First Midterm Stuff

Front 1

Convergence

Back 1

a sequence an converges to A if for every E>0 there exists N such that for all n with n≥N we have |an-A|<E

Front 2

Archimedean Principle

Back 2

for every real number r, there exists n with n>r

Front 3

Neighborhood

Back 3

Q is a neighborhood of A if there exists E>0 with
(A-E,A+E)cQ

Front 4

Upper bound

Back 4

M is an upper bound of S iff for all sinS, s<=M

Front 5

lower bound

Back 5

m is a lower bound of S iff for all sinS s>=m

Front 6

Bounded

Back 6

S is bounded if it has a lower and upper bound

Front 7

Cauchy sequence

Back 7

a sequence an is Cauchy iff for every E>0 there exists and N such that for all n,m>=N we have |an-am|<E

Front 8

8. Supremium: SupS

Back 8

least upper bound
every nonempty set in R has one

Front 9

Infumum: InfS

Back 9

greatest lower bound
S has one if ScR and S is nonempty and s is bounded from below

Front 10

Accumulation point
(actual)

Back 10

A is an accu point of S iff there exists a nbhd Q of A such that QINTS is infinite

Front 11

Accumulation point
(alternative)

Back 11

A is an accu point of S iff there exists a nbdh Q of A such that QINTS\{A}not=emptyset

Front 12

Let (an) be a sequence and A in R. TFRE
a.(an) converges to A
b.For ever neighborhood Q of A, there exists a finite subset F c N such that an in Q for all n in N\F
a=>b

Back 12

let an converge to A. let Q:=(A-E,A+E), where E>0 thus Q is a nghbd of A. by one there there exists an N such that for all n>=N we have |an-A|<E. thus anin(A-E,A+E). take F={1, . . .,N-1}. then F is finite and from anin N\F it follows that n>=N.

Front 13

Let (an) be a sequence and A in R. TFAE
a.(an) converges to A
b.For ever neighborhood Q of A, there exists a finite subset F c N(natural num) such that an in Q for all n in N\F

b=>a

Back 13

Suppose 2 holds true. let E>0. then Q:=(A-E,A+E) is nbhd of A. by 2 there exists a finite FcNat s.t an for all ninN\F. take(for instance) N=1+max(F). now let n>=N. then n>max(F) and hence nnotinF, hence ninNat\F. therefore anin(A-E,A+E) by choice of F. thus |an-A|<E (an converges to A) for all n s.t n>=N

Front 14

If a sequence converges to A and B, A=B

Back 14

Let an be a convergence sequence that converges to A and B. now suppose Anot=B and WLOG A>B. then for every E=(A-B)/2>0, there exists N1 such that for all n>=N1 we have |an-A|<E and a N2 such that for all n>N2, we have |an-B|<E. let N=max(N1,N2) so |an-A|<E and |an-B|<E gives an<A+E=(A+B)/2=B-E<an. an<an is a contradiction, thus A=B.

Front 15

Let (an) be a convergent sequence, then an is bounded

Back 15

let A=liman and E=1000, so E>0. then since an is convergent, there exists N with n>=N such that |an-A|<1000. let M=max(A+1000,a1, . . . .aN-1), then an<=M. now let m=min(A-1000,a1,. . . . .,aN-1), then an>=m. thus M is and upper bound and m=lower bound, so an is bounded

Front 16

every convergent sequence is Cauchy

Back 16

let an be a sequence converging to A. then for all E>0, let E'=E/2. there exists and N such that for all n>=N we have |an-A|<E'. now suppose m>=N. then |am-A|<E' is also true. thus |an-am|=|an-A+A-am|<=|an-A|+|A-am|=|an-A|+|am-A|<E/2+E/2=E. thus an is Cauchy.

Front 17

every Cauchy sequence is bounded?????

Back 17

let an be a Cauchy sequence. then for every E>0 there exists an N such that n,m>=N we have |an-am|<E.

Front 18

Let ScR, A in R. Then TFAE
a. A is an accumulation point of S
b.Whenever Q is a nbhd of A, the set of Q ᴒ (S\{A})does not = the empty set-->Q contains at least one element of which is diff from A
a=>b

Back 18

suppose a holds. let E>0 let Q:=(A-E,A+E). then Q is a neighborhood of A. by 1, QINTS=infinite. thus Qint(S\{A}) is still infinite, and thus non-empty

Front 19

6. Let ScR, A in R. Then TFAE
a.A is an accumulation point of S
b.Whenever Q is a nbhd of A, the set of Q ᴒ (S\{A})does not = the empty set-->Q contains at least one element of which is diff from A
b=>a

Back 19

suppose b holds. i will proceed by contraposition,(if not a then not b). assume A is not an accu point of S. then QINTS is finite. if it is the empty set then Q ᴒ (S\{A}) is also the empty set and we are done.
suppse Q ᴒ (S\{A})={a1,. . . ,am}. since a cant=A for all i=1,. . . .m, |ai-A|>0. now let E:=min{|ai-A|:i,. . . .. m}=1,. . . .,m}>0. thus H:(A-E,A+E) is a nbhd of A. then SINT(H\{A})=empty set since by construction ai isnt in Q=H for i=1, . . . .m. SINT(H\{A})cSINT(Q\{A}), since (H\{A})cQ LAST LINE NEEDED????

Front 20

The set of all accu points of Q is R

Back 20

let xinR and H be a nbhd of x. then E>0 s,t (x-E,x+E)cH. since there are infinitely many rational numbers between x and x+E, thus H contains infinitely many rational numbers. thus HINTR=infinite. so x is an accu point of Q

Front 21

The set of all accu points of R\Q is R

Back 21

let xinR. let E>0nthen H:=(x-E,x+E) is a nbhd of x. since there are infinitely many irrational numbers between x and x+E, H is infinite. thus HINTR is infinite, and thus x is an accu point of R\Q

Front 22

Bolzano-Weirstiass

Back 22

every infinite bounded set of R numbers has an accu point

Front 23

every Cauchy seq of real #s converges

Back 23

let an be Cauchy and let S:=|an:ninN}.
Case 1:s is finite. then |s|=m. then S={s1,. . . .sm} w/ s1=si for inot=1.Let E=min|Si-Sj|>0. since an is Cauchy there exists an N s.t for n,m>N we have
|an-am|<E. in particular, |aN-an|<E for all n>N since aN,an<n>N belongs to S. thus by choice of E, aN=an for all n>N. so an=(a1, . . .aN-1,aN,. . . .N). vii. Thus {an:n in N} converges to aN.
Case 2. suppose S is infinite. now since an is cauchy, s is bounded. then by 8. Bolzano-Weirstiass, S has an accupoint. let E>0 and Q:=(A-E/2,A+E/2) be a nbhd of A. by definition of Cauchy, there exists and N such that for all n,m>N we have |an-am|<(E/2). now since A is an accu point, QINTS is infinite. so there exists M with M>N s.t aM is in QINTS. thus for all |A-an|=|A-am+am-aN|, by the traingle inequality |A-am+am-aN|<|A-am|+|am-aN|<E/2+E/2=E