The real numbers. Supremum and Infimum of a set.




ЗНАЕТЕ ЛИ ВЫ?

The real numbers. Supremum and Infimum of a set.



Some set theory.

Answer:

Definition: Let S and T be set:

a) S contains T, and we write S c T or T c S, if every number of T is also in S. in this case, T is a subset of S.

b) S\T is the set of elements that are in S, but not in T.

c) S equals T, and we write: S=T, is S contains T, and T contains S, thus, S=T if and only if S and T have the same members.

d) The compecment of S denoted by Sc , is the set of elements in the universal set that are not in S.

e) The union of S and T, denoted by S T, is the set of elements in at least one of S and T.

f) The intersection of Sand T , denoted by S T is the set of element in both S and T. If S T 0(the empty set), then S and T are disjoint sets.

Definition If x0 is real number and, ε>0, then the open interval (x0-ε,x0+ε) in an neighborhood of x0.

If the set S contains an ε-neighborhood of x0 , then S is a neighborhood of x0, and x0 is an interior point of S.

Definition. Let S be subset of R. Then (a) x0 is a limit-point of S, if every deleted neighborhood of x0 contains a point of S.

(b) x0 is boundary point of S, if every neighborhood of x0 contains a least one point in S and one not in S. The set of boundary points of S denoted by S

The closure of S, denoted by =S S

(c)x0 is exterior to S, of x0 is in the interior of Sc

Theorem. A set Sis closed, if and only if no point of Scis a limit point of S.

Corollary. A set is closed, if and only if it contains all its limit points.

Definition . A collection H of open sets, is an open covering of a set S, if every point in S is contained

in a set H belonging to H, that is, if Sc {H|HcH}

 

Heine-borel Theorem. Bolzano-Weierstrass Theorem.

Answer:

If H is an open covering of a closed and bounded subset S of the real line, then S has an open covering H consisting of finitely many open sets belonging to H.

Proof: Since S bounded, it has an infimumα and supremum β , and, since S is closed,α and β belong to S. Define St=S [α;t] for ≥α.

and let F={t|α<t<β and finitely many sets form H cover St}

Since Sβ =S1 the theorem will be proved, if we can show that β F. To do this, we use the completeness of the reals.

Since α S, Sαis the singleton set {α}, which is contained in some open set Hαform H, because H covers S; therefore, αcF.

Since, F is nonempty and bounded above by β, it has supremumλ. First, we wish to show that λ=β by definition of F , it suffices to rull out the possibility that α<β.

We consider two cases:

Case 1. Suppose that α<β and λ S. Then, since S is closed λ is not a limitpointof S. Consequently, there is an ε>0 such that [λ-ε, λ+ε] S= , so Sλ=Sλ+ε

However, the definition of λ implies that Sλ-ε has a finite subcovering from H, while Sλ+εdoes not. This is a contradiction.

3.2 Case 2.Suppose that λ<β and λ S. Then there is an open set Hλ in H that contains λ and along with λ, an interval [λ-ε, λ+ε] for some positive ε. Since Sλ-ε has a finite covering.{H1,…,Hn} of sets from H, it follows, that Sλ+ε has the finite covering{H1,….,Hn,Hλ}. This contradicts the definition of λ.

Now we know that α=β, which is in S. Therefore, there is an open set Hβ in H that contains β and along with, an interval of the from [β-ε,β+ε], for some positive ε.

Since Sβ-ε is covered by a finite collection of set {H1,…,Hk}, Sp is covered by the finite collection {H1,…,Hk,Hβ}. Since Sβ=S1 .

Henceforth , we will say a closed and bounded set is compact.

As an application of the Heine-Borel theorem, we prove the following theorem of Bolzano andWeierstrass.

Bolzano-Weierstrass theorem. Every boumded infinite set of real numbers has at least one limit point.

Proof: We will show that a bounded nonempty set without a limit point can contain only a finite number of points.

If S has no limitpoints,then S is closed and every point x of S has an open neighborhood Nx that contains no point of other than x. The collection H={Nx| x S} is an open covering for S. Since S is also bounded Heine-Borel theorem implies that S can be covered by a finite collection of sets from H1,…. say Nx1,…,Nxn. Since these set contain only x1,x2,…xn from S, it follows that S={x1,…,xn}

 

Taylor’s formula.

Answer:

A polynomial is of the form P(x)=Q0+Q1(x-x0)+…+an(x-x0)n (1)

where Q0,Q1,…Qn and x0 are constants.

In particular, a constant polynomial P(x)=Q0 is of degree zero, if Q0 0

If f is differentiable all x0 , then f(x)=f(x0)+f’(x0)(x-x0)+ (x-x0)

Where =0

The polynomial T1(x)=f(x0)+f’(x0)(x-x0) which is of degree 1 and satisfies T1(x)=f(x0) T1(x0)=f’(x0), approximates f so well near x0 that

=0 (2)

Now suppose that f has n derivatives at x0 and Tn is the polynomial of degree n such that

Tn(r)(x0)=f(r)(x0) 0 r n (3)

Tn is polynomial of degree n;

Tn(x)=a0+a1(x-x0)+…+an(x-x0)n (4)

Differentiating (4) yields Tn(r)(x0)=r!ar

So (3) determines ar uniquely as ar=f(r)(x0)/r!

Therefore,

Tn(x)=f(x0)+ (x-x0) + (x-x0)2 + …+ (x-x0)n = (x-x0)r

We callTn the n-th Taylor polynomial of f about x0 (xxx дегенимиз x3)

1. sinx=x – +– -– +….

2. cosx=1 – – + – – – + – - …

3. ln(1+x)=x – x2/2 + x3/3 – x4/4 + x5/5 - …

4. ex= 1+ x + x2/2! +x3/3! + x4/4! + …

5. (1+x)a=1 + ax + a(a-1)x2/2! + a(a-1)(a-2)x3/3! + … + xa

The real numbers. Supremum and Infimum of a set.

Answer:

The real number system is first of all a set {a,b,c,…} on which the operation of addition and multiplication a defined that every pair of real numbers has a unique sum and product, both real numbers,with the following properties,

A) a+b=b+a and ab=ba (commutative laws)

B) (a+b)+c=a+(b+c) (ab)c=a(bc) (associative laws)

C) a(b+c)=ab+ac (distributive law)

D) There are distinct real numbers 0 and 1 such that a+0=a and a1=a for all a

E) For each (a) there is a real number –a such that a+(-a)=0 and if a 0 there is a real number such that a =1

A set on which two operations are difined so as to have properties (A)-(E) called a field. The simplest possible field consist of two elements which we denote by 0 and 1, with addition defined by 0+0=1+(-1)=0, 1+0=0+1=1 (1) and multiplication defined by 0×0=0×1=1×0=0, 1×1=1 (2)

The order Relation

The real number system is ordered by relation <, which has the following properties:

F) For each pair of real numbers (a) and (b) exactly one of the following is true: a=b ,a<b or a>b

G) If a<b and b<c, then a<c (The relation < is transitive)

H) If a<b,thena+c<b+c for any c and if c>0, then a×c<b×c.

A field with an order relation satisfying (F)-(H) in an ordered field.

Theorem 1 (The Triangle Inequality). If aandb are any two real numbers then:

|a+b|≤|a|+|b| (3)

Corollary 2. If a and b are any two real numbers then |a-b|≥||a|-|b|| (4) and |a+b|≥||a|+|b|| (5) supremum of a set:

A set S of real numbers is bounded above if there is a real number b such that x≤b whenever x S. In this case, b is an upper bound of S. If b is an upper bound of S, then so is any large number, because of property (6)

If is an upper bound of S, but no number less than β is, then β is a supremum of S and we write β=supS

Example: If S is the set of negative numbers, then any nonnegative number is an upper bound of S,and sup=0

1.2 If S1 is the set of negative integers,then any number a such that a≥-1 is an upper bound of S1 , sup=-1

This example shows that a supremumof a set may or may not be in the set, since S1 , contains its supremum, but S does not.

A nonempty set is a set that has at least one number. The empty set, denoted by , is the set that has no numbers.

The Completeness Axiom.

(I) If a nonempty set of reak numbers is bounded above, then it has a supremum.

Property (I) is called conpleteness and we say that the real number system is a complete ordered field.

Thorem 3. If a nonempty set S of real numbers is bounded above, then sups is the unique real number β such that (a) x≤β for all x in S.

(b) if ε>0 there is an x0 in S such that x0>β-ε

Proof: we first show that β=supS has properties (a) and (b). Since β is an upper bound of it must satisfy (a). Since any real number a less than β can be written as β-ε with ε=β-a>0, (b) is just another way of saying that no number less than β is an upper bound of S, β=supS (satisfies (a) and (b))

Now we show that, there cannot be more than one real number with properties (a) and (b). Suppose thatβ12 and β2 has property (b) thus, if ε>0, there is an x0 in S such that x0> 2-ε. Then by taking ε=β21 , we see that there is an x0 in S such that x02-(β21)=β1,

So β1 cannot have property (a). Therefore, there cannot be more than one real number that satisfies both (a) and (b).

Some Natation: “x is a member of S”=>x S “x is not a member of S” => x S.

Theorem 4. (The ArchemedeanProperty ). If p and ε are positive, then nε>p for some integer n.

Proof: The proof is by contradiction. If the statement is false, p is an upper bound of the set S={| x | x = nε, n is a integer}. Therefore, S has a supremum β, by property (I). Therefore, nε≤β (b)

1.3For all integer whenever n is , (b) implies that (n+1)ε≤β and therefore nε≤β-ε for all integers n. Hence, β-ε is an upper bound of S. Since, β-ε<β, this contradicts the definition of β.

Infimum of a set.

A set S of real numbers is bounded below if there is a real number a such that x≥a, whenever x S. In this case, a is a lower bounded of S.

but no number greater than α is, then α is an infimum of S, and we write: α=infS.

Theorem 8. If nonempty set S of real numbers is bounded below, then infS is the unique real number α such that:

(a) x≥α for all x in S.

(b) If ε>0, there is an x0 in such that x0< α+ε

Proof: A set S in bounded, of there are numbersa and b such that a≤x≤b for all x in S.

A bounded nonempty set has a unique supremum and a unique infimum, and: infS≤supS.

A nonempty set S of real numbers is unbounded above if it no upper bound, or unbounded below, if it has no lower bound - < x <+ (7)

We call points of infinity. If (S)is a nonempty set of real numbers, we write: supS= (8)

to indicate that S is unbounded above, and infS=- (9)

to indicate that S is unbounded below.

(a) If a is any real number, then: a+ = +a= , a- =- +a= - , = =0

(b) If a>0, then: a = a= , a(- )=(- )a=-

(c) If a<0 , then: a = a= , a(- )=(- )a=

We also define: + = =(- )(- )= and - - = (- )=-

Finally, we define | |=|- |= .

 

 

Some set theory.

Answer:

Definition: Let S and T be set:

a) S contains T, and we write S c T or T c S, if every number of T is also in S. in this case, T is a subset of S.

b) S\T is the set of elements that are in S, but not in T.

c) S equals T, and we write: S=T, is S contains T, and T contains S, thus, S=T if and only if S and T have the same members.

d) The compecment of S denoted by Sc , is the set of elements in the universal set that are not in S.

e) The union of S and T, denoted by S T, is the set of elements in at least one of S and T.

f) The intersection of Sand T , denoted by S T is the set of element in both S and T. If S T 0(the empty set), then S and T are disjoint sets.

Definition If x0 is real number and, ε>0, then the open interval (x0-ε,x0+ε) in an neighborhood of x0.

If the set S contains an ε-neighborhood of x0 , then S is a neighborhood of x0, and x0 is an interior point of S.

Definition. Let S be subset of R. Then (a) x0 is a limit-point of S, if every deleted neighborhood of x0 contains a point of S.

(b) x0 is boundary point of S, if every neighborhood of x0 contains a least one point in S and one not in S. The set of boundary points of S denoted by S

The closure of S, denoted by =S S

(c)x0 is exterior to S, of x0 is in the interior of Sc

Theorem. A set Sis closed, if and only if no point of Scis a limit point of S.

Corollary. A set is closed, if and only if it contains all its limit points.

Definition . A collection H of open sets, is an open covering of a set S, if every point in S is contained

in a set H belonging to H, that is, if Sc {H|HcH}

 





Последнее изменение этой страницы: 2016-08-26; Нарушение авторского права страницы

infopedia.su Все материалы представленные на сайте исключительно с целью ознакомления читателями и не преследуют коммерческих целей или нарушение авторских прав. Обратная связь - 54.208.73.179 (0.017 с.)