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### 23 Cards in this Set

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 Convergence 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|r Neighborhood Q is a neighborhood of A if there exists E>0 with (A-E,A+E)cQ Upper bound M is an upper bound of S iff for all sinS, s<=M lower bound m is a lower bound of S iff for all sinS s>=m Bounded S is bounded if it has a lower and upper bound Cauchy sequence 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|b 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|=N. 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 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|=N If a sequence converges to A and B, A=B 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|N2, we have |an-B|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 every convergent sequence is Cauchy 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|=N. then |am-A|0 there exists an N such that n,m>=N we have |an-am|Q contains at least one element of which is diff from A a=>b 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 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 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???? The set of all accu points of Q is R 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 The set of all accu points of R\Q is R 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 Bolzano-Weirstiass every infinite bounded set of R numbers has an accu point every Cauchy seq of real #s converges 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|N since aN,anN 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|