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 S^c , 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 S^c

Theorem. A set Sis closed, if and only if no point of S^cis 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 S_t=S [α;t] for ≥α.

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

Since S_β =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, S_αis the singleton set {α}, which is contained in some open set H_αform 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 S_λ-ε=S_λ+ε

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

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

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

Since S_β-ε is covered by a finite collection of set {H₁,…,H_k}, S_p is covered by the finite collection {H₁,…,H_k,H_β}. Since S_β=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 N_x that contains no point of other than x. The collection H={N_x| 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 N_x1,…,N_xn. Since these set contain only x₁,x₂,…x_n from S, it follows that S={x₁,…,x_n}

 

Taylor’s formula.

Answer:

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

where Q₀,Q₁,…Q_n 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) T^’₁(x₀)=f’(x₀), approximates f so well near x₀ that

=0 (2)

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

T_n^(r)(x₀)=f^(r)(x₀) 0 r n (3)

T_n is polynomial of degree n;

T_n(x)=a₀+a₁(x-x₀)+…+a_n(x-x₀)ⁿ (4)

Differentiating (4) yields T_n^(r)(x₀)=r!a_r

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

Therefore,

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

We callT_n 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. e^x= 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! + … + x^a

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 thatβ₁<β₂ and β₂ has property (b) thus, if ε>0, there is an x₀ in S such that x₀> ₂-ε. Then by taking ε=β₂-β₁ , we see that there is an x₀ in S such that x₀>β₂-(β₂-β₁)=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 S^c , 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 S^c

Theorem. A set Sis closed, if and only if no point of S^cis 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}