Exams: Defs and Thms

Question 1

Closed Set in Rn

Answer 1

A set S⊆Rn is closed if Rn-S (the complement) is open

Question 2

Adherent Point

Answer 2

Let S⊆Rn and x∈S. Then x is an adherent point if ∀r∈R+, B(x,r)∩S ≠∅

Question 3

Accumulation Point

Answer 3

Let S⊆Rn and x∈Rn. Then x is an accumulation point of s if ∀r∈R+, B(x,r)∩(S{x})≠∅

Question 4

Bounded in Rn

Answer 4

Let S⊆Rn. Then S is bounded if ∃x∈Rn and r>0 s.t. S⊆B(x,r)

Question 5

Bolzano-Weierstrass Thereom

Answer 5

Every infinite bounded subset of Rn has an accumulation point.

Question 6

Cantor’s Intersection Theorem

Answer 6

Let {Qi⊆Rn:i∈R+}:
1) Qi ≠ ∅
2) Qi is closed
3) Qi is bounded
4) Qi+1 ≤ Qi (subscript)
Then ∩(i=1->infinity)Qi ≠∅

Question 7

Covering

Answer 7

Let S⊆Rn, Let F be a family of sets in Rn. We say F is a covering of S if ∪(A∈F)A⊇S:
-If ∀A∈F, A is an open cover
-If F is a countable collection then F is a countable cover
-If F is finite then F is a finite cover
-If F and G are covers of S and G⊆F then G is a subcover

Question 8

Lindeloff Covering Theorem

Answer 8

Let A⊆Rn and F is an open cover of A. Then there exists a countable subcover G of A

Question 9

Heine-Borel Theorem

Answer 9

If S⊆Rn which is closed and bounded and F is an open cover of S, then F has a finite subcover (closed and bounded subsets of Rn are compact)

Question 10

Compact

Answer 10

Let S⊆Rn. We say S is compact if every open cover F has a finite subcover. (There exists G⊆F s.t. G is a finite and G covers S)

Question 11

Metric Space

Answer 11

A metric space is a pair (M,d) where M is a set and d:MxM -> R is called a metric (or distance) such that∀x,y,z∈M:
1) d(x,x) = 0
2) d(x,y) > 0 if x≠y
3) d(x,y) = d(y,x)
4) d(x,y) + d(y,z) ≥ d(x,z)

Question 12

Open Ball in Metric Spaces

Answer 12

Let (M,d) be a metric space and x∈M. The open ball centered at x of radius r>0 is B(x,r)={y∈M:d(x,y)<r}

Question 13

Interior Point

Answer 13

If S⊆M then x∈S is an interior point if ∃r>o s.t. B(x;r)⊆S

Question 14

Open Set in Metric Spaces

Answer 14

A set S is open if every point of S is an interior point

Question 15

Closed Set in Metric Spaces

Answer 15

A set S is closed if M\S (complement) is open

Question 16

Bounded in Metric Spaces

Answer 16

We say S is bounded if for x∈M ∃r>0 s.t. S⊆B(x;r)

Question 17

Compact is Metric Spaces

Answer 17

Let (M,d) be a metric space and S⊆M. The set S is compact if every open cover has a finite subcover

Question 18

Boundary

Answer 18

Let S⊆M, then the boundary of S is ∂ = S\∩(M-S)\ (put line over the elements to represent complement)

Question 19

Convergence

Answer 19

Let (S,d) be a metric space and {xn} a sequence of points in S. We say that this sequence converges to a point p∈S if ∀Ɛ>0, ∃N∈Z+ s.t. if n≥N then d(xn,p)<Ɛ. We write xn -> p

Question 20

Cauchy Sequence

Answer 20

A sequence {xn} in S is a Cauchy Sequence if ∀Ɛ>0 ∃N∈Z s.t. if m,n≥N then d(xm,xn)<Ɛ

Question 21

Cauchy Complete

Answer 21

The metric space (S,ds) is Cauchy complete if every Cauchy Sequence is convergent

Question 22

Limit

Answer 22

Let A⊆S and f:A->T a function if p∈T then limx->p f(x)=b if ∀Ɛ>0 ∃δ>0 s.t if 0<ds(x,p)<S then dt(f(x),b)<Ɛ. We write f(x)->b as x->p

Question 23

Continuous (Vector Valued Functions)

Answer 23

Let f:S->T Be a function and p∈S then f is continuous at p if∀Ɛ>0 ∃δ>0 s.t if ds(x,p)<δ the dt(f(x),f(p)<Ɛ. We say f is continuous on A⊆S if f is continuous at every point in A

Question 24

Triangle Inequality

Answer 24

or all x, y ∈ R we have
|x + y| ≤ |x| + |y|

Question 25

Open Cover in Rn

Answer 25

Let S⊆Rn. An open cover of S is a collection C of open sets such that S C. The collection C of open sets is said to cover the set S

Question 26

Field (And Field Axioms)

Answer 26

R is a field meaning R is equipped with two binary operations: +, * ∀ a,b,c∈R:
1) a+b=b+a, ab=ba (commutativity)
2) a+(b+c)=(a+b)+c,a(bc)=(ab)c (associativity)
3) ∃0,1∈R s.t. 0≠1 and a+0=a,a1=a
4) ∀ a∈R s.t. a≠0 ∃-a∈R s.t. a+(-a)=0 and ∃a^(-1)∈R s.t. a(a^(-1))=1
4) a(b+c)=ab+ac (distributivity)

Question 27

Ordered Field (And Ordered Field Axioms)

Answer 27

An ordered field F is a field equipped with a binary relation < s.t. ∀ a,b,c∈F:
1) a>b,b<a, or a=b
2) If a<b and b<c then a<c (transitivity)
3) If a<b then a+c<b+c
4) If 0<a and 0<b then 0<ab

Question 28

Upperbound/Lowerbound

Answer 28

Let S⊆R and x∈R then we say x is an upperbound for S if for all s∈S, s ≤ x (x is to the right of s). Similarly for lowerbound, s ≥ x (x is to the left of s)

Question 29

Least Upperbound (supremum)/Most Lowerbound (infemum)

Answer 29

Let S⊆R and x∈R then we say x is a least upperbound for S if x is an upperbound for S and for every upperbound y for S, we have x≤y (smallest upperbound). Similarly for greatest lowerbound, x ≥ y

Question 30

Complete Ordered Field

Answer 30

An ordered field F is (Dedekind) complete if every S⊆F which has an upperbound has a least upperbound. R is the unique complete ordered field.

Question 31

Inductive

Answer 31

A set S⊆R is inductive if:
a) 1∈S
b) If x∈S then x + 1∈S

Question 32

Fundamental Theorem of Arithmetic

Answer 32

Let n > 1 be an integer. Then n can be represented as a product of prime numbers in a unique way, apart from the order of the factors.

Question 33

Dedekind Completeness Axiom

Answer 33

For any S⊆R s.t. S has an upperbound, S has a least upperbound.

Question 34

Approximation Property

Answer 34

Let S⊆R where S≠∅ with b=sup(s). For every a∈R s.t. a < b there exists some x∈S s.t. a<x≤b

Question 35

Set and Elements of a Set

Answer 35

A set is a collection of objects. The objects in a set are called elements.

Question 36

Cartesian Products of Sets X and Y

Answer 36

X x Y = {(x,y) : x∈X, y∈Y}

Question 37

Comparison Property

Answer 37

Suppose that S,T⊆R s.t. ∀s∈S, ∀t∈T. We have s≤t. Suppose S and T are bounded from above then sup(s)≤sup(t)

Question 38

Function

Answer 38

A relation f⊆X x Y is a function if:
-∀x∈X ∃ y∈Y s.t. (x,y)∈f (f(x)=y)
-∀x∈X and ∀y1,y2∈Y if. (x,y1)∈f and (x,y2)∈f that implies y1=y2

Question 39

Surjective (onto)

Answer 39

A function is surjective if f(x) = y (codomain(f) = range(f))

Question 40

Injective (one-to-one)

Answer 40

A function is injective if ∀x1,x2∈X and x1≠x2 then f(x1)≠f(x2)

Question 41

The Archimedean Property of R

Answer 41

If x,y∈R and x>0 then there exists a positive integer n∈Z+ s.t. nx>y

Question 42

Bijective

Answer 42

A function that is both injective and surjective

Question 43

Composition of functions

Answer 43

Let f : X → Y and g : Y → Z be functions. The composition of g and f , written g ◦ f , is the function of type X → Z given by (g ◦ f)(x) := g(f(x)).

Question 44

Sequence

Answer 44

Let f : Z+ → X for some X {f(1), f(2),…} is a sequence and we write f(1) = fi so {f1, f2,…..} = {fn}

Question 45

Equinumerous

Answer 45

Let X and Y be sets. We say X and Y are equinumerous (or similar) id there exists some function f: X → Y which is a bijection. We write X~Y

Question 46

Finite

Answer 46

We say a set is finite if X~{1,…,n} for some n in Z+

Question 47

Countably Infinite

Answer 47

A set X is countably infinite if X~Z+

Question 48

Order Preserving

Answer 48

Let K: Z+ → Z+ we say that K is order preserving (or increasing) if ∀m,n∈Z+ if m<n then K(m)<K(n)

Question 49

Subsequence

Answer 49

If {fn} is a sequence and K: Z+ →Z+ is order preserving then f◦K is a subsequence and is denoted {fK(n)}

Question 50

Complement

Answer 50

The complement of a set in mathematics is the collection of elements not in the set.

Question 51

Closed Set

Answer 51

A set S⊆Rn is closed if Rn complement S is open

Question 52

Representational Theorem of Open Sets in R

Answer 52

Every open set in R is a countable union of disjoint open intervals.

Question 53

Standard Basis Vector

Answer 53

ui∈Rn where ui=(0,…,0,1,0,….,0) (where 1 is the ith coordinate_ then {ui : 1≤i≤n} are the standard basis vectors

Question 54

Open Ball in Rn

Answer 54

Let a∈Rn and let r∈R+ then B(a;r):{x∈Rn : ||x-a||<r} is the open ball centered a of radius r.

Question 55

Component Interval

Answer 55

Let I be an open set in Rn. We say that I is a component interval of S if I is an open interval and there exists no open J⊆S s.t I⊊J

Question 56

S ¯ (line over top)

Answer 56

{adherent points of S}. Closure of S

Question 57

S’

Answer 57

{accumulation points of S}. Derived Set of S

Question 58

Disconnected/Connected

Answer 58

A metric space (S,d) is disconnected if ∃ A,B∈S s.t. A∩B=∅, A,B≠∅, A∪B=S, and A,B are open. S is connected if it is not disconnected

Question 59

Two-Valued Function

Answer 59

A two-valued function of f on D is a continuous function f:S→{0,1}

Question 60

Components

Answer 60

Let F be the collection of containment wise maximal connected subsets of S. Then the elements of F are called components of S and any two components are disjoint.

Question 61

Disjoint

Answer 61

A and B are disjoint sets if and only if A ∩ B = ∅

Question 62

Arcwise Connected

Answer 62

Let S⊆Rn then S is arcwise connected if ∀a,b∈S, ∃ f:[0,1]→S which is continuous and f(0)=a and f(1)=b