a group with the identity element I_{X}. G is commutative if and only if \left| X \right|\le 2.

PROOF:

DEFINITION10: Let X be a non-empty set and f:X\to{X}. Then we can define as,

{{f}^{0}}={{I}_{X}}, {{f}^{1}}=f, {{f}^{2}}=f\circ f and

{{f}^{n}}=\underbrace{f\circ f\circ \cdots \circ f}_{n \text{ times}}.

Where n is a natural number.

]]> http://www.academicmaths.com/analysis/function.html/feed 0 Equivalence Relation http://www.academicmaths.com/analysis/equivalence-relation.html http://www.academicmaths.com/analysis/equivalence-relation.html#comments Sat, 18 Sep 2010 19:45:35 +0000 ufukkaya http://www.academicmaths.com/?p=34 DEFINITION1: Let X be a set and R\subset{X\times{X}}. If the relation R is reflexive, symmetric and transitive, then the relation R is called an “equivalence relation” and denoted by R=\sim in general.

DEFINITION2: Let \sim be an equivalence relation over a set X and a\in{X}. The set defined as \{x\in{X}\:|\:a\sim{x}\}\subset{X} is called the “equivalence class” of a under \sim and denoted by \overline{a}, [a] or [a]_{\sim}. Since a\sim{a} for every a\in{X}, then a\in{\overline{a}}. So the equivalence class \overline{a} is non-empty for every a\in{X}. The family of all the equivalence classes of the relation \sim is called the “quotient set” of X by \sim and denoted by {^X}/{_\sim}

I.e.,

{^X}/{_\sim}=\{\overline{a}\:|\:a\in{X}\}\subset{\mathbf{P}(X)}.

DEFINITION3: Let \sim be an equivalence relation over a set X and a,b\in{X}. If b\in{\overline{a}}, then b is called a “representative class” of the equivalence class \overline{a}.

PROPOSITION1: Let \sim be an equivalence relation over a set X and a,b\in{X}. Then,

a\sim{b}\Leftrightarrow{a\in{\overline{b}}}\Leftrightarrow{\overline{a}=\overline{b}}\Leftrightarrow{b\in{\overline{a}}}.

PROOF:

THEOREM1: Let \sim be an equivalence relation over a set X. Then the quotient set {^X}/{_\sim} is a partition of X.

PROOF:

EXAMPLE1: Define R\subset{\mathbb{Z}\times{\mathbb{}Z}} as xRy\Leftrightarrow{n\mid{x-y}}\Leftrightarrow{\exists{k\in{\mathbb{Z}}}: x-y=kn}. (Where n is an arbitrary constant in \mathbb{N}).

(i) R is reflexive because \forall{x}\in{\mathbb{Z}}, x-x=0=0.n\Rightarrow{xRx}.

(ii) R is symmetric because

xRy\Rightarrow{\exists{k}\in{\mathbb{Z}}: x-y=k.n}\Rightarrow{y-x=(-k).n\land{-k\in{\mathbb{Z}}}}\Rightarrow{yRx}.

(iii) R is transitive because

xRy\land{yRz}\Rightarrow{\exists{k,l}\in{\mathbb{Z}}: x-y=k.n\land{y-z=l.n}} \Rightarrow{x-z=(x-y)+(y-z)=k.n+l.n=(k+l)n\land{k+l\in{\mathbb{Z}}}}\Rightarrow{xRz}.

Consequently R is an equivalence relation over \mathbb{Z}. We write R=\sim_{n}. Let’s examine the equivalence classes of this equivalence relation in case of n=2: For this, we will separately find the equivalence classes of the numbers 0 and 1. According to Theorem1, \overline{0}\cap{\overline{1}}=\varnothing because 0\nsim_{2}1. Let x be an even integer. Since \exists{k}\in{\mathbb{Z}}: x-0=2k, then x\sim_{2}0. Hence, x\in{\overline{0}}. Now, let x be an odd integer. Since \exists{k}\in{\mathbb{Z}}: x-1=2k, then x\sim_{2}1. Hence, x\in{\overline{1}}. If 2\mathbb{Z} and 2\mathbb{Z}+1 denote the even integers and the odd integers respectively, then \overline{0}=2\mathbb{Z} and \overline{1}=2\mathbb{Z}+1. There is no another equivalence class of this equivalence relation because any integer is either even or odd. Hence, according to Theorem1, 2\mathbb{Z}\cap{2\mathbb{Z}+1}=\varnothing and 2\mathbb{Z}\cup{2\mathbb{Z}+1}=\mathbb{Z}. More generally, all the equivalence classes of the relation \sim_{n} are n\mathbb{Z}, n\mathbb{Z}+1, n\mathbb{Z}+2,\dots, n\mathbb{Z}+(n-1). Consequently, according to Theorem1, for all distinct {k,l}\in{\{0,1,\dots,n-1\}},

n\mathbb{Z}+k\cap{n\mathbb{Z}+l}=\varnothing

and the following equality are true:

\displaystyle{\bigcup_{k=1}^{n-1}(n\mathbb{Z}+k)=\mathbb{Z}}

EXAMPLE2: Let X be the set of all the lines lying on the plane. Since a line isn’t orthogonal to itself, the orthogonality isn’t an equivalence relation among all the lines.

EXAMPLE3: Let R be the parallelism relation over the set of all the lines lying on the plane. I.e., R=//

(i) R is reflexive because every line is parallel to itself.

(ii) R is symmetric because l_{1}//l_{2}\Rightarrow{ l_{2}//l_{1}}.

(iii) R is transitive because l_{1}//l_{2}\land{ l_{2}//l_{3}}\Rightarrow{ l_{1}//l_{3}}.

(Where l_{1}, l_{2}, l_{3} are arbitrary lines on the plane)

Hence, the parallelism is an equivalence relation among all the lines.

EXAMPLE4: Let f be a function from X to Y. Define R\subset{X\times{X}} as aRb\Leftrightarrow{f(a)=f(b)}.

(i) R is reflexive because \forall{a}\in{X}, f(a)=f(a)\Rightarrow{aRa}.

(ii) R is symmetric because aRb\Rightarrow{f(a)=f(b)}\Rightarrow{f(b)=f(a)}\Rightarrow{bRa}.

(iii) R is transitive because

aRb\land{bRc}\Rightarrow{f(a)=f(b)\land{f(b)=f(c)}}\Rightarrow{f(a)=f(c)}\Rightarrow{aRc}.

Hence, R is an equivalence relation over the domain of the function f.

Theorem1 shows that all the equivalence classes of an equivalence relation over a non-empty set is a partition of this set. Now, we will show the opposite. I.e., we will prove that when it’s given a partition of a non-empty set, there is one and only one equivalence relation, the equivalence classes of which are the parts of the partition.

THEOREM2: Let \{A_{i}\}_{i\in{I}} be a partition of a set X. (Where I\ne{\varnothing}, X\ne{\varnothing} and \forall{i\in{I}}, A_{i}\ne{\varnothing}) Define R\subset{X\times{X}} as aRb\Leftrightarrow{\exists{i\in{I}}: a,b\in{A_{i}}}. Then R is an equivalence relation, the equivalence classes of which are the sets of the family \{A_{i}\}_{i\in{I}}.

PROOF:

]]> http://www.academicmaths.com/analysis/equivalence-relation.html/feed 0 Partial Order Relation http://www.academicmaths.com/analysis/partial-order-relation.html http://www.academicmaths.com/analysis/partial-order-relation.html#comments Sat, 18 Sep 2010 19:27:20 +0000 ufukkaya http://www.academicmaths.com/?p=29 DEFINITION1: Let X be a set and R\subset{X\times{X}}. If the relation R is reflexive, antisymmetric and transitive, then the relation R is called a “partial order relation” and denoted by R=\le in general. If “\le” is a partial order relation over a set X, then (X,\le) is called “partially ordered set” or shortly “poset”.

DEFINITION2: Let x and y are elements of a partially ordered set X. If it holds “x\le{y}\lor{y\le{x}}”, then x and y are called “comparable”. Otherwise they are called “incomparable”.

DEFINITION3: If x and y are comparable for all x,y in a partially ordered set (X,\le), then the relation \le is called a “total order” and the set X is called a “totally ordered set” or “linearly ordered set”.

DEFINITION4: Let (X,\le) be a partially ordered set and A\subset{X}. If (A,\le) is a totally ordered set, then A is called a “chain” in X.

DEFINITION5: Let (X,\le) be a partially ordered set and A\subset{X}. If there exists an element a^{*}\in{A} satisfying a\le{a^{*}} for all a\in{A}, then a^{*} is called the maximum of A, and if there exists an element a_{*}\in{A} satisfying a_{*}\le{a} for all a\in{A}, then a_{*} is called the minimum of A. The minimum and the maximum of A are denoted by \min{A} and \max{A} respectively.

DEFINITION6: If every non-empty subset of a partially ordered set (X,\le) has a minimum, then \le is called a “well order”, and (X,\le) is called a “well ordered set”.

PROPOSITION1: Every well ordered set is a totally ordered set.

EXAMPLE1: The set of the real numbers with well-known “at-most relation” (\le) is totally ordered but not well ordered because non-empty subset (0,1) has no a minimum. Similarly, the set of the integers is totally ordered but not well ordered because this set has no a minimum.

EXAMPLE2: The set of natural numbers is well ordered. Note that any non-empty subset of a well ordered set is also well ordered.

EXAMPLE3: Let E be a set and X=\mathbf{P}(E). Hence, the set X with the inclusion relation “\subset” is partially ordered set. The inclusion relation is, reflexive because every set is a subset of itself, antisymmetric because A\subset{B}\land{B\subset{A}}\Rightarrow{A=B} and transitive because A\subset{B}\land{B\subset{C}}\Rightarrow{A\subset{C}} (A,B,C\subset{E}). Consequently, the inclusion is a partial order. Assume the set E has two distinct elements such as a, b and choose A=\{a\}, B=\{b\}. Then, (\mathbf{P}(E),\subset) isn’t totally ordered because A\nsubseteq{B} and B\nsubseteq{A}. I.e., if the set E has more than one element, then \mathbf{P}(E) has at least two incomparable elements. Besides, (\mathbf{P}(E),\subset) is totally ordered if and only if \arrowvert{E}\arrowvert\le{1}. Choose E=\mathbb{N}. Since \mathbb{N} has infinitely many elements, \mathbf{P}(\mathbb{N}) isn’t totally ordered. However, we can give an example of chain in \mathbf{P}(\mathbb{N}): Choose A_{0}=\varnothing, A_{1}=\{1\}, A_{2}=\{1,2\}, A_{3}=\{1,2,3\}, \dots, A_{n}=\{1,2,\dots,n\}, \dots. If we define F=\{A_{n}\:|\:n\in{\mathbb{N}}\}, then F is a chain of \mathbf{P}(\mathbb{N})  because A_{n}\subset{A_{m}}\Leftrightarrow{n\le{m}}.

EXAMPLE4: For n,m\in{\mathbb{N}}, define nRm\Leftrightarrow{n\arrowvert{m}}. The relation R is called “division”. (n\arrowvert{m}\Leftrightarrow m can be divided by n \Leftrightarrow{\exists{k}\in{\mathbb{N}}: m=k.n}). The division is, reflexive because every natural number can be divided by itself, antisymmetric because n\arrowvert{m}\land{m\arrowvert{n}}\Rightarrow{m=n} and transitive because n\arrowvert{m}\land{m\arrowvert{k}}\Rightarrow{n\arrowvert{k}}. Consequently, the division is a partial order. Since 2\nmid{3} and 3\nmid{2}, the numbers 2 and 3 are incomparable each other, i.e., \mathbb{N} isn’t a totally ordered set with the division relation. If we want to define a chain in \mathbb{N}, we can examine the subset F=\{n^k\:|\:k\in{\mathbb{N}}\} for the arbitrary constant n>1. It is clear that the subset F is a chain of \mathbb{N}.

DEFINITION7: Let (X,\le) be a partially ordered set, A\subset{X} and x_{0}\in{X}. If a\le{x_{0}} for all a\in{A}, then the element x_{0} is called an upper bound of the set A and if x_{0}\le{a} for all a\in{A}, then the element x_{0} is called a lower bound of the set A. If a subset of X has an upper bound and a lower bound, then this set is called a “bounded set”.

Note that x_{0} don’t have to be chosen in A. An upper bound or a lower bound of a set don’t have to be included by this set. If an upper bound or a lower bound of a set were included by this set, then we would use the term “maximum” instead of the term “upper bound” and the term “minimum” instead of the term “lower bound”. This definitions shouldn’t be confused with the definitions of the maximum and the minimum. If there is a minimum or maximum of a set, then it’s unique. However, the number of all the lower bounds or all the upper bounds of a set may be more than one, even infinity. Besides, if there is the maximum of a set, then it’s already an upper bound of this set. However, there may be one or more upper bounds of a set although there isn’t the maximum of this set. The same of the last truth is also valid for the minimum. We will explain what we say above by the help of an example:

EXAMPLE5: Let X=\mathbb{R} and A=(0,1). There isn’t the maximum of (0,1). However, 1 is an upper bound of (0,1) because a\le{1} for all a\in{(0,1)}. Similarly, 12 is also an upper bound of (0,1) because a\le{12} for all a\in{(0,1)}. Let A^{*} denote the set of all the upper bounds of (0,1) and A_{*} also denote the set of all the lower bounds. It is clear that A^{*}=[1,+\infty) and A_{*}=(-\infty,0]. As is seen, (0,1) has infinitely many upper bounds and lower bounds although there isn’t its minimum and maximum.

DEFINITION8: Let (X,\le) be a partially ordered set and A\subset{X}. A^{*}=\{x\in{X}\:|\:\forall{a}\in{A}, a\le{x}\} is the set of all the upper bounds of A and A_{*}=\{x\in{X}\:|\:\forall{a}\in{A}, x\le{a}\} is the set of all the lower bounds of A. If there is the minimum of A^{*}, then this minimum is called the “supremum” of A and if there is the maximum of A_{*}, then this maximum is called the “infimum” of A. If there is a supremum or infimum of A, then it’s obviously unique. The supremum and the infimum of a set A are denoted by \sup{A} and \inf{A} respectively. The supremum of a set is the least upper bound of this set and the infimum of a set is the greatest lower bound of this set.

THE PROPERTIES OF SUPREMUM AND INFIMUM:

1) If there is a supremum or infimum of a set, then it’s unique.

2) If there is the maximum of a set, then the maximum is also the supremum of this set. Similarly, if there is the minimum of a set, then the minimum is also the infimum of this set. However, the opposite of this isn’t true in general. I.e., there may not be the maximum of a set although there is the supremum of this set. Similarly, there may not be the minimum of a set although there is the infimum of this set. (See: example6)

3) Any subset of the real numbers has the infimum and the supremum.

4) Let X=\mathbb{R}, \varnothing\ne{A}\subset{\mathbb{R}} and l,L\in{\mathbb{R}}. Then,

(i) \sup{A}=L \iff

a) \forall{a}\in{A}, a\le{L},

b) \forall{\varepsilon}>0, \exists{a_{\varepsilon}}\in{A}: L-\varepsilon<a_{\varepsilon}.

(ii) \inf{A}=l \iff

a) \forall{a}\in{A}, l\le{a},

b) \forall{\varepsilon}>0, \exists{a_{\varepsilon}}\in{A}: a_{\varepsilon}<l+\varepsilon.

5) Let X=\mathbb{R}, \varnothing\ne{A}\subset{\mathbb{R}} and l,L\in{\mathbb{R}}. Then,

(i) \sup{A}=L \iff

a) \forall{a}\in{A}, a\le{L},

b) \displaystyle{\exists{(x_{n})\subset{A}}: \lim_{n\to{\infty}}x_{n}=L}

(ii) \inf{A}=l \iff

a) \forall{a}\in{A}, l\le{a},

b) \displaystyle{\exists{(x_{n})\subset{A}}: \lim_{n\to{\infty}}x_{n}=l}

6) Let X=\mathbb{R}, \varnothing\ne{A}\subset{\mathbb{R}} and l,L\in{\mathbb{R}}. We can give the following property for the people who know the topology:

(i) \sup{A}=L \iff

a) \forall{a}\in{A}, a\le{L},

b) L\in{\overline{A}}.

(ii) \inf{A}=l \iff

a) \forall{a}\in{A}, l\le{a},

b) l\in{\overline{A}}.

(Where \overline{A} denotes the closure of A).

EXAMPLE6: It was mentioned that (0,1)\subset{\mathbb{R}} has no minimum and maximum in example5. The set of all the upper bounds of (0,1) is A^{*}=[1,+\infty). Hence, the equality \sup{A}=\min{A}^{*}=\min{[1,+\infty)}=1 is true. Similarly, the equality \inf{A}=0 is also true.

EXAMPLE7: Let E be a set and X=\mathbf{P}(E). As it is well known, (\mathbf{P}(E),\subset) is a partially ordered. S,T\subset{E}, A=\{S,T\}\subset{\mathbf{P}(E)}. Now, let’s analyse \sup{A}=\sup\{S,T\} and \inf{A}=\inf\{S,T\}. I.e., we will find the supremum and the infimum of two sets S and T. Since S\subset{S\cup{T}} and T\subset{S\cup{T}}, the set S\cup{T} is an upper bound of the set \{S,T\}. Assume that U is another upper bound of \{S,T\}. Then, S\subset{U} and T\subset{U}. So, S\cup{T}\subset{U} is obtained. Consequently, the least upper bound of \{S,T\} is the union of S and T i.e., S\cup{T}. Similarly, one can easily prove the equality \inf\{S,T\}=S\cap{T}. Let I be an index set and \{S_{i}\}_{i\in{I}} be a family of sets of \mathbf{P}(E). Hence, the following equalities are true:

\displaystyle{\sup_{i\in{I}}S_{i}=\bigcup_{i\in{I}}S_{i}} and \displaystyle{\inf_{i\in{I}}S_{i}=\bigcap_{i\in{I}}S_{i}}.

The proof of above equalities is the same of the proof for two sets.

EXAMPLE8: We know that the natural numbers \mathbb{N} is a partially ordered set with the division. Let’s show that \sup\{n,m\}=\text{lcm}\{n,m\}=[n,m] and \inf\{n,m\}=\text{gcd}\{n,m\}=(n,m) for n,m\in{\mathbb{N}} (lcm: least common multiple, gcd: greatest common divisor). First, we will prove for the supremum: Since n\mid{[n,m]} and m\mid{[n,m]}, the number [n,m] is an upper bound of the set \{n,m\}. Assume that the natural number k is another upper bound of \{n,m\}. Then, n\mid{k} and m\mid{k}. So, [n,m]\mid{k}. Consequently, the least upper bound of two natural numbers is their least common multiple. Now, we will prove for the infimum: Since (n,m)\mid{n} and (n,m)\mid{m}, the number (n,m) is a lower bound of the set \{n,m\}. Assume that the natural number k is another lower bound of \{n,m\}. Then, k\mid{n} and k\mid{m}. So, k\mid{(m,n)}. Consequently, the greatest lower bound of two natural numbers is their greatest common divisor.

DEFINITION9: Let (X,\le) a partially ordered set and A\subset{X}.

(i) M\in{A} is called a maximal element of A if A has no an element being greater than M.

(ii) m\in{A} is called a minimal element of A if A has no an element being lower than m.

THE PROPERTIES OF MAXIMAL AND MINIMAL ELEMENTS:

1) Another expression for the maximal element: Let M be a maximal element of A and a\in{A}. Hence, M and a are incomparable each other or a\le{M}. Another expression for the minimal element: Let m be a minimal element of A and a\in{A}. Hence, m and a are incomparable each other or m\le{a}.

2) Any maximal or minimal element of A is in A.

3) Although a set has a maximal element, there may not be the maximum element of this set. Similarly, although a set has a minimal element, there may not be the minimum element of this set.

4) The number of minimal or maximal elements of a set may be more than one.

5) If a^{*} is the maximum element of a set, then a^{*} is also a maximal element of this set and this set has no another maximal element. Similarly, if a_{*} is the minimum element of a set, then a_{*} is also a minimal element of this set and this set has no another minimal element.

6) Let A be a chain of a partially ordered set. If M is a maximal element of A, then M is also the maximum element of A and A has no another maximal element. Similarly, if m is a minimal element of A, then m is also the minimum element of A and A has no another minimal element.

EXAMPLE9: We know that the natural numbers \mathbb{N} is a partially ordered set with the division. Choose A=\{2,3,4,5,12,15\}\subset{\mathbb{N}}. Since A has no element divided by 12 except for 12, the number 12 is a maximal element of A. Similarly, the number 15 is also a maximal element of A. Since A has no element dividing 2 except for 2, the number 2 is a minimal element of A. Similarly, the numbers 3 and 5 are also two minimal elements of A. A has no element being neither a maximal element nor a minimal element except for the number 4. See the following diagram:

Besides, since \text{lcm}A=60 and \text{gcd}A=1, \sup{A}=60 and \inf{A}=1.

ZORN’S LEMMA: Every partially ordered set in which every chain has an upper bound contains at least one maximal element.

]]> http://www.academicmaths.com/analysis/partial-order-relation.html/feed 0 Relation http://www.academicmaths.com/analysis/relation.html http://www.academicmaths.com/analysis/relation.html#comments Sat, 18 Sep 2010 18:34:22 +0000 ufukkaya http://www.academicmaths.com/?p=22 DEFINITION1: Let X and Y be two sets. Any subset of the cartesian product X\times{Y} is called a relation with domain X and codomain Y. While some sources are giving the definition of relation, they assume X,Y\ne{\varnothing} and it’s said the emptyset being a subset of X\times{Y} isn’t a relation. However, the assumption “the emptyset is a relation” is not a problem for any branch of the mathematics. On the contrary, the assumption “the emptyset is a relation” plays an important role in some branch of the mathematics.

If X and Y are two sets with n and m elements respectively, then the cartesian product X\times{Y} has n.m elements. Since a relation with domain X and codomain Y is an element of the power set \mathbf{P}(X\times Y) and the number of the elements of the power set of a set with k elements is 2^{k}, then the number of all the relations with domain X and codomain Y is 2^{n.m}. If X,Y\ne{\varnothing} and at least one of X and Y is infinite set, then the number of all the relations with domain X and codomain Y is also infinity.

Let R\subset{X\times{Y}} be a relation not being the emptyset. The statement (x,y)\in{R} is read “x is R-related to y” and is denoted by xRy or R(x)=y.

EXAMPLE1: Let X=\{a,b,c\} and Y=\{1,2\}. Since X has 3 elements and Y has 2 elements, the number of all the relations with domain X and codomain Y is 2^{2.3}=2^{6}=64. We can give some of these 64 relations:

R_1=\emptyset,

R_2=\{(a,1)\},

R_3=\{(b,1),(b,2),(c,2)\},

R_4=X\times{Y}.

EXAMPLE2: Let X=Y=\mathbb{R} and R=\{(x,y)\in{\mathbb{R}\times\mathbb{R}}\:|\:x^2=y^2\}. Since (-1)^2=1^2, then (-1,1)\in{R}. Let’s show all the elements of this relation on the cartesian coordinate plane:

DEFINITION2: Let X, Y be two sets and R\subset{X\times{Y}}. The relation defined as

R^{-1}=\{(y,x)\:|\:(x,y)\in{R}\}\subset{Y\times X}

is called the inverse or converse relation of R. If R=\varnothing, then the inverse of R is defined by R^{-1}=\varnothing.

EXAMPLE3: Find the inverses of the relations in Example1:

R_1^{-1}=\emptyset,

R_2^{-1}=\{(1,a)\},

R_3^{-1}=\{(1,b),(2,b),(2,c)\},

R_4^{-1}=Y\times{X}.

DEFINITION3: Let X,Y,Z be three sets and R\subset{X\times{Y}}, S\subset{Y\times{Z}} be two relations. The relation defined as

S\circ R=\{(x,z)\:|\:\exists y\in{Y}: (x,y)\in{R}\land (y,z)\in{S}\}\subset{X\times{Z}}

is called the composition of the relations R and S.

EXAMPLE4: X=\{a,b,c\}, Y=\{1,2,3\}, Z=\{x,y,z\}, R_1=\{(a,1),(a,2),(b,3)\}, R_2=\{(b,2),(b,3)\}, S_1=\{(1,x),(1,z)\} and S_2=\{(2,z),(3,x)\}. Hence:

S_1\circ{R_1}=\{(a,x),(a,z)\},

S_2\circ{R_1}=\{(a,z),(b,x)\},

S_1\circ{R_2}=\varnothing,

S_2\circ{R_2}=\{(b,z),(b,x)\}.

DEFINITION4: Let X be a set and R\subset{X\times{X}}. (Note that the domain and the codomain of the relation R are equal. The following definitions can be given for the relations over a set X. It can’t be given for the relations between two difference sets)

a) If xRx for all x\in{X}, then the relation R is called “reflexive”.

b) If it holds “xRy\Rightarrow{yRx}” for x,y\in{X}, then the relation R is called “symmetric”.

c) If it holds “xRy\land{yRx}\Rightarrow{x=y}” for x,y\in{X}, then the relation R is called “antisymmetric”.

d) If it holds “xRy\land{yRz}\Rightarrow{xRz}” for x,y,z\in{X}, then the relation R is called “transitive”.

DEFINITION5: Let X be a set and R\subset{X\times{X}}. If the relation R is reflexive, antisymmetric and transitive, then the relation R is called a “partial order relation” and denoted by R=\le in general. If “\le” is a partial order relation over a set X, then (X,\le) is called “partially ordered set” or shortly “poset”.

DEFINITION6: Let X be a set and R\subset{X\times{X}}. If the relation R is reflexive, symmetric and transitive, then the relation R is called an “equivalence relation” and denoted by R=\sim in general.

]]> http://www.academicmaths.com/analysis/relation.html/feed 0 Index Set http://www.academicmaths.com/analysis/index-set.html http://www.academicmaths.com/analysis/index-set.html#comments Sat, 18 Sep 2010 18:17:56 +0000 ufukkaya http://www.academicmaths.com/?p=18 Let X be a set. A set is called an “index set” of the set X if X and that set are equipollent. I.e., a set I is called an index set of X if there exists a bijective function (injective and surjective) between I and X. As is clear from the definition, the number of all the index sets of a set may be more than one, even infinite. Besides, we can say from the definition: there exists at least one index set of any set. Because, the function I_{X}:X\to{X} is bijective. This is a trivial example since the set X is indexed by itself. If we indexed every set with itself, the indexing operation would be meaningless. We must give a reinforcement example: Let be X=\{\diamondsuit, \heartsuit, \clubsuit, \spadesuit\}. We will give three index sets for this set: The sets I_1=X, I_2=\{1,2,3,4\} and I_3=\{a,b,c,d\} can be given as index sets of X. (Can be given more index sets for this set). The type of index set that is widely used by the mathematicians is I_{2} since it one by one counts the elements of X. In general, I=\{1,2,\dots,n\} can be chosen as an index set for a set with n elements. An index set of a set is directly associated with the cardinality of that set. Since the cardinality relation on any family of sets is an equivalence relation, an index set of a set can be actually considered as the most reasonable “representation of class” of the equivalence class of a set. For example, the most reasonable index set of any countable set is naturally the set of the natural numbers. \mathbb{R}, [0,1] or (0,1) is widely used as an index set for a set that is equipollent with the set of the real numbers.

]]> http://www.academicmaths.com/analysis/index-set.html/feed 2 Russell’s Paradox http://www.academicmaths.com/analysis/russells-paradox.html http://www.academicmaths.com/analysis/russells-paradox.html#comments Sat, 18 Sep 2010 18:09:35 +0000 ufukkaya http://www.academicmaths.com/?p=13 By the end of the 19th century, the mathematicians had called “set” a collection of  any objects. For example, the set of the natural numbers, the set of integers, the set of even numbers, the set of the real numbers, the set of any sets, the set of all the sets. We can give further similar examples. All the mathematicians had no doubt about that the unique condition to be a set was to gather any objects by  the time Bertrand Russell’s paradox emerged. Russell had proved that when the term “set” is defined as “a collection of any objects”, there emerges a paradox in the set theory . Now, let’s examine Russell’s paradox and its proof:  Assume that a collection of any objects is a set. In that case, the collection of all the sets is a set. We denote this set by X. Hence, any set is an element of the set X i.e., if A is a set, then A\in{X}. Since X is also a set, then X\in{X}. Let’s construct a subset of X:

Y=\{A\in{X}\:|\:A\notin{A}\}.

Which proposition of the two is the true one Y\in{Y} or Y\notin{Y}?

i) Let’s assume that the proposition Y\in{Y} is the true one. In that case, since any element of Y is a set that is not an element of itself, the proposition Y\notin{Y} is true.

ii) Let’s assume that the proposition Y\notin{Y} is the true one. In that case, according to the definition of Y, the proposition Y\in{Y} is true. As a result, the following proposition has been proved:

Y\in{Y}\Leftrightarrow{ Y\notin{Y}}.

This is obviously a contradiction.

]]> http://www.academicmaths.com/analysis/russells-paradox.html/feed 1 Set Theory http://www.academicmaths.com/analysis/set-theory.html http://www.academicmaths.com/analysis/set-theory.html#comments Sat, 18 Sep 2010 12:55:09 +0000 ufukkaya http://www.academicmaths.com/?p=6 The concept of the “set” is one of the basic concepts of mathematics. However, in spite of this fact, there is no definition agreed on by the authority. Some mathematicians define sets as “the class of objects that have certain properties”. Although this definition is widespread, there are deficiencies.

Objects that form the set are called “elements”. The sets are represented with capital letters such as A, B, C, X, Y, and the elements of sets are represented with lower-case letter such as a, b, c, x, y. If a is an element of A, this case is denoted by a\in A, if a is not an element of A, this case is denoted by a\notin A. There are three types of representations to display the sets:

1. List method: In this representation, the elements of set are written into the curly braces, by putting the commas between the elements. In a set, an element cannot be written twice. As an example, A=\{a,b,c,d,e\} can be given.

2. Venn diagram: In this representation, the elements of set are written inside a circle or rectangle. Let’s show the above example, A=\{a,b,c,d,e\}, by using Venn diagram:

Although the two representations above seem useful, their uses are limited because these representations can only and only be used for finite sets. Because it is generally studied on the infinite sets in mathematics, the following third representation is used in common. However, there are cases that the two representations above are also useful. For example, the first representation is useful to show the finite sets and, the second representation is useful to show relations and functions between two finite sets.

3. Common properties method: In this representation, the elements providing a specific proposition or propositions are collected in a set, is denoted by \{x\:|\: x, \text{provides that the proposition} \: p\}. (Sometimes the form \{x\:{:}\: x, \text{provides that the proposition} \: p\} is also used) Let X be a set. The elements which belong to X and provide the proposition p is denoted by \{x\in{X}\:|\: x, \text{provides that the proposition} \: p\}. Of course other letters can be used instead of x. For example, the set of odd integers that is greater than 100 is denoted by \{ n \in \mathbb{Z}\:|\: n \: \text{is odd and} n>{100} \}. This set can also be denoted by \{ n \in \mathbb{Z} : n>{100} \land \exists k \in \mathbb{Z} : n=2k+1 \}.

DEFINITION1: The set that has no element is called the emptyset. The emptyset is denoted by \varnothing or \{\}.

DEFINITION2: Let A and B be two sets. If each element of A is also an element of B, namely; \forall x \in{A}, x \in{B}, then A is called a subset of B. This case is denoted by A\subset{B} (or A\subseteq{B}). The emptyset is subset of any set. (\varnothing \subset{A}) Furthermore, any set is subset of itself. (A\subset{A}) If A is a subset of B and B has the elements that they are not in A, then A is called a proper (or strict) subset of B. If A is a proper subset of B, this case is denoted by A \varsubsetneqq{B}.

DEFINITION3: Let A and B be two sets. If A\subset{B} and B\subset{A}, A and B are called equal sets. This case is denoted by A=B.

DEFINITION4: Let A and B be two sets. Now, let’s form new sets by using A and B.

The set defined as

A \cup{B}=\{ x\:|\: x \in{A}\lor x \in{B} \}

is called the union of A and B.

The set defined as

A \cap{B}=\{ x\:|\: x \in{A}\land x \in{B} \}

is called the intersection of A and B.

The set defined as

A \setminus{B}=\{ x\:|\: x \in{A}\land x \notin{B} \}

is called the difference of A and B.

The set defined as

A \bigtriangleup{B}=(A \setminus{B}) \cup{(B \setminus{A})}

is called the symmetric difference of A and B.

If A \cap{B}=\varnothing, then A and B are called disjoint sets.

Now, let’s mention a few elementary features of these operations:

1) In the union, intersection and symmetric difference operations, A and B can be replaced by each other (A\cup{B}=B\cup{A}, A\cap{B}=B\cap{A}, A\bigtriangleup{B}=B\bigtriangleup{A}). But, in the difference operation, A and B may not be replaced by each other. As it is clear from the definition of difference operation, A \setminus{B} set contains the elements that are in A and not in B. Now, let’s give an example that shows the case in which A and B may not be replaced by each other: We take A=\{a,b,c\} and B=\{b,c,d\}. Because A\setminus{B}=\{a\} and B\setminus{A}=\{d\}, we obtain A\setminus{B} \ne B\setminus{A}.

2) The union, intersection and symmetric difference operations are associative:

(A\cup{B})\cup{C}=A\cup{(B\cup{C})},

(A\cap{B})\cap{C}=A\cap{(B\cap{C})},

(A\bigtriangleup{B})\bigtriangleup{C}=A\bigtriangleup{(B\bigtriangleup{C})}.

So, in the union, intersection and symmetric difference of three or more sets, the order of sets and which two sets enter the operation are not important.

3) If A is a set, then A\cup{A}=A, A\cap{A}=A, A\setminus{A}=\varnothing and A\bigtriangleup{A}=\varnothing.

4) If A and B are two sets, then A\subset{A\cup{B}}, B\subset{A\cup{B}}, A\cap{B}\subset{A} and A\cap{B}\subset{B}.

DEFINITION5 (FAMILY OF SETS): Let I be an index set and \forall i \in{I}, A_i be sets. So \mathcal{A}=\{A_i : i \in{A} \} is called a family of sets. Generally, a family of sets is denoted by a shorter form of \mathcal{A}=\{A_i\}_{i\in{I}}. Actually, a family of sets is a set that its elements are sets, i.e. it is a set of sets. But, because using the statement of “set of sets” may cause confusion, the statement of “family of sets” is used in general. Let’s give an example: Let be I=\{1,2,3,4\}. (Because this is an example, I have chosen the index set that has four elements. But the index set can be chosen as a set that has fewer elements, more elements, finite elements, infinite elements or can even be chosen the emptyset) Let be A_1=\mathbb{N}, A_2=\mathbb{Z}, A_3=\mathbb{Q} and A_4=\mathbb{R}. Then \mathcal{A}=\{ \mathbb{N}, \mathbb{Z}, \mathbb{Q}, \mathbb{R} \}. There is a confusing problem: \mathcal{A} is a family of sets and has four elements. (I.e. it is a finite set) \mathbb{R} \in{\mathcal{A}}. The set of real number being infinite doesn’t make \mathcal{A} infinite. We should consider that the set of real numbers is an ordinary element of \mathcal{A}. If the set of real numbers was a subset of \mathcal{A}, we would say “\mathcal{A} is infinite”. But, because this is not true, \mathcal{A} is a finite set.

DEFINITION6 (THE UNION AND INTERSECTION OF FAMILY OF SETS): Let \{A_i\}_{i\in{I}} be a family of sets.

The set defined as

\displaystyle{\bigcup_{i\in{I}}A_{i}=\{x\:|\: \exists{i}\in{I}:\:x\in{A_{i}}\}}

is called the union of \{A_i\}_{i\in{I}} and the set of defined as

\displaystyle{\bigcap_{i\in{I}}A_{i}=\{x\:|\: \forall{i}\in{I},\:x\in{A_{i}}\}}

is called the intersection of \{A_i\}_{i\in{I}}. One can easily see this result from the definitions: The elements that are included in at least one of the sets of {A_{i}}_{i\in{I}} form the set of union, and the elements that are included in all the sets of \{A_i\}_{i\in{I}} form the set of intersection. In the event of I=\mathbb{N}, the set of union of \{A_i\}_{i\in{I}} is denoted by

\displaystyle{\bigcup_{i=1}^{\infty}A_{i}}

and the set of intersection of \{A_i\}_{i\in{I}} is denoted by

\displaystyle{\bigcap_{i=1}^{\infty}A_{i}}.

Let’s give a generalization of property 4th for union and intersection operations: Let be \{A_i\}_{i\in{I}} a family of sets. Then,

\displaystyle{A_{k}\subset\bigcup_{i=1}^{\infty}A_{i}}

and

\displaystyle{\bigcap_{i=1}^{\infty}A_{i}\subset{A_{k}}}

for any k\in{I}.

DEFINITION7: Let \{A_i\}_{i\in{I}} be a family of sets. If A_i \cap{A_j}=\varnothing for any i\ne{j}, then, \{A_i\}_{i\in{I}} is called pairwise disjoint or mutually disjoint. For example, if X\ne{\varnothing} is a set, the family of \{\{x\}\:|\:x\in{X}\} is pairwise disjoint. A more concrete example: The family of \{ [n,n+1) \}_{n=1}^{\infty} is pairwise disjoint because [n,n+1) \cap{[m,m+1)}=\varnothing for any m\ne{n} in \mathbb{R}.

DEFINITION8: Let X be a set. The set \mathbf{P}(X)=\{A\:|\:A\subset{X}\} is called the power set of X. I.e. the power set of a set is the family of sets formed all the subsets of this set. The power set of any set is always non-empty. Because the emptyset is a subset of any set (\varnothing\subset{X}), the emptyset is an element of \mathbf{P}(X) for any set X. For example, if X=\varnothing then \mathbf{P}(X)=\{\varnothing\}, if X=\{1\} then \mathbf{P}(X)=\{\varnothing,\{1\}\}, if X=\{1,2\} then \mathbf{P}(X)=\{\varnothing,\{1\},\{2\},\{1,2\}\} and, if X=\{1,2,3\} then \mathbf{P}(X)=\{\varnothing,\{1\},\{2\},\{3\},\{1,2\},\{1,3\},\{2,3\},\{1,2,3\}\}.

DEFINITION9: Let A be a set and \{A_i\}_{i \in{I}}\subset{\mathbf{P}}(A) be a family of the subsets of A. If the following two conditions are provided, the family \{A_i\}_{i \in{I}} is called a partition of A:

i) \displaystyle{\bigcup_{i\in{I}}A_{i}=A}

ii) The family \{A_i\}_{i \in{I}} is pairwise disjoint.

There exists at least one partition of any set, because the family \{\{x\}\:|\:x\in{A}\} is a family providing the conditions (i) and (ii) for any set A. This is a trivial example. Let’s give two concrete examples:

EXAMPLE1: If A=\mathbb{R} and I=\mathbb{Z}, then, the family A_n=[n,n+1), n\in{\mathbb{Z}} is a partition of \mathbb{R}, because,

i) \displaystyle{\bigcup_{n=-\infty}{\infty}[n,n+1)=\mathbb{R}} and,

ii) [n,n+1)\cap{[m,m+1)}=\varnothing for any n\ne{m} in \mathbb{Z}.

EXAMPLE2: Let n\in{\mathbb{N}} be an arbitrary constant. Let r(m) denote the remainder when dividing m by n. We consider A_k= n\mathbb{Z}+k=\{m\in{\mathbb{Z}} \:|\: r(m)=k\} for 0\le k \le n-1. Because,

i) \displaystyle{\bigcup_{k=0}^{n-1}\left( n\mathbb{Z}+k \right)=\mathbb{Z}} and,

ii) A_k \cap{A_l}=\varnothing for any k \ne l in \{0,1,\dots,n-1\}, the family \{n\mathbb{Z}+k\:|\:k=\overline{0,n-1}\} is a partition of \mathbb{Z}.

DEFINITION10 (UNIVERSAL SET): The set including all the sets is called universal set and denoted by \mathbb{E} in general. Actually, Russell’s paradox proofs that there is not a set including all the sets. So, in spite of the collection denoted by the form of \mathbb{E} is universal, we can consider that it is not a set. It’s true that \mathbb{E} is not a set. However, \mathbb{E} can be turned into a set by restricting the definition of the universal set. Let’s change the definition of the universal set for restricting: “The set including all the sets that we study on is called universal set”. What is the meaning of this? Let’s explain the definition by giving some examples: For example, we consider that you are studying on sequences in the set of the real numbers. Therefore, you aren’t going out of the set of the real numbers. I.e. the largest set that you are studying on is the set of the real numbers and any other set that you are studying on is a subset of the set of the real numbers. In that case, your universal set is the set of the real numbers. Let’s give another example: we consider that you are studying on the prime numbers. In that case, your universal set is the set of the integers. As is seen, actually, the universal set depends on the topics that you are studying on at that time, i.e, depends on your choosing. The choosing of the universal set is rather important for the concept of the “complement”. Now, let’s give the definition of the complement:

DEFINITION11: Let A be a set. Then, the set A^{C}=\mathbb{E}\setminus{A} is called the complement of A. (Where \mathbb{E} is the universal set.) I.e. A^{C} is formed the \mathbb{E}’s elements that are not in A. The complement of A is denoted by A' in some sources. Let’s explain the importance of the choosing of the universal set by the help of an example:

EXAMPLE3: Let A=\{1,2\}. If \mathbb{E}=\mathbb{N}, then, A^C=\{2,3,\dots\}, and if \mathbb{E}=\mathbb{R}, then, A^C=(-\infty,0)\cup{(0,1)\cup{(1,+\infty)}}.

PROPOSITION1: Let A\subset{\mathbb{E}} and \{A_i\}_{i\in{I}}\subset{\mathbb{E}}. Then, the following five propositions are true:

i) \varnothing^C=\mathbb{E},

ii) \mathbb{E}^C=\varnothing,

iii) (A^C)^C=A,

iv) \displaystyle{\left( \bigcup_{i\in{I}}A_{i} \right)^{C}=\bigcap_{i\in{I}}(A_{i})^{C}},

v) \displaystyle{\left( \bigcap_{i\in{I}}A_{i} \right)^{C}=\bigcup_{i\in{I}}(A_{i})^{C}}.

(iv) and (v) are called De Morgan’s laws. The law (iv) turn into (A\cup{B})^C=A^C\cap{B^C} and the law (v) turn into (A\cap{B})^C=A^C\cup{B^C} in the case I=\{1,2\}.

PROOF:

DEFINITION12: Let X,Y\ne\varnothing be two sets, x\in{X} and y\in{Y}. The set defined as (x,y)=\{x,\{x,y\}\} is called an ordered pair. The following “equality of two ordered pairs” is true for x_1,x_2\in{X}, y_1,y_2\in{Y}:

(x_1,y_1)=(x_2,y_2)\Leftrightarrow x_1=x_2\land{y_1=y_2}.

This equality can be easily proved by using the equality of two sets.

The equality (x,y)=(y,x) is not true in general. By using the equality of two ordered pairs, the proposition

(x,y)=(y,x)\Leftrightarrow x=y

can be easily obtained.

DEFINITION13: Let X,Y be two sets. The set defined by X\times{Y}=\{(x,y) \:|\: x\in{X}, y\in{Y}\} is called the Cartesian product of X and Y. In general X\times{Y}\ne Y\times{X}. More generally, the following proposition is provided:

X\times{Y}= Y\times{X}\Leftrightarrow X=Y\lor X=\varnothing\lor Y=\varnothing.

If X has n elements and Y has m elements, X\times{Y} has n.m elements.

PROBLEMS AND SOLUATIONS ON THE SET THORY

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