Infi-1 2

1a

Prove or disprove: For every non-empty subset of real numbersthere is at least one upper bound or at least one lower boundDisproof:Let S=RS≠∅,S⊆R∀M∃x=M+1:x∈S∧x>M⟹∄M:∀x∈S:x≤M∀m∃x=m−1:x∈S∧x<m⟹∄m:∀x∈S:x≥m

1b

Prove or disprove: For every non-empty subset of real numbers,if there is at least one upper bound, then there is at least one lower boundDisproof:Let S={x∈R∣x<0}S≠∅,S⊆R∃M=0:∀x∈S:x≤M∀m∃x=m−1:x∈S∧x<m⟹∄m:∀x∈S:x≥m

1c

Prove or disprove: For every non-empty subset of real numbers,if there is at least one lower bound, then there is at least one upper boundDisproof:Let S={x∈R∣x>0}S≠∅,S⊆R∃m=0:∀x∈S:x≥m∀M∃x=M+1:x∈S∧x>M⟹∄M:∀x∈S:x≤M

1d

Prove or disprove: For every non-empty subset of rational numberswith a maximum, there is a rational upper bound.Proof:Let S≠∅,S⊆QLet MS=max(S)MS∈S,S⊆Q⟹M∈QLet M=MS, note that M∈QM=max(S)⟹∀x∈S:x≤M⟹∃M∈Q:∀x∈S:x≤M

1e

There exists a non-empty subset of real numberswhose upper bound is equal to its lower boundProof:Let S={0}S≠∅,S⊆RLet M=m=0∀x∈S:0=m≤x≤M=0⟹∃m=M:∀x∈S:m≤x≤M

1f

There exists a non-empty subset of real numberswhose upper bound is equal to its lower bound and which has no maximumDisproof:Let S≠∅,S⊆R∃m,M=m:∀x∈S:m≤x≤M⟹∃m,M=m:∀x∈S:x=m=M⟹M∈S⟹∃M∈S:∀x∈S:x≤M⟺∃max(S)

2

Given: A⊆R,A has an upper-boundProve or disprove: B={−a∣a∈A}⟹B has a lower-bound,inf(B)=−sup(A)Proof:A has an upper-bound⟹∃sup(A)Let sup(A)=M⟺∀ε>0∃M:(∀a∈A:a≤M)∧(∃a∈A:a>M−ε)⟺∀ε>0∃m=−M:(∀a∈A:m≤−a)∧(∃a∈A:−a<m+ε)⟺∀ε>0∃m=−M:(∀b∈B:m≤b)∧(∃b∈B:b<m+ε)⟺∃inf(B),inf(B)=−M=−sup(A)∃inf(B)⟹B has a lower-bound

3

Given: A,B⊆R have upper-boundsProve or disprove: A+B={a+b∣a∈A,b∈B}⟹A+B has an upper-bound,sup(A+B)=sup(A)+sup(B)Proof:A has an upper-bound⟹∃sup(A)⟺∀ε>0∃MA:(∀a∈A:a≤MA)∧(∃a∈A:a>MA−ε)B has an upper-bound⟹∃sup(B)⟺∀ε>0:∃MB:(∀b∈B:b≤MB)∧(∃b∈B:b>MB−ε)Let sup(A)=MA,sup(B)=MB(∃sup(A)∧∃sup(B))⟺∀ε>0∃MA,MB:(∀a∈A,∀b∈B:a≤MA∧b≤MB)∧(∃a∈A,b∈B:a>MA−ε∧b>MB−ε)⟺∀ε>0∃MA,MB:(∀a∈A,∀b∈B:a+b≤MA+MB)∧(∃a∈A,b∈B:a+b>MA+MB−2ε)⟺∀ε1=2ε>0∃M=MA+MB:(∀a+b∈A+B:a+b≤M)∧(∃a+b∈A+B:a+b>M−ε1)⟺∃sup(A+B),sup(A+B)=MA+MB=sup(A)+sup(B)∃sup(A+B)⟹A+B has an upper-bound

4a

A={14−(13)n∣n∈N}={14−13n∣n∈N}Let us prove that sup(A)=14∀n∈N:13n>0⟹∀n∈N:14−13n≤14∀n∈N:3n>n,∀r∈R:∃n∈N:n>r⟹∀ε>0∃n∈N:3n>n>1ε⟺∀ε>0∃n∈N:14−13n>14−ε⟹sup(A)=14Let us also prove that ∄max(A)∃max(A)⟺∃M∈A:∀a∈A:a≤MLet ∃max(A)=x,x∈AThen ∃n∈N:x=14−13nLet m=n+1m∈N⟹14−13m∈A14−13m=14−13n+1>14−13n⟹∃m∈N:am∈A∧am>x⟹x≠max(A)−Contradiction!⟹∄max(A)Let an=14−13n∀n∈N:an+1>an⟺14−13n+1>14−13n⟺−13n+1>−13n⟺13⋅3n<13n⟺13<1⟹Sequence an is monotonically ascending⟹∃min(an)=a1a1=14−13=−112⟹∃min(an)=−112min(an)=min(A)=−112∃min(A)⟹inf(A)=min(A)=−112

4b

B={3n−1n∣n∈N}={3−1n∣n∈N}Let bn=3−1n∀n∈N:bn+1>bn⟺3−1n+1>3−1n⟺1n+1<1n⟺n+1>n⟺1>0⟹ Sequence bn is monotonically ascending⟹limn→∞bn=sup(bn)=sup(B)limn→∞bn=limn→∞3−1n=limn→∞3−limn→∞1n=3−0=3⟹sup(bn)=sup(B)=3Let us also prove that ∄max(B)∃max(B)⟺∃M∈B:∀b∈B:b≤MLet ∃max(B)=x,x∈BThen ∃n∈N:x=3−1nLet m=n+1m∈N⟹3−1m∈A3−1m=3−1n+1>3−1n⟹∃m∈N:bm∈B∧bm>x⟹x≠max(B)−Contradiction!⟹∄max(B)bn is a monotonically ascending sequence⟹∃min(bn)=b1b1=3−11=2⟹∃min(bn)=2min(bn)=min(B)=2∃min(B)⟹inf(B)=min(B)=2

4c

C={(−1)n⋅n∣n∈N}Let us prove that C has no upper-bound and thus ∄max(C),∄sup(C)C has an upper-bound⟺∃M:∀c∈C:c≤MLet x be an upper-bound of C∀r∈R∃n∈N:n≥⌈r⌉+1>r⟹∃n,m=2n∈N:(−1)m⋅m=2n>n≥⌈x⌉+1>x⟹∃c∈C:c>x⟹x is not an upper-bound - Contradiction1⟹C has no upper-bound⟹∄max(C),∄sup(C)Let us also prove that C has no lower-bound and thus ∄min(C),∄inf(C)C has a lower-bound⟺∃m:∀c∈C:c≥mLet x be a lower-bound of C∀r∈R∃n∈N:−n≤⌈r⌉−1<r⟹∃n,m=2n+1∈N:(−1)m⋅m=−2n−1<−n≤⌈x⌉−1<x⟹∃c∈C:c<x⟹x is not a lower-bound - Contradiction1⟹C has no lower-bound⟹∄min(C),∄inf(C)

4d

D={n+1m∣n,m∈N}Let us prove that D has no upper-bound and thus ∄max(D),∄sup(D)D has an upper-bound⟺∃M:∀d∈D:d≤MLet x be an upper-bound of D∀r∈R∃n∈N:n≥⌈r⌉+1>r⟹∃n,m=1∈N:n+1m>n≥⌈x⌉+1>x⟹∃d∈D:d>x⟹x is not an upper-bound - Contradiction!⟹D has no upper-bound⟹∄max(D),∄sup(D)Let us prove that ∄min(D),inf(D)=1Let x∈DThen ∃n,m∈N:x=n+1mx=n+1m>n+12m∈D⟹∀x∈D∃d∈D:d<x⟹∄min(D)Let ∃inf(D)=1Then ∀ε>0:(∀d∈D:d≥1)⏟True,n≥1,1m>0∧(∃d∈D:d<1+ε)Let n=1,m=⌈1ε⌉+1n∈N,m∈N⟹d=n+1m∈Dd=1+1⌈1ε⌉+1≤1+11ε+1<1+11ε=1+ε⟹∀ε>0∃d∈D:d<1+ε⟺inf(D)=1