Linear Algebra and Real Analysis I: Week 6
Fall 2024
Homework Problems
1. More theorems about limits of sequences
Consider a sequence a⃗1 , a⃗2 , ... in Rn that converges to ⃗a
Consider a sequence b⃗1 , b⃗2 , ... in Rn that converges to ⃗b
(a) Prove that the sequence of lengths |b⃗1 |, |b⃗2 |, ... in R is bounded:
∃K such that ∀n, |b⃗m | < K
Proof. WTS: ∃K s.t. ∀m, |b⃗m | < K
|b⃗m | = |b⃗m − ⃗b + ⃗b| ≤ |b⃗m − ⃗b| + |⃗b| by triangle inequality
∃ϵ > 0 where ∃M ∈ N s.t. ∀m ≥ M, |b⃗m − ⃗b| < ϵ since b⃗m converges to ⃗b
Let ϵ = 1, then |b⃗m | < ϵ + |⃗b| ⇒ |b⃗m | < 1 + |⃗b|
Choose K = max{|b⃗1 |, |b⃗2 |...|b⃗m |, 1 + |⃗b|} for all m < M
Since some K is able to be chosen that will bound |b⃗m |, K exists.
(b) Define the sequence of dot products: cm = a⃗m · b⃗m
Prove that this sequence c1 , c2 , ... converges to ⃗a · ⃗b
Proof. WTS: ∀ϵ > 0, ∃M ∈ N s.t. ∀m > M |a⃗m · b⃗m − ⃗a · ⃗b| < ϵ
|a⃗m · b⃗m − ⃗a · ⃗b| = |a⃗m · b⃗m − ⃗a · b⃗m + ⃗a · b⃗m − ⃗a · ⃗b|
⇒ |(a⃗m − ⃗a) · b⃗m + (b⃗m − ⃗b) · ⃗a|
⇒ ≤ |(a⃗m − ⃗a) · b⃗m | + |(b⃗m − ⃗b) · ⃗a| by triangle inequality
⇒ ≤ |a⃗m − ⃗a||b⃗m | + |b⃗m − ⃗b||⃗a| by Cauchy Schwartz inequality
Since a⃗m converges to ⃗a, ∀ϵ > 0, ∃Ma s.t. ∀m > Ma , |a⃗m − ⃗a| < ϵ and from
part (a), we know |b⃗m | < K. ⇒ |a⃗m − ⃗a||b⃗m | < |a⃗m − ⃗a|K
Let ϵ = ϵK
, so that |a⃗m − ⃗a|K < ϵK
⇒ |a⃗m − ⃗a| < 2ϵ
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Since b⃗m converges to ⃗b, ∀ϵ > 0, ∃Mb s.t. ∀m > Mb , |b⃗m − ⃗b| < ϵ and from
part (a), we know |a⃗m | < K1 . ⇒ |b⃗m − ⃗b||a⃗m | < |a⃗m − ⃗a|K1
Let ϵ = ϵK2 1 , so that |b⃗m − ⃗b|K1 < ϵK2 1 ⇒ |b⃗m − ⃗b| < 2ϵ
This shows if we let M = max{Ma , Mb }, ∀ϵ > 0 and ∀m > M , |a⃗m · b⃗m − ⃗a · ⃗b| <
ϵ
+ 2ϵ ⇒ |a⃗m · b⃗m − ⃗a · ⃗b| < ϵ
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Therefore, since a M was found and by definition of convergence, c1 , c2 , ... converges to ⃗a · ⃗b
2. Consider a set with just 3 elements X = {a, b, c}
(a) I want to define the following sets as open, and no others.
∅, {a}, {a, b}, {a, b, c}
Check the axioms of topology to verify whether this is a valid topology
Solution: The first axiom states the empty set and the set X are open. In this
example, since the empty set is included, and X consists of the elements {a, b, c}
which has been defined as an open set, these sets meet the first axiom. The second
axiom states the union of any finite or infinite collection of open sets is open.
∅ ∪ {a} is {a} which is an open set.
∅ ∪ {a, b} is {a, b} which is an open set.
∅ ∪ {a, b, c} is {a, b, c} which is an open set.
{a} ∪ {a, b} is {a, b} which is an open set.
{a} ∪ {a, b, c} is {a, b, c} which is an open set.
{a, b} ∪ {a, b, c} is {a, b, c} which is an open set.
Since all unions of the sets are also open, second axiom is met. The third axiom
states the intersection of two open sets is open.
∅ ∩ {a} is ∅ which is an open set.
∅ ∩ {a, b} is ∅ which is an open set.
∅ ∩ {a, b, c} is ∅ which is an open set.
{a} ∩ {a, b} is {a} which is an open set.
{a} ∪ {a, b, c} is {a} which is an open set.
{a, b} ∩ {a, b, c} is {a, b} which is an open set.
Since all intersections of the sets are also open, the third axiom is met. Therefore
this is a valid topology.
(b) I want to define the following sets as open, and no others:
∅, {a, c}, {a, b}, {a, b, c}
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Check the axioms of topology to verify whether this is a valid topology
Solution: The first axiom states the empty set and the set X are open. In this
example, since the empty set is included, and X consists of the elements {a, b, c}
which has been defined as an open set, these sets meet the first axiom. The second
axiom states the union of any finite or infinite collection of open sets is open.
∅ ∪ {a, c} is {a, c} which is an open set.
∅ ∪ {a, b} is {a, b} which is an open set.
∅ ∪ {a, b, c} is {a, b, c} which is an open set.
{a, c} ∪ {a, b} is {a, b, c} which is an open set.
{a, c} ∪ {a, b, c} is {a, b, c} which is an open set.
{a, b} ∪ {a, b, c} is {a, b, c} which is an open set.
Since all unions of the sets are also open, second axiom is met. The third axiom
states the intersection of two open sets is open.
∅ ∩ {a, c} is ∅ which is an open set.
∅ ∩ {a, b} is ∅ which is an open set.
∅ ∩ {a, b, c} is ∅ which is an open set.
{a, c} ∩ {a, b} is {c} which is NOT an open set.
{a, c} ∪ {a, b, c} is {a, c} which is an open set.
{a, b} ∩ {a, b, c} is {a, b} which is an open set.
Since there was an intersection of two open sets that did not result in an open
set, this does not meet the third axiom and these sets are, therefore, not a valid
topology.
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