Intro to Real Analysis Test 1

Every nonempty subset of N has a least element.

(A1)a+b=b+a
(A2)(a+b)+c=a+(b+c)
(A3)0+a=a
(A4)a+(-a)=0
(M1)a*b=b*a
(M2)(a*b)*c=a*(b*c)
(M3)1*a=a
(M4)a*(1/a)=1
(D)a*(b+c)=(a*b)+(a*c)

(1)If a,b belong to P, then A+B belongs to P.
(2)If a,b belong to P, then ab belongs to P.
(3)Trichotomy Property

If a belongs to R, then one of the following holds:
a∈P, a=0, -a∈P

Let a,b be elements of R.
(a)If a-b∈P, then a>b or b<a
(b)If a-b∈P∐{0}, then a≥b or b≤a

1)If a>b and b>c then a>c
2)If a>b then a+b>b+c
3)If a>b and c>0 then ca>cb.
4)If a>b and c<0 then ca<cb.

. a if a>0
. 0 is a=0
. -a if a<0

If a,b∈R, then ∣a+b∣≤∣a∣+∣b∣

. ∣∣a∣-∣b∣∣≤∣a-b∣
. ∣a-b∣≤∣a∣+∣b∣

(a) ∣ab∣=∣a∣∣b∣ ∀ a,b∈R
(b) ∣a∣^2 = a^2 ∀ a∈R
(c) If c≥0, then ∣a∣≤c iff -c≤a≤c
(d) -∣a∣≤a≤∣a∣

is the set Vε(a):={x∈R:∣x-a∣<ε}

-ε<x-a<ε and a-ε<x<a+ε

then x=a

∃ a number u∈R s.t. s≤u ∀s∈S.
u is called an upper bound of S.

∃ a number w∈R s.t. w≤s ∀s∈S.
w is called a lower bound of S.

it is both bounded above and bounded below.

it is not bounded.

(1)u is an upper bound of S, and
(2)if v is any upper bound of S, then u≤v

sup S

(1)w is a lower bound of S, and
(2)if t is any lower bound of S, then t≤w

inf S

IF x∈R, then ∃ nₓ∈N s.t. x<nₓ

(a,b):={x∈R:a<x<b}

[a,b]:={x∈R:a≤x≤b}

If S is a subset of R that contains at least two points and has the property,
(1)if x,y∈S and x<y, then [x,y]⊆S, then S is an interval.

I(1)⊇I(2)⊇....I(n)⊇I(n+1)⊇...

If Iₓ=[aₓ,bₓ], x∈N is a nested sequence of closed bounded intervals, then ∃ a number έ∈R s.t έ∈Iₓ for all x∈N

defined on the set N={1,2,...._ of natural numbers whose range is contained in the set R of real numbers.

Then the sequence X'=x(nₓ) given by {x(n₁),...,x(nₓ),..} is a subsequence of X.

If a sequence X=x(n) of real numbers converges to a real number x, then any subsequence X'=x(nₓ) of X also converges to x.

(1)X has 2 convergent subsequences X'=x(nₓ) and X"=x(rₓ) whose limits are not equal.
(2)X is unbounded.

If X=x(n) is a sequence of real numbers, then there is a subsequence of X that is monotone.

A bounded sequence of real numbers has a convergent subsequence.

- intervals are nested, then I(n)⊆I(1) for all n∈N so a(n)≤b(n) for all n∈N
- there is a nonempty set {a(n):n∈N} that is bounded above
- let ξ be its supremum
- then a(n)≤ξ for all n∈N
- claim ξ≤b(n) for all n∈N
Case1: If n≤k, with I(n)⊇I(k), then a(k)≤b(k)≤b(n)
Case2: If k<n, with I(k)⊇I(n), then a(k)≤a(n)≤b(n)
-therefore, a(k)≤b(n) for all k, so b(n) is upper bound for set {a(k):k∈N}
-ξ≤b(n) for each n∈N
-a(k)≤ξ≤b(n) for all n which means ξ∈I(n) for all n∈N

If a-b∈P and b-c∈P then (a-b)+(b-c)=(a-c)∈P
therefore a>c

If a-b∈P then (a+c)-(b+c)=(a-b)∈P

If a-b∈P and c∈P then ca-cb=c(a-b)∈P
therefore ca>cb when c>0

If c<0, then -c∈P so that cb-ca=(-c)(a-b)∈P
therefore cb>ca when c<0

Every nonempty set of real numbers that has an upper bound also has a supremum in R.

A sequence X=x(n) in R is said to converge to x∈R, or x is a limit of x(n), if for every ε>0 ∃ a natural number K(ε) s.t. for n≥K(ε), the terms x(n) satisfy ∣x(n)-x∣<ε

A sequence in R can have at most 1 limit

Suppose lim(x(n))=X' and X". For each ε>0 ∃ K' s.t. ∣x(n)-X'∣<ε/2 for all n≥K' and ∃ K" s.t. ∣x(n)-X"∣<ε/2 for all n≥K". Let K be the larger of K' and K". Then for n≥K, with the Triangle Inequality,
∣X'-X"∣=∣X'-x(n)+x(n)-X"∣
≤∣X'-x(n)∣+∣x(n)-X"∣
<ε/2+ε/2=ε
Since ε>0 then X'-X"=0

sequence X+Y converges to x+y

sequence X-Y converges to x-y

sequence X*Y converges to xy

sequence cX converges to cx

Suppose X=x(n), Y=y(n) and Z=z(n) are sequences of real numbers s.t. x(n)≤y(n)≤z(n) for all n∈N and the lim(x(n))=lim(z(n)).
Then Y=y(n) is convergent and lim(x(n))=lim(y(n))=lim(z(n))

X=x(n) is a sequence then X is monotone if it either increases or decreases

If an infinite set S is contained in a finite interval [a,b] then the set S has at least 1 limit point A in the interval [a,b]

A number x s.t. for all ε>0, ∃ a member of set y, different from x, s.t. ∣y-x∣<ε