(Partially and Totally Ordered Sets.) (1) the Ordered Pair (X, Ρ) Is Called a Partially Ordered Set If X Is a Set and Ρ Is a Partial Order Relation in X

Home , Bounded set, Partially ordered set

1. ORDERED SETS

Deﬁnitions. (Partially and totally ordered sets.) (1) The ordered pair (X, ρ) is called a partially ordered set if X is a set and ρ is a partial order relation in X. (2) The ordered pair (X, ρ) is called a totally ordered set (or linear ordered set) if X is a set and ρ is a total (linear) order relation in X .

Let (X, ρ) be a (partially or totally) ordered set and S ⊂ X .

Deﬁnitions. (Upper and lower bounds, bounded set.) (1) u ∈ X is called an upper bound for S if ∀x ∈ S (x, u) ∈ ρ . (2) v ∈ X is called a lower bound for S if ∀x ∈ S (v, x) ∈ ρ . (3) S is called bounded above if ∃u ∈ X upper bound for S . (4) S is called bounded below if ∃v ∈ X lower bound for S . (5) S is called bounded if it is bounded both above and below.

Deﬁnitions. (Maximal and minimal elements.) (1) M ∈ S is called a maximal element of S if there is no x ∈ S such that x 6= M and (M, x) ∈ ρ . (2) m ∈ S is called a minimal element of S if there is no x ∈ S such that x 6= m and (x, m) ∈ ρ .

Deﬁnitions. (Greatest and least elements.) (1) M ∈ S is called the greatest element of S if ∀x ∈ S (x, M) ∈ ρ . (2) m ∈ S is called the least element of S if ∀x ∈ S (m, x) ∈ ρ .

Definitions. (Supremum and infimum.) (1) If S is bounded above and the set of all upper bounds for S has a least element, we call it the least upper bound or the supremum for S , and denote it by sup S . (I.e., sup S ∈ U := { u ∈ X : ∀x ∈ S (x, u) ∈ ρ } and ∀u ∈ U (sup S, u) ∈ ρ .) (2) If S is bounded below and the set of all lower bounds for S has a greatest element, we call it the greateast lower bound or the infimum for S , and denote it by inf S . (I.e., inf S ∈ V := { v ∈ X : ∀x ∈ S (v, x) ∈ ρ } and ∀v ∈ V (v, inf V ) ∈ ρ .)

Example. Let ρ be the divisibility in N+. Then (N+, ρ) is a partially ordered set. 1 is the only minimal (and least) element of N+, 1 is also the inﬁmum for N+. The primes are minimal elements of N+\{1}, there is no least element of N+\{1}, 1 is the inﬁmum for N+\{1}.

1 2. REAL NUMBERS

The system of axioms for the ordered ﬁeld R of real numbers is given, as follows:

R is a set whose elements are called real numbers. There exists distinguished elements 0 ∈ R and 1 ∈ R . Two operations, called addition and multiplication are deﬁned in R , for which the notations (x, y) 7→ x+y and (x, y) 7→ x · y are used, respectively. A total order relation, denoted by ≤, is deﬁned in R, (i.e. (R, ≤) is a totally ordered set). (We use the notation x ≤ y instead of (x, y) ∈ ≤ , and write x < y if x ≤ y and x 6= y.)

The following axioms are satisﬁed:

(1) ∀ x, y, z ∈ R (x + y) + z = x + (y + z), (the addition is associative) (2) ∀ x ∈ R x + 0 = x, (3) ∀ x ∈ R ∃u ∈ R x + u = 0, ( −x := u , y − x := y + (−x)) (4) ∀ x, y ∈ R x + y = y + x, (the addition is commutative) (5) ∀ x, y, z ∈ R (x · y) · z = x · (y · z), (the multiplication is associative) (6) ∀ x ∈ R x · 1 = x, (7) ∀x ∈ R\{0} ∃u ∈ R x · u = 1, ( 1/x := u , y/x := y · 1/x ) (8) ∀ x, y ∈ R x · y = y · x, (the multiplication is commutative) (9) ∀ x, y, z ∈ R x · (y + z) = x · y + x · z, (the multiplication is distributive with respect to the addition)

(10) ∀ x, y, z ∈ R x ≤ y ⇒ x + z ≤ y + z, (11) ∀ x, y, z ∈ R ( x ≤ y and 0 ≤ z ) ⇒ x · z ≤ y · z,

(12) If a nonempty set S ⊂ R is bounded above, then the set of all upper bounds for S has a least element, (least upper bound for S). I.e., ∃ sup S ∈ R, (i) ∀x ∈ S x ≤ sup S , (ii) ∀u ∈ R ( ∀x ∈ S x ≤ u ) ⇒ sup S ≤ u . ( ∀y ∈ R y < sup S ⇒ ∃ x ∈ S y < x . )

Remarks. (i) Axioms (1)–(9) are the axioms for a ﬁeld . Axioms (1)–(11) are the axioms for an ordered ﬁeld . Axiom (12) is usually called as axiom of least upper bound .

(ii) We show how some elementary laws of algebra and inequalities follow from the axioms (1)–(11) :

2 Proposition 1. ∀x ∈ R x · 0 = 0 . Proof. x·0 = x·0+0 = x·0+(x·0−x·0) = (x·0+x·0)−x·0 = x·(0+0)−x·0 = x·0−x·0 = 0 . ♠ Proposition 2. ∀x, y ∈ R x · y = 0 ⇒ x = 0 or y = 0 . Proof. x 6= 0 ⇒ y = 1 · y = (1/x · x) · y = 1/x · (x · y) = 1/x · 0 = 0 . ♠ Proposition 3. ∀x ∈ R − (−x) = x . Proof. −(−x) = −(−x) + 0 = −(−x) + (−x + x) = (−(−x) + (−x)) + x = 0 + x = x . ♠ Proposition 4. ∀x, y ∈ R − (x + y) = (−x) + (−y) = −x − y . Proof. −(x + y) = −(x + y) + 0 = −(x + y) + (0 + 0) = −(x + y) + ((−x + x) + (−y + y)) = = −(x+y)+((x+y)+(−x−y)) = (−(x+y)+(x+y))+(−x−y) = 0+(−x−y) = −x−y. ♠ Proposition 5. ∀x ∈ R\{0} 1/x 6= 0 and 1/(1/x) = x . Proof. (i) 1/x = 0 ⇒ 1 = x · 1/x) = x · 0 = 0, a contradiction. (ii) 1/(1/x) = 1/(1/x) · 1 = 1/(1/x) · (1/x · x) = (1/(1/x) · 1/x) · x = 1 · x = x . ♠ Proposition 6. ∀x, y ∈ R\{0} 1/(x · y) = 1/x · 1/y . Proof. 1/(x · y) = 1/(x · y) · 1 = 1/(x · y) · (1 · 1) = 1/(x · y) · ((1/x · x) · (1/y · y)) = = 1/(x · y) · ((x · y) · (1/x · 1/y)) = (1/(x · y) · (x · y)) · (1/x · 1/y) = 1 · (1/x · 1/y) = 1/x · 1/y . ♠ Proposition 7. ∀x, y ∈ R x · (−y) = −(x · y). Proof. x · (−y) = x · (−y) + 0 = x · (−y) + (x · y − (x · y)) = (x · (−y) + x · y) − (x · y) = x · (−y + y) − (x · y) = x · 0 − (x · y) = 0 − (x · y) = −(x · y). ♠ Proposition 8. ∀x, y ∈ R (−x) · (−y) = x · y . Proof. (−x) · (−y) = −((−x) · y) = −(y · (−x)) = −(−(y · x)) = y · x = x · y . ♠ Proposition 9. ∀x ∈ R (i) 0 ≤ x ⇒ −x ≤ 0 (ii) x ≤ 0 ⇒ 0 ≤ −x . Proof. (i) 0 ≤ x ⇒ 0 + (−x) ≤ x + (−x) ⇒ −x ≤ 0 (ii) x ≤ 0 ⇒ x + (−x) ≤ 0 + (−x) ⇒ 0 ≤ −x . ♠

3 Proposition 10. ∀x, y, u, v ∈ R ( x ≤ y and u ≤ v ) ⇒ x + u ≤ y + v . Proof. x ≤ y ⇒ x + u ≤ y + u , u ≤ v ⇒ y + u ≤ y + v . Therefore, x + u ≤ y + v . ♠ Proposition 11. ∀x, y, u, v ∈ R ( 0 ≤ x ≤ y and 0 ≤ u ≤ v ) ⇒ x · u ≤ y · v . Proof. ( x ≤ y and 0 ≤ u ) ⇒ x · u ≤ y · u , and ( u ≤ v and 0 ≤ y ) ⇒ y · u ≤ y · v . Therefore, it follows that x · u ≤ y · v . ♠ Proposition 12. 0 < 1 . Proof. 1 < 0 ⇒ 0 < −1 ⇒ 0 = (−1) · 0 < (−1) · (−1) = 1 · 1 = 1 , a contradiction. ♠

Theorem (2.1). The following propositions are equivalent: (1) If a nonempty subset of R is bounded above, then it has a least upper bound in R. (2) If a nonempty subset of R is bounded below, then it has a greatest lower bound in R. (3) If A and B are nonempty subsets of R such that ∀(a, b) ∈ A × B a ≤ b , then ∃ c ∈ R ∀(a, b) ∈ A × B a ≤ c ≤ b . Proof. (1) ⇒ (2) : Let ∅ 6= S ⊂ R be bounded below. Since the set of all lower bounds for S, V := {v ∈ R : ∀x ∈ S v ≤ x } is nonempty and bounded above, we know by (1) that ∃ sup V ∈ R . (i) ∀x ∈ S x is an upper bound for V , therefore sup V ≤ x , thus sup V is a lower bound for S. (ii) If y ∈ R is a lower bound for S, then y ∈ V , thus y ≤ sup V . Hence, sup V is the greatest lower bound for S, i.e. inf S = sup V . (2) ⇒ (3) : Let A and B are nonempty subsets of R such that ∀(a, b) ∈ A × B a ≤ b . Since B is nonempty and bounded below, we know by (2) that ∃ inf B ∈ R , and since ∀a ∈ A a is a lower bound for B, we have a ≤ inf B . Therefore, with c := inf B we obtain ∀(a, b) ∈ A × B a ≤ c ≤ b . (3) ⇒ (1) : Let ∅ 6= S ⊂ R be bounded above. Since the set of all upper bounds for S, U := {u ∈ R : ∀x ∈ S x ≤ u } is nonempty, we know by (3) that ∃ c ∈ R , such that ∀(x, u) ∈ S × U x ≤ c ≤ u . (i) c is an upper bound for S, (ii) c is less or equal than any upper bound for S, therefore, c ∈ U is the least upper bound for S, i.e. ∃ sup S = c . ♠

4 Remark. The propositions (2) and (3) of the theorem (2.1) are known as the ”axiom” of greatest lower bound and Dedekind’s ”axiom”, respectively.

Deﬁnition. (Inductive sets.) A set S ⊂ R is said to be inductive if (i) 1 ∈ S , (ii) x ∈ S implies x + 1 ∈ S .

Deﬁnition. (Positive integers.) The set of all positive integers, denoted by the symbol N+, is deﬁned by T N+ := { S ⊂ R : S is inductive } (⊂ R).

Theorem (2.2). (Principle of mathematical induction.) If S ⊂ N+ is an inductive set, then S = N+. Proof. Since S is an inductive set, we know from the definition of N+ that N+ ⊂ S , thus we have S = N+ . ♠ Definition. (Modified inductive sets.) A set S ⊂ R has the modified inductive property (with respect to m ∈ R) if (i) m ∈ S , (ii) x ∈ S implies x + 1 ∈ S .

Theorem (2.3). (Modified form of mathematical induction.) If S ⊂ N+ is a modified inductive set with respect to m ∈ N+, then ∀n ∈ N+ n ≥ m implies n ∈ S . Proof. Let S˜ := { n ∈ N+ : n < m } ∪ S . Evidently S˜ ⊂ N+ is an inductive set, thus S˜ = N+. Therefore, we have {n ∈ N+ : n ≥ m } ⊂ S˜, and so {n ∈ N+ : n ≥ m } ⊂ S must be also satisfied. ♠

Example. Prove that ∀n ∈ N+ n ≥ 3 ⇒ nn+1 > (n+1)n . Solution. Let S := {k ∈ N+ : k ≥ 3 and kk+1 > (k+1)k}. (i) 3 ∈ S, since 34 > 43. (ii) If k−1 ∈ S, then (k−1)k > kk−1, therefore kk+1 = kk+1 · (k+1)k · (k−1)k / ((k+1)k · (k−1)k ) > > (k+1)k · kk+1 · kk−1 / ((k2 −1)k ) = (k+1)k · ( k2/(k2 −1) )k > (k+1)k . Hence S is an inductive set with respect to 3, thus {n ∈ N+ : n ≥ 3 } ⊂ S. ♠

5 Definitions. (Natural numbers, integers, rational numbers, irrational numbers.) (1) The set of all natural numbers, denoted by N , is defined by N := N+ ∪ {0} . (2) The set of all integers, denoted by Z , is defined by Z := N ∪ { −n : n ∈ N+ } . (3) The set of all rational numbers, denoted by Q , is defined by Q := { x ∈ R : ∃(p, q) ∈ Z × N+ x = p/q } . (4) The set R\Q is called the set of all irrational numbers.

Theorem (2.4). (Archimedes’ theorem.) The set N+ of all positive integers is not bounded above. Proof. Suppose that N+ is bounded above. Then ∃ sup N+ ∈ R, ∀n ∈ N+ n ≤ sup N+. Since sup N+ − 1 is not an upper bound for N+, ∃ m ∈ N+ sup N+ − 1 < m , thus we have sup N+ < m+1 ≤ sup N+, a contradiction. ♠ Remarks. The following propositions are obvious consequences of Archimedes’ theorem: (i) If p ∈ R+ and x ∈ R , then ∃ n ∈ N+ such that x < n · p . (ii) The set Z of all integers is not bounded below.

Theorem (2.5). For all x ∈ R there exists a unique integer, denoted by [x] and called the largest integer in x, such that [x] ≤ x < [x]+1 . Proof. Let x ∈ R and deﬁne the set K := { z ∈ Z : z ≤ x }. K 6= ∅ is bounded above (x is an upper bound for K), thus ∃ sup K ≤ x . We show that sup K ∈ Z : (i) Suppose that sup K/∈ Z , i.e. ∀z ∈ K z < sup K . (ii) Since sup K−1 is not an upper bound for K, ∃ w ∈ K sup K−1 < w < sup K . We have sup K < w+1 , thus w+1 ∈/ K , i.e. x < w+1 . Therefore, ∀z ∈ Z z > w implies z∈ / K , thus we have ∀z ∈ K z ≤ w , a contradiction. Deﬁne [x] := sup K ∈ Z . Evidently, we have [x] ≤ x < [x]+1 . ♠ Remarks. The next two propositions follow immediately from the theorem (2.5). (i) If ∅ 6= K ⊂ Z is bounded above, then sup K ∈ K is the greatest element of K . (ii) If ∅ 6= K ⊂ Z is bounded below, then inf K ∈ K is the least element of K . The following propositions are also consequences of the theorem (2.5).

6 Proposition (2.6). If a, b ∈ R and b−a > 1 , then ∃ z ∈ Z such that a < z < b . Proof. a+1 < b ⇒ a < [a]+1 ≤ a+1 < b . ♠

Proposition (2.7). If a, b ∈ R and a < b , then ∃ r ∈ Q such that a < r < b . Proof. b−a > 0 ⇒ ∃ n ∈ N+ n > 1 / (b−a) ⇒ n · b − n · a > 1 ⇒ ⇒ ∃ z ∈ Z n · a < z < n · b ⇒ a < z/n < b . ♠

Proposition (2.8). If a, b ∈ R and a < b , then ∃ x ∈ R\Q such that a < x < b .

Proof. √ √ √ √ √ a < b ⇒ a · 2 < b · 2 ⇒ ∃ r ∈ Q a · 2 < r < b · 2 ⇒ a < r / 2 < b . ♠

Definition. (The set R of extended real numbers.) The set R is defined by R := R ∪ {−∞, +∞} , where the symbols −∞ and +∞ are introduced such that (i) −∞ < +∞ , (ii) ∀x ∈ R − ∞ < x < +∞ . (Hence (R, ≤) is a totally ordered set, but we emphasize that R is not a field.)

Deﬁnitions. (Intervals in R .) Let a, b ∈ R and a ≤ b . The following subsets of R are called intervals in R with endpoints a and b : (1) (a, b) := { x ∈ R : a < x < b } (open interval) (2) [ a, b ] := { x ∈ R : a ≤ x ≤ b } (closed interval) (3) (a, b ] := { x ∈ R : a < x ≤ b } (semiclosed interval) (4) [ a, b) := { x ∈ R : a ≤ x < b } (semiclosed interval)

The following subsets of R are unbounded intervals in R : (5) (−∞, b) := { x ∈ R : x < b } (open interval) (6) (a, +∞) := { x ∈ R : a < x } (open interval) (7) (−∞, b ] := { x ∈ R : x ≤ b } (closed interval) (8) [ a, +∞) := { x ∈ R : a ≤ x } (closed interval) (We sometimes write (−∞, +∞) instead of R .)

Deﬁnition. (Dense subsets of R.) A set S ⊂ R is said to be dense in R if ∀x ∈ R and ∀ε ∈ R+ the set ( x−ε , x+ε ) ∩ S is not empty. Remarks. The propositions (2.7) and (2.8) show that Q and R\Q are dense in R.

7 The next three propositions follow immediately from the deﬁnitions of bounded sets, supremum and inﬁmum in ordered sets.

Proposition (2.9). Every set S ⊂ R is bounded and has supremum and inﬁmum in R.

Proposition (2.10). Let S ⊂ R ⊂ R be a nonempty set. (1) If S is bounded above in R, then sup S in R equals sup S in R. (2) If S is bounded below in R, then inf S in R equals inf S in R. (3) If S is not bounded above in R, then sup S in R equals +∞. (4) If S is not bounded below in R, then inf S in R equals −∞. Remark. If S ⊂ R ⊂ R is any set, sup S in R and inf S in R will be denoted simply by sup S and inf S, respectively. (The same notations as in R, provided they exist in R.)

Proposition (2.11). Let S ⊂ R ⊂ R be a set. Then (1) sup S = +∞ ⇔ S is not bounded above in R. (2) inf S = −∞ ⇔ S is not bounded below in R. (3) sup S = −∞ ⇔ inf S = +∞ ⇔ S = ∅ .

Definition. (Absolute value.) If x ∈ R, the absolute value of x , denoted by | x | , is defined by n x if x ≥ 0 , | x | := −x if x < 0 . Theorem (2.12). For all x, y ∈ R the following properties are satisfied: (1) | x | ≥ 0 and | x | = 0 ⇔ x = 0 . (2) | x · y | = | x | · | y | . (3) | x + y | ≤ | x | + | y | . Proof. (1) and (2) are simple consequences of the definition of absolute value. To prove (3) observe that − | x | ≤ x ≤ | x | and − | y | ≤ y ≤ | y | . Adding these inequalities we get −(| x | + | y |) ≤ x + y ≤ (| x | + | y |), which is equivalent to (3). ♠ Remark. (2.13). For all x, y ∈ R | | x | − | y | | ≤ | x − y | . Proof. | x | = | (x−y) + y | ≤ | x − y | + | y | ⇒ | x | − | y | ≤ | x − y | . Similarly, we have | y | − | x | ≤ | y − x | = | x − y | . Therefore, we obtain | x | − | y | ≤ | x − y | and −(| x | − | y |) ≤ | x − y | , from which the conclusion follows. ♠

8 3. INFINITE SEQUENCES

Definition. (Infinite sequences.) A sequence in X is defined to be a function c : N ⊃→ X. If the domain of the sequence is a finite subset of N, we say the sequence is finite. Otherwise, the sequence is called infinite. In the following we study infinite sequences in R (simply said sequences in R ) with domains Nm := { n ∈ N : n ≥ m } , m ∈ N, (usually m=0 or m=1 ).

The most common notation is to write n 7→ cn instead of n 7→ c(n). The elements of the range { cn : n ∈ Nm } are called the terms of the sequence.

In general, we write the sequence as (cn)n∈Nm , or simply (cn) when the domain of the sequence is understood from the context, or is not relevant to the discussion. Sometimes the notation cm, cm+1, . . . , cn,... is also used.

Theorem (3.1). (”Nested intervals theorem”.) If (an) and (bn) are sequences in R , and + (i) ∀n ∈ N an ≤ an+1 ≤ bn+1 ≤ bn , + + (ii) ∀ε ∈ R ∃ n ∈ N bn −an < ε , + then ∃ unique c ∈ R such that ∀n ∈ N an ≤ c ≤ bn . Proof. (1) The existence of c follows from Dedekind’s ”axiom”, with the sets + + + A := { an : n ∈ N } and B := { bn : n ∈ N }, since ∀ m, n ∈ N (m ≤ n) ⇒ am ≤ an ≤ bn and (n < m) ⇒ am ≤ bm ≤ bn , + thus we have ∀ m, n ∈ N am ≤ bn . + (2) Suppose ∃ c1 < c2 such that ∀n ∈ N an ≤ c1 < c2 ≤ bn . + Then from (ii) we know ∃ n ∈ N such that bn −an < c2 −c1 , thus we get bn −an < c2 −c1 ≤ bn −an , a contradiction. ♠ Remarks. (1) If we omit the condition (ii), the theorem can be formulate as follows: If (Jn) is a sequence of closed and bounded intervals such that + T∞ Jn+1 ⊂ Jn for all n ∈ N , then n=1 Jn 6= ∅ . (2) If the intervals are not closed or bounded, their intersection can be empty. E.g. Jn := ( 0, 1/n ) or Jn := [ n, +∞ ).

Deﬁnition. (Bounded sequences.) A sequence (cn) in R is said to be bounded if the range of (cn) is a bounded set in R.

Deﬁnitions. (Monotone sequences.) Let (cn) be a sequence in R . (1) We say that (cn) is an increasing sequence if ∀n ∈ N cn ≤ cn+1 ; we call it strictly increasing if ∀n ∈ N cn < cn+1 . (2) We say that (cn) is a decreasing sequence if ∀n ∈ N cn ≥ cn+1 ; we call it strictly decreasing if ∀n ∈ N cn > cn+1 . (cn) is called monotone if it is either increasing or decreasing.

9 Deﬁnition. (Convergent and divergent sequences.) Let (cn) be a sequence in R .

(1) We say that (cn) is convergent and converges to the limit C ∈ R + if ∀ε ∈ R ∃ M ∈ N such that ∀n ∈ NM | cn − C | < ε .

(2) We say that (cn) is divergent if it does not converge to any limit in R .

(3) We say that (cn) tends to the limit +∞ (as n tends to +∞ ) if ∀K ∈ R ∃ M ∈ N such that ∀n ∈ NM K < cn .

(4) We say that (cn) tends to the limit −∞ (as n tends to +∞ ) if ∀K ∈ R ∃ M ∈ N such that ∀n ∈ NM cn < K .

If α∈R is the limit of (cn), we write lim(cn)=α, lim cn =α, or cn →α as n→+∞. n→ +∞ Remarks. (3.2). (i) Evidently, the limit of a sequence is uniquely determined, provided it exists. (ii) By (1) it follows that a sequence (cn) converges to the limit C ∈ R if and only if + ∀ε ∈ R the set { n ∈ N : | cn − C | ≥ ε } is ﬁnite. (iii) By (3)–(4) it follows that a sequence (cn) tends to +∞ (−∞) if and only if ∀K ∈ R the set { n ∈ N : cn ≤ K (K ≤ cn) } is ﬁnite.

Theorem (3.3). If (cn) is a convergent sequence in R , then it is a bounded sequence. Proof. Let ε := 1 , C := lim (cn) . Then { cn : | cn −C | ≥ 1 } is bounded (since it is ﬁnite), thus { cn : n∈N } = { cn : | cn −C | < 1 } ∪ { cn : | cn −C | ≥ 1 } is also bounded. ♠ Theorem (3.4). Let (cn) be a monotone sequence in R .

(1) If (cn) is bounded, then (cn) is convergent, and lim (cn) = sup {cn : n ∈ N} , provided (cn) is increasing, or lim (cn) = inf {cn : n ∈ N } , provided (cn) is decreasing.

(2) If (cn) is not bounded, then (cn) is divergent, and (cn) tends to +∞ , provided (cn) is increasing, or (cn) tends to −∞ , provided (cn) is decreasing. Proof. + (1) Let ε ∈ R , S := {cn : n ∈ N} , A := sup S , and B := inf S . Then, since A−ε is not an upper bound for S, ∃ p ∈ N A−ε < cp , thus, if (cn) is increasing, then ∀n ∈ Np A−ε < cp ≤ cn ≤ A . Similarly, since B+ε is not a lower bound for S, ∃ p ∈ N cp < B+ε , thus, if (cn) is decreasing, then ∀n ∈ Np B ≤ cn ≤ cp < B+ε .

(2) By the theorem (3.3) it follows that (cn) is divergent. Let K ∈ R . If (cn) is increasing, then ∃ p ∈ N K < cp and ∀n ∈ Np K < cp ≤ cn . If (cn) is decreasing, then ∃ p ∈ N cp < K and ∀n ∈ Np cn ≤ cp < K . ♠

10 Remark. Theorem (3.4) is equivalent to the statement that every monotone sequence in R has a limit in R: (1) lim (cn) = sup {cn : n ∈ N} , provided (cn) is increasing, and (2) lim (cn) = inf {cn : n ∈ N } , provided (cn) is decreasing.

Deﬁnition. (Rearrangement of a sequence.) Let (cn) be a sequence in R . A sequence (bn) is said to be a rearrangement of (cn) if there exists a bijection ϕ : N → N such that ∀k ∈ N bk = cϕ(k) . Deﬁnition. (Subsequences of a sequence.) Let (cn) be a sequence in R . A sequence (bn) is said to be a subsequence of (cn) if there exists a strictly increasing sequence (nk) in N such that ∀k ∈ N bk = cnk .

Theorem (3.5). Let (cn) be a sequence in R and suppose that ∃ lim (cn) ∈ R .

(1) If (an) is a rearrangement of (cn) , then ∃ lim (an) = lim (cn). (2) If (bn) is a subsequence of (cn) , then ∃ lim (bn) = lim (cn). Proof. Both (1) and (2) are immediate consequences of the propositions (ii) and (iii) of remarks (3.2). ♠ Theorem (3.6). Every sequence in R has a monotone subsequence. Proof. Let (cn) be a sequence in R and P := { n ∈ N : ∀m ∈ Nn\{n} cm < cn }.

(1) If P is an inﬁnite set, we deﬁne a sequence (bn) recursively, as follows:

(i) Let n0 be any element of P (e.g. the least one), and deﬁne b0 := cn0 .

(ii) If nk ∈ P is chosen for k ∈ N, and bk := cnk is deﬁned, then we choose for k+1 an element nk+1 ∈ {n∈P : n > nk} (e.g. the least one),

and deﬁne bk+1 := cnk+1 .

Since nk+1 > nk and nk ∈ P, we have cnk+1 < cnk , thus bk+1 < bk . Therefore, (bn) is a strictly decreasing subsequence of (cn).

(2) If P is a ﬁnite set, then ∃M ∈ N ∀n ∈ NM n∈ / P , thus ∀n ∈ NM ∃ m ∈ NM such that cn ≤ cm .

(i) Let n0 be any element of NM (e.g. n0 := M), and deﬁne b0 := cn0 .

(ii) If nk ∈ NM is chosen for k ∈ N, and bk := cnk is deﬁned, then we choose for k+1 an element nk+1 ∈ {n ∈ NM : n > nk}

such that cnk ≤ cnk+1 , and deﬁne bk+1 := cnk+1 . Hence, (bn) is an increasing subsequence of (cn). ♠

11 Theorem (3.7). (Bolzano-Weierstrass theorem.) Every bounded sequence in R has a convergent subsequence. Proof. This is an immediate consequence of the theorems (3.6) and (3.4.): Let (cn) be a bounded sequence in R and (bn) be a monoton subsequence of (cn). Since (bn) is also bounded in R ,(bn) is convergent. ♠

Deﬁnition. (Cauchy sequences.) Let (cn) be a sequence in R . We say that (cn) is a Cauchy sequence if + ∀ε ∈ R ∃ M ∈ N such that ∀ n, m ∈ NM | cn −cm |< ε .

Theorem (3.8). (Cauchy criterion for convergence.) A sequence in R is convergent if and only if it is a Cauchy sequence. Proof. + (1) Let (cn) be a convergent sequence in R , C := lim (cn) and ε ∈ R . Then ∃ M ∈ N ∀n ∈ NM | cn −C | < ε/2 . Thus, ∀ n, m ∈ NM we have | cn −cm | = | (cn −C) + (C−cm) | ≤ | (cn −C) | + | (C−cm) | < ε/2 + ε/2 = ε . + (2) Let (cn) be a Cauchy sequence in R and ε ∈ R .

Then ∃ M1 ∈ N ∀ n, m ∈ NM1 | cn −cm | < ε/2 . (i) Since ∃ L ∈ N ∀n ∈ NL | cn −cL | < 1 and {cn : n < L} is bounded, (cn) is a bounded sequence.

(ii) Therefore, ∃ (cnk ) a convergent subsequence of (cn) , and since (nk) is a strictly increasing sequence in N , ∀ k ∈ N nk ≥ k is satisﬁed.

Let C := lim (cnk ) . Then ∃ M ∈ NM1 ∀k ∈ NM | cnk −C | < ε/2 . Hence, we obtain that ∀k ∈ NM

| ck −C | = | (ck −cnk ) + (cnk −C) | ≤ | (ck −cnk ) | + | (cnk −C) | < ε/2 + ε/2 = ε . ♠ Theorem (3.9). (Limits and order.) Let (an) and (bn) be sequences in R having limits in R, A := lim (an), B := lim (bn).

(1) If A < B , then ∃ M ∈ N such that ∀n ∈ NM an < bn ; (2) If ∃ M ∈ N such that ∀n ∈ NM an ≤ bn , then A ≤ B . Proof. (1) Let C ∈ R such that A < C < B holds. Then by the deﬁnition of the limits ∃ M ∈ N ∀n ∈ NM an < C < bn . (2) Suppose B < A . Then it follows from (1) that ∃ L ∈ N such that ∀n ∈ NL bn < an , a contradiction. ♠ Theorem (3.10). ( ”Sandwiching” theorem.) Let (an), (bn), (xn) be sequences in R such that ∃ lim (an) = lim (bn) =: A ∈ R, and ∃ M ∈ N ∀n ∈ NM an ≤ xn ≤ bn . Then ∃ lim (xn) = A . Proof. Let ε ∈ R+. ∗ Then ∃ M ∈ NM such that ∀n ∈ NM ∗ A − ε < an ≤ xn ≤ bn < A + ε . ♠

12 Deﬁnition. (Null sequences.) Let (cn) be a sequence in R . We say that (cn) is a null sequence if ∃ lim (cn) = 0 . Proposition (3.11). (Convergence and null sequences.) Let (cn) be a sequence in R and C ∈ R. Then lim (cn) = C if and only if (cn − C) is a null sequence. Proof. Let ε ∈ R+. lim (cn − C) = 0 ⇐⇒ ∃M ∈ N ∀n ∈ NM | (cn −C) − 0 | < ε ⇐⇒ ⇐⇒ ∃M ∈ N ∀n ∈ NM | cn −C | < ε ⇐⇒ lim (cn) = C . ♠ Proposition (3.12). (Null sequences and algebraic operations.) Let (an) and (bn) be sequences in R.

(1) If (an) and (bn) are null sequences, then (an +bn) is a null sequence. (2) If (an) is bounded in R , and (bn) is a null sequence , then (an · bn) is a null sequence. Proof. Let ε ∈ R+.

(1) ∃ M1 ∈ N ∀n ∈ NM1 | an | < ε / 2 , and ∃ M2 ∈ N ∀n ∈ NM2 | bn | < ε / 2 ⇒ ∀n ≥ max {M1,M2} | an +bn | ≤ | an | + | bn | < ε / 2 + ε / 2 = ε. + (2) ∃ K ∈ R ∀n ∈ N | an | ≤ K , and ∃ M ∈ N ∀n ∈ NM | bn | < ε / K ⇒ ∀n ∈ NM | an · bn | = | an | · | bn | < K · ε / K = ε . ♠ Theorem (3.13). (Convergence and algebraic operations.) If (an) and (bn) are convergent sequences in R, A := lim (an), B := lim (bn), then

(1) (an + bn) and (an · bn) are convergent sequences, lim (an + bn) = A + B and lim (an · bn) = A · B .

(2) If B 6= 0 and ∀n ∈ N bn 6= 0 , then (1 / bn) is convergent, and lim (1 / bn) = 1 /B. Proof. (1) (an − A) and (bn − B) are null sequences ⇒ ⇒ ((an −A) + (bn −B)) = ((an +bn) − (A+B)) is a null sequence ⇒ ⇒ lim (an +bn) = A+B .

∀n ∈ N (an · bn) − (A · B) = (an −A) · bn + A · (bn −B); (bn) and (A) are bounded, (an −A) and (bn −B) are null sequences ⇒ ⇒ ((an · bn) − (A · B)) is a null sequence ⇒ lim (an · bn) = A · B . 2 (2) ∀n ∈ N 1 / bn − 1 /B = (B − bn) / (bn · B) ; lim (bn · B) = B ⇒ 2 2 ⇒ ∃ M ∈ N ∀n ∈ NM B / 2 < bn · B ⇒ 0 < 1 / (bn · B) < 2 /B ⇒ ⇒ (1 / (bn · B)) is bounded (and (B − bn) is a null sequence) ⇒ ⇒ ((B − bn) · (1 / (bn · B)) ) is a null sequence ⇒ ⇒ (1 / bn − 1 /B) is a null sequence ⇒ lim (1 / bn) = 1 /B. ♠ Proposition (3.14). Let (an) be a convergent sequence in R, and A := lim (an). Then ∃ lim(| an |) = | A | . + Proof. Let ε ∈ R . ∃ M ∈ N ∀n ∈ NM | | an |−| A | | ≤ | an −A | < ε . ♠

13 Deﬁnition. (Limes superior, limes inferior.) Let (cn) be a sequence in R .

(1) If (cn) is bounded above in R , let an := sup {ck : k ∈ Nn} for all n ∈ N, if (cn) is bounded below in R , let bn := inf {ck : k ∈ Nn } for all n ∈ N.

Clearly, (an) is a decreasing sequence in R,(bn) is an increasing sequence in R. By the theorem (3.4) (an) and (bn) have limits in R.

The limit of (an) is called the limes superior or upper limit of the sequence (cn), and is denoted by the symbol lim sup (cn).

The limit of (bn) is called the limes inferior or lower limit of the sequence (cn), and is denoted by the symbol lim inf (cn). It follows from the theorem (3.4) that

if (cn) is bounded above in R , then lim sup (cn) = inf {an : n ∈ N } = inf { sup {ck : k ∈ Nn } : n ∈ N } ,

if (cn) is bounded below in R , then lim inf (cn) = sup {an : n ∈ N } = sup { inf {ck : k ∈ Nn } : n ∈ N } .

(2) If (cn) is not bounded above in R , we deﬁne lim sup (cn) := +∞ , and if (cn) is not bounded below in R , we deﬁne lim inf (cn) := −∞ .

Theorem (3.15). Let (cn) be a sequence in R and A ∈ R . The following propositions are equivalent:

(1) lim sup (cn) = A . (2) For all G ∈ R , if G > A then {n ∈ N : cn ≥ G} is a finite set, and for all L ∈ R , if L < A then {n ∈ N : cn > L} is an infinite set. Proof. (1) ⇒ (2) Let G, L ∈ R such that G > A (if A 6= +∞) and L < A (if A 6= −∞). Then ∃ M ∈ N aM = sup {ck : k ∈ NM } < G , thus ∀k ∈ NM ck < G , and ∀m ∈ N am = sup {ck : k ∈ Nm} ≥ A > L , hence ∀m ∈ N ∃ n ∈ Nm cn > L . (2) ⇒ (1) If G ∈ R and G > A , then {n ∈ N : cn ≥ G} is a finite set, thus ∃ M ∈ N ∀k ∈ NM ck < G , and hence aM = sup {ck : k ∈ NM } ≤ G . Therefore, lim sup (cn) ≤ G for all G > A , which implies that lim sup (cn) ≤ A .

If L ∈ R and L < A , then {n ∈ N : cn > L} is an infinite set, thus ∀m ∈ N ∃ n ∈ Nm cn > L , and hence ∀m ∈ N am = sup {ck : k ∈ Nm} > L . Therefore, lim sup (cn) ≥ L for all L < A , which implies that lim sup (cn) ≥ A. ♠ Remark. (3.16). The corresponding characterization of the lower limit can be formulated as follows: lim inf (cn) = B ∈ R if and only if for all L ∈ R , L < B implies that {n ∈ N : cn ≤ L} is a finite set, and for all G ∈ R , G > B implies that {n ∈ N : cn < G} is an infinite set.

14 Proposition (3.17). Let (cn) be a sequence in R and (dn) be a subsequence of (cn). If ∃ lim (dn) ∈ R , then lim inf (cn) ≤ lim (dn) ≤ lim sup (cn). Proof. Suppose that lim sup (cn) < lim (dn) . Then ∃ D ∈ R lim sup (cn) < D < lim (dn). By the theorem (3.15), {n ∈ N : D ≤ cn} ⊃ {n ∈ N : D ≤ dn} is a ﬁnite set, which is impossible, since D < lim (dn) implies that ∃ M ∈ N ∀n ∈ NM D < dn .

On the other hand, if lim (dn) < lim inf (cn) , then ∃ K ∈ R lim (dn) < K < lim inf (cn). By the remark (3.16), {n ∈ N : cn ≤ K} ⊃ {n ∈ N : dn ≤ K} is a ﬁnite set, which is impossible, since lim (dn) < K implies that ∃ M ∈ N ∀n ∈ NM dn < K . ♠ Proposition (3.18). Let (cn) be a sequence in R . There exist (dn) and (sn), subsequences of (cn), such that ∃ lim (dn) = lim sup (cn) and ∃ lim (sn) = lim inf (cn). Proof. (1) Let A := lim sup (cn). We deﬁne a subsequence (dn) of (cn) , using the theorem (3.15), as follows: If A ∈ R :

∃ n0 ∈ N A − 1 < cn0 < A + 1 . Deﬁne d0 := cn0 .

Suppose that nk ∈ N has been already chosen for k ∈ N, and dk := cnk is deﬁned.

∃ nk+1 ∈ N nk+1 > nk and A−1/(k+1) < cnk+1 < A+1/(k+1). Deﬁne bk+1 := cnk+1 . Evidently, (dn) is a convergent subsequence of (cn) and lim (dn) = A . If A = +∞ :

∃ n0 ∈ N 1 < cn0 . Deﬁne d0 := cn0 .

Suppose that nk ∈ N has been already chosen for k ∈ N, and dk := cnk is deﬁned.

∃ nk+1 ∈ N nk+1 > nk and (k+1) < cnk+1 . Deﬁne bk+1 := cnk+1 . Evidently, (dn) is a subsequence of (cn) and ∃ lim (dn) = +∞ . If A = −∞ :

∃ n0 ∈ N cn0 < −1 . Deﬁne d0 := cn0 .

Suppose that nk ∈ N has been already chosen for k ∈ N, and dk := cnk is deﬁned.

∃ nk+1 ∈ N nk+1 > nk and cnk+1 < −(k+1) . Deﬁne bk+1 := cnk+1 . Evidently, (dn) is a subsequence of (cn) and ∃ lim (dn) = −∞ .

(2) Let B := lim inf (cn) . In an analogous way, using the remark (3.16), we can deﬁne a subsequence (sn) of (cn) , so that ∃ lim (sn) = B . ♠ Proposition (3.19). Let (cn) be a sequence in R . ∃ lim (cn) ∈ R if and only if lim inf (cn) = lim sup (cn), and then lim (cn) = lim sup (cn). Proof. This is an immediate consequence of the theorems (3.18), (3.15) and remark (3.16). ♠