Math 341 Lecture Notes Chapter 3. Basic Topology of R

Math 341 Lecture Notes – Chapter 3

Basic Topology of R

§3.1 – The Cantor Set

In Chapter 3 we study the topology (special kinds of subsets) of R. The four most important kinds of sets in this class are open sets, closed sets, connected sets, and compact sets.

The Cantor set will be a very curious example of a compact set. Understanding it will help us avoid certain possible misconceptions about compact sets while deepening our understanding of compact sets.

First, we’ll define the Cantor set. Let

C0= [0, 1]

be the closed interval from 0 to 1.

Define C1by removing the middle third open interval as follows:

C1= C0\(¹

3,²₃) = [0,¹₃] ∪ [²₃, 1]

Define C2as the set obtained by removing the middle thirds of the two intervals whose union forms C1 as follows:

C2= [0,¹₉] ∪ [²₉,¹₃]
∪ [²₃,⁷₉] ∪ [⁸₉, 1]

In general, having constructed Cn−1as the union of 2ⁿ⁻¹closed intervals of length 1/3ⁿ⁻¹, we form Cnby removing the middle third of each interval to obtain Cnas the union of 2ⁿclosed intervals of length 1/3ⁿ. Observe that we have a nested sequence of sets:

C0⊋ C1⊋ C2⊋ C3⊋ · · ·

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Definition of the Cantor Middle Thirds Set

The Cantor set is the set C defined by

C =

∞

\

n=0

Cn.

Length of Cantor Set?

Cnconsists of 2ⁿintervals of length 1/3ⁿ. Adding these lengths gives (2/3)ⁿ, but 2ⁿ

3ⁿ → 0 as n → ∞.

It seems like C is a very ‘small’ set. In a more advanced real analysis course, you would study

‘measurable’ sets. The Cantor set has Lebesgue measure 0. However, in terms of cardinality,

|C| = |R|.

To understand the Cantor set, it is very useful to understand the base 3 representation.

Decimals in base b, where b ≥ 2 and b ∈ N. Use digits 0, 1, 2, . . . , b − 1.

• In base 10,

534.9801 . . . = 5 · 10²+ 3 · 10¹+ 4 · 10⁰+ 9 10¹ + 8

10² + 0 10³ + 1

10⁴ + · · ·

• In base 2,

110.101101 . . . = 1 · 2²+ 1 · 2¹+ 0 · 2⁰+ 1 2¹ + 0

2² + 1 2³+ 1

2⁴ + 0 2⁵ + 1

2⁶ + · · ·

• In base 3,

121.02012 . . . = 1 · 3²+ 2 · 3¹+ 1 · 3⁰+ 0 3¹ + 2

3²+ 0 3³ + 1

3⁴ + 2 3⁵+ · · ·

• If b ≥ 2 is the base, write a = b − 1. Then

1.00000000 . . . = 0.aaaaaaaa . . . The last formula is just a geometric series!

0.aaaaaaa . . . = a b + a

b² + a

b³+ · · · = a/b

1 − 1/b=(b − 1)/b 1 − 1/b = 1.

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In the construction of Cn, we obtained Cnas the union of 2ⁿclosed intervals each of length 1/3ⁿ. Let Lndenote the set of left endpoints of the closed intervals whose union forms Cn.

L0= {0}

L1= {0,²₃}

L2= {0,²₉,²₃,⁸₉} =n

0

3¹+₃⁰₂, ₃⁰₁ +₃²₂, ₃²₁ +₃⁰₂, ₃²₁+₃²₂ o ...

By considering the process of creating Cnfrom Cn−1, we can describe Lnin a simple way.

Theorem (Left Endpoints of C

)

For each n ∈ N, the set Lnof the left endpoints of the intervals comprising Cnconsists of all sums of the form

b1

3¹+b2

3² + · · · +bn

3ⁿ where bi∈ {0, 2} for each i.

† Proof.

We’ll prove this by induction on n. Since L₁= {0,²₃}, the theorem in true when n = 1. Now assume the theorem is true when n = k. The set C_kis the union of certain intervals of length 1/3^kof the form

ha 3^k, a

3^k+ 1 3^k i

.

The left endpoint is a/3^k. Removing the middle third results in two closed intervals of length 1/3^k+1: ha

3^k+ 1 3^k+1

i

and ha

3^k+ 2 3^k+1, a

3^k+ 1 3^k+1

i . We obtain two left endpoints: ^a

3k and ^a

3k+ ²

3k. This is the original left endpoint along with another point obtained by adding ²

3k the first left endpoint. So, for each left endpoint in L_kof the form b1

3¹ +b2

3² + · · · +b_k 3^k we obtain exactly two corresponding points in the set Lk+1which are

b1

3¹+b2

3²+ · · · +bk

3^k+ 0

3^k+1 and b1

3¹+b2

3²+ · · · +bk

3^k+ 2 3^k+1. Thus every point of L_k+1is of the form

b1

3¹ +b2

3² + · · · +bk

3^k+bk+1

3^k+1 where bi∈ {0, 2} for each i. This proves the theorem.

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Theorem (Description of C

)

The set Cnmay be written as the sum of the left endpoint set Lnand the closed interval [0, 1/3ⁿ]. That is,

Cn= Ln+h 0, 1

3ⁿ i

= {x + y : x ∈ Ln, y ∈ [0, 1/3ⁿ]}.

Proof.

From the theorem describing the left endpoints of the intervals in the set Cn, we know that a number t belongs to Cnif and only if

t =b1

3¹ +b2

3² + · · · +bn

3ⁿ+ y where each bi∈ {0, 2} and y ∈ [0, 1/3ⁿ].

Theorem (Description of C

in base 3)

t ∈ Cnif and only if t has a tertiary expansion of the form

t = b1

3¹ +b2

3²+ · · · + bn

3ⁿ

| {z }

b_i∈ {0, 2} for 1 ≤ i ≤ n

+bn+1

3ⁿ⁺¹+bn+2

3ⁿ⁺²+ · · ·

| {z }

b_i∈ {0, 1, 2} for i > n

Proof.

Since y ∈ [0, 1/3ⁿ] if and only if y has a tertiary expansion of the form

y = bn+1

3ⁿ⁺¹+bn+2

3ⁿ⁺²+ · · · =

∞

X

k=n+1

bk

3^k where bk∈ {0, 1, 2}, this is merely a restatement of the previous theorem.

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Theorem (Base 3 Description of the Cantor Set)

A real number z belongs to the Cantor set if and only if it may be expressed as a series

z =b1

3¹ +b2

3² + · · · +bn

3ⁿ+ · · · (1)

where bi∈ {0, 2} which, of course, is the same thing as writing the base 3 decimal expansion z = 0.b1b2b3. . .

using only the digits 0 and 2.

As a consequence of this representation of C, we can fairly easily prove that |C| = |R|.

† Proof.

Let z belong to the Cantor set C =T∞

n=1Cn. We must show that z may be expressed as in equation (1). Since z ∈ C, z ∈ Cnfor each n. From the theorem describing Cn, we may write

z = xn+ yn

where

xn= b1

3¹ +b2

3² + · · · +bn

3ⁿ where bi∈ {0, 2} and 0 ≤ y_n≤ 1/3ⁿ.

With this description, how are xnand xn+1related? If z belongs to the first third of the closed interval of length 1/3ⁿwith left endpoint xn, then xn+1= xn. If z belongs to the last third of the interval, then xn+1= xn+_3n+1² . These two cases are illustrated in the picture below:

xn= xn+1

z

xn+₃¹n

xn xn+1 xn+₃¹n

z

(continued on next page)

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† Proof (continued).

So, x1≤ x2≤ x3≤ x4≤ · · · In other words, (xn) is a monotone increasing sequence which is bounded above by 1. Since 0 ≤ yn≤ 1/3ⁿ, we have lim(yn) = 0. Thus

z = lim(z) = lim(xn+ yn) = lim(xn) + lim(yn) = lim(xn).

This gives us a description of z as the infinite series in equation (1).

Conversely, suppose z is a real number of the form

z =

∞

X

k=1

bk

3^k, bi∈ {0, 2},

as in equation (1). Then, for any n ∈ N, we may write z = xn+ ynwhere

xn= b1

3¹+b2

3² + · · · +bn

3ⁿ ∈ Ln

and

0 ≤ yn=

∞

X

k=n+1

bk

3^k ≤

∞

X

k=n+1

2 3^k = 1

3^k.

By the theorem describing Cn, we have z ∈ Cn. Since z ∈ Cnfor every n, z ∈ C. This completes the proof.

Theorem (Cardinality of Cantor Set)

|C| = |R|.

Proof.

Since C ⊆ [0, 1], we have |C| ≤ |[0, 1]| = |R|.

On the other hand, write each element of the open interval (0, 1) in base 2 as 0.a1a2a3. . .

where each an∈ {0, 1} and we avoid any representation with infinitely many repeated 1’s. Define the function g : (0, 1) → C by

g(0.a1a2a3. . .

| {z }

base 2

) = 0.b1b2b3. . .

| {z }

base 3

where

bn= (

0 if an= 0, 2 if an= 1.

This mapping is an injection and so |R| = |(0, 1)| ≤ |C|.

Since |C| ≤ |R| and |R| ≤ |C|, the Schr¨oder-Bernstein Theorem tells us that |C| = |R|.

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Exercise 3.3.7 (A Very Interesting Exercise!)

If C is the Cantor set prove that C + C = [0, 2].

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End of §3.1

Next: §3.2 – Open and Closed Sets

§3.2 – Open and Closed Sets

Definition of Open Set

1 Given a ∈ R and ϵ > 0, the ϵ-neighborhood of a is the set

Vϵ(a) = {x ∈ R : |x − a| < ϵ} = (a − ϵ, a + ϵ).

2 A set O ⊆ R is open if, for every a ∈ O, there is an ϵ-neighborhood a ∈ Vϵ(a) ⊆ O.

Examples

1 R and ∅ are open.

2 Any open interval

(c, d) = {x ∈ R : c < x < d}

is an open set. If x0∈ (c, d), let ϵ = min{x0− c, d − x0}, then x0∈ Vϵ(x0) ⊆ (c, d).

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Theorem (Unions and Intersections of Open Sets)

1 Let {O_λ: λ ∈ Λ} be any collection of open sets. ThenS

λ∈ΛO_λis open.

2 Let {O1, . . . , On} be a finite collection of open sets, thenTn

k=1O_kis open.

Note: The indexing set Λ in the previous theorem is a nonempty set and there is no restriction on its cardinality.

Proof of 1.

Let x ∈ ∪_λ∈ΛO_λ. Then x ∈ O_λ0 for some λ0∈ Λ, where O_λ0 is open. So for some ϵ0> 0, x ∈ Vϵ₀(x) ⊆ Oλ₀⊆ [

λ∈Λ

Oλ.

Thus,S

λ∈ΛO_λis open.

Proof of 2.

Let x ∈Tn

k=1Ok. Then for each k = 1, . . . , n there exists ϵksuch that x ∈ Vϵ_k(x) ⊆ O_k.

Set ϵ = min{ϵ1, . . . , ϵn}. Then

x ∈ Vϵ(x) ⊆

n

\

k=1

Ok,

and soTn

k=1O_kis open.

Definition of Limit Point

A point x is a limit point of a set A if for all ϵ > 0 the intersection Vϵ(x) ∩ A

contains some point a ∈ A with a ̸= x.

Some Terminology

Different authors use the alternative names:

“limit point”=“cluster point” = “accumulation point”.

Examples

1 A = {¹₁,¹₂,¹₃, . . . }. No point of A is a limit point of A, but 0 is a limit point of A.

2 B = [0, 1]. The point x is a limit point of B ⇐⇒ x ∈ B.

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Theorem about Limit Points

A point x is a limit point of a set A if and only if x = lim anfor some sequence (an) contained in A satisfying an̸= x for all n ∈ N.

Proof.

(⇒) Let x be a limit point of the set A. Then for each n ∈ N there exists an∈ A such that an∈ V_1/n(x) ∩ A and an̸= x.

Let ϵ > 0 be given. Choose N ∈ N such that N > 1/ϵ. If n ≥ N , then

0 < |an− x| <1 n≤ 1

N < ϵ.

Thus, (an) is a sequence in A with an̸= x for all n ∈ N and lim an= x.

(⇐) Assume there is a sequence (an) in A with an̸= x for all n ∈ N and lim an= x. From the definition of convergence of a sequence, for any ϵ > 0, there exists N ∈ N such that an∈ V_ϵ(x) whenever n ≥ N . In particular,

a_N∈ Vϵ(x).

So, x is a limit point of A.

Definition of Isolated Point

A point a ∈ A is an isolated point of A if it is not a limit point of A.

Example

A = [−1, 0] ∪ {1,¹₂,¹₃,¹₄, . . . }

• If −1 ≤ x ≤ 0, then x is a limit point of A.

• If x = ¹_n for n ∈ N, then x is an isolated point of A.

Definition of Closed

A set F is closed (in French, ferm´e) if it contains all of its limit points.

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Closed Sets and Cauchy Sequences

A set F ⊆ R is closed if and only if every Cauchy sequence contained in F has a limit that is also in F .

Proof.

(⇒) Assume F is closed. By definition, F contains all of its limit points. Now let (xn) be a Cauchy sequence in F . By Cauchy’s Criterion, (xn) is also a convergent sequence. Write lim xn= x. There are two cases:

1 For some N ∈ N, xN= xN +1= xN +2= · · · . In other words, the sequence is eventually constant. Then lim xn= xN∈ F .

2 The sequence (xn) contains infinitely many distinct values of xn. Then we can form a subsequence (yn) by removing all occurrences of the limit x from the original sequence (xn).

Since a subsequence of a convergent sequence converges to the same limit as the sequence, x = lim ynwhere yn̸= x for all n ∈ N and yn∈ F . Then x is a limit point of F and so x ∈ F since F is closed.

(⇐) If every Cauchy sequence in F has a limit that is in F , then F contains all of its limit points, and so F is closed.

Example (Closed Intervals)

Every closed interval [c, d] = {x ∈ R : c ≤ x ≤ d} is a closed set.

Proof.

Easy exercise.

Example (Limit Points of Q)

What is the set of all limit points of the set Q?

Answer: R. If x ∈ R is any real number, for any ϵ > 0, the neighborhood Vϵ(x) = (x − ϵ, x + ϵ)

contains a rational number aϵ̸= x.

Density of Q in R

For every y ∈ R, there exists a sequence of rational numbers converging to y.

Both open and closed

The sets R and ∅ are both open and closed. They are the only subsets of R with this property.

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Definition of Closure

Given a set A ⊆ R, let L be the set of all limit points of A. The closure of A is defined to be A := A ∪ L.

Examples

1 Q = R

2 {¹

1,¹₂,¹₃,¹₄, . . . } = {0,¹₁,¹₂,¹₃,¹₄, . . . }

3 (a, b) = [a, b]

Question

When we form the closure A = A ∪ L, is possible that A has additional limit points that do not belong to A? In other words, is it possible that A ⊊ A?

Theorem (Property of the Closure)

A is closed and it is the smallest closed set containing A.

Smallest Closed Set Containing A

What is the only reasonable meaning of the phrase “A is the smallest closed set containing A”?

Answer: A is the smallest closed set containing A means that if F is any closed set containing A, then A ⊆ A ⊆ F .

Notice how this is similar to the definition of the least upper bound on a set of real numbers: The number s is the least upper bound on a set B if for any upper bound M on the set B we have x ≤ s ≤ M for all x ∈ B.

In one case we are using set inclusion ⊆ and in the other case we are using the numerical inequality ≤.

Would it make sense to talk about the largest open set A^◦contained in a set A?

Answer: Let T be the collection of all open sets O ⊆ A. Then A^◦= [

O∈T

O.

Any union of open sets is open, and so this is clearly the largest open set contained in A.

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Theorem (Property of the Closure)

A is closed and it is the smallest closed set containing A.

Proof.

Let L be the set of limit points of A. We’ll show that L is a closed set. Let (xn) be any sequence in L that converges. Write x = lim xn. Let ϵ > 0 be given. There exists N ∈ N such that if n ≥ N , then xn∈ V_ϵ/3(x). Since xN is a limit point of A, there exists some a ∈ A with a ∈ V_ϵ/3(xN). But then a ∈ V_2ϵ/3(x) ⊆ Vϵ(x). This shows that x is a limit point of A, and therefore lim xn= x ∈ L.

Next, suppose x is a limit point of A ∪ L. Then there is a convergent sequence (xn) in A ∪ L such that lim xn= x but x ̸= xnfor all n. Let (an) be the subsequence of (xn) consisting only of points in A, and let (bn) be the subsequence of (xn) consisting only of points in L. At least one of these sequences is infinite, and it also converges to x. If the sequence (an) is infinite, then lim an= x ∈ L. By the first paragraph, if the sequence (bn) is infinite, then lim bn= x ∈ L.

This proves that the set ¯A = A ∪ L contains all of its limit points. Hence, ¯A is closed.

Furthermore, if B is any closed set containing A, then B contains every limit point of A. That is B ⊇ L. Hence B ⊇ A, which shows that A is the smallest closed set containing A.

Definition of Complement

The complement of the set A ⊆ R is the set A^c:= {x ∈ R : x ̸∈ A}.

Theorem

1 A set O is open ⇐⇒ O^cis closed.

2 A set F is closed ⇐⇒ F^cis open.

Proof.

Let O be an open set and x ∈ O. Then there exists ϵ > 0 such that Vϵ(x) ⊆ O. Hence Vϵ(x) contains no points of O^cand so x is not a limit point of O^c. Since no point of O is a limit point of O^c, it follows that O^ccontains all of its limit points and is therefore a closed set.

Conversely, assume O^cis closed. Then it contains all of its limit points. Hence, if x ∈ O, then x is not a limit point of O^c. This means that for some ϵ > 0, Vϵ(x) ∩ O^c= ∅. In other words, Vϵ(x) ⊆ O. Since x ∈ O was arbitrary, this shows that O is open.

Part 2 is a consequence of Part 1 and the fact that for any set S we have (S^c)^c= S.

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De Morgan’s Laws for Sets

Let {E_λ: λ ∈ Λ} be a collection of sets.

1

 [

λ∈Λ

Eλ

c

= \

λ∈Λ

E_λ^c.

2

 \

λ∈Λ

Eλ

c

= [

λ∈Λ

E_λ^c.

Proof.

x ∈

 [

λ∈Λ

Eλ

c

⇐⇒ x ̸∈ [

λ∈Λ

Eλ x ∈

\

λ∈Λ

Eλ

c

⇐⇒ x ̸∈ \

λ∈Λ

Eλ

⇐⇒ x ̸∈ E_λfor each λ ∈ Λ ⇐⇒ x ̸∈ E_λfor some λ ∈ Λ

⇐⇒ x ∈ E_λ^c for each λ ∈ Λ ⇐⇒ x ∈ E^c_λfor some λ ∈ Λ

⇐⇒ x ∈ \

λ∈Λ

E_λ^c ⇐⇒ x ∈ [

λ∈Λ

E_λ^c

Theorem (Unions and Intersections of Closed Sets)

1 Let {F_λ: λ ∈ Λ} be any collection of closed sets. ThenT

λ∈ΛF_λis closed.

2 Let {F1, . . . , Fn} be a finite collection of closed sets. ThenSn

k=1Fkis closed.

This is very similar to a previously proved theorem about open sets:

Theorem (Unions and Intersections of Open Sets)

1 Let {Oλ: λ ∈ Λ} be any collection of open sets. ThenS

λ∈ΛO_λis open.

2 Let {O1, . . . , On} be a finite collection of open sets, thenTn

k=1O_kis open.

Proof of 1.

Recall that a set if open if and only if its complement is closed. For every λ ∈ Λ, the set F_λ^cis open. Hence, by the theorem about open sets, the following set is closed:

 [

λ∈Λ

F_λ^c

c

= \

λ∈Λ

F_λ^c
c

= \

λ∈Λ

Fλ.

Proof of 2.

By the theorem about open sets, the following set is closed:

\ⁿ

k=1

F_k^c

c

=

n

[

k=1

F_k^c
c

=

n

[

k=1

Fk.

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Example (Infinite Intersection of Closed Sets)

Recall that the Cantor set was defined as

C =

∞

\

n=0

Cn

where each Cnis a closed set consisting of the union of 2ⁿintervals of length 1/3ⁿand C0⊋ C1⊋ C2⊋ · · · ⊋ Cn⊋ · · · .

By the theorem on the previous page, C is closed.

Example (Infinite union of closed sets)

By the theorem on the previous page, the finite union of closed sets is closed. However, infinite unions of closed sets are not necessarily closed.

∞

[

n=3

₁

n, 1 −¹_n = (0, 1), which is open.

Finally, we’ll mention one more important subset of a given set.

Definition of Interior

Let E ⊆ R. The interior of E denoted by E^◦is the set

E^◦:= {x ∈ E : there exists Vϵ(x) ⊆ E}.

Open and Closed Unit Disks

The interior of the closed unit disk

{(x, y) ∈ R²: x²+ y²≤ 1}

is the open unit disk

{(x, y) ∈ R²: x²+ y²< 1}.

Examples

1 R^◦= R

2 Q^◦= ∅

3 [a, b]^◦= (a, b)

4 {0,¹₁,¹₂,¹₃,¹₄, . . .}^◦= ∅

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Exercise 3.2.14 (Closure and Interior)

1 E is closed if and only if E = E.

2 E is open if and only if E = E^◦.

Proof of 1.

(⇒) Assume E is closed. Then E contains the set L of its limit points. Hence E = E ∪ L = E.

(⇐) Assume E = E = E ∪ L where L is the set of limit points of E. Then L ⊆ E, and so E is closed.

Proof of 2.

(⇒) Assume E is open. If x ∈ E, there exists ϵ > 0 such that Vϵ(x) ⊆ E. Hence, x ∈ E^◦ showing that E ⊆ E^◦. But from its definition E^◦⊆ E. Hence, E = E^◦.

(⇐) Assume E = E^◦. Then for every x ∈ E, we have x ∈ E^◦and so for some ϵ > 0, Vϵ(x) ⊆ E.

But this means that E is open.

Description of the interior E

E^◦is the largest open set contained in E. If S is the collection of all open subsets O of E, then E^◦=S

O∈SO.

Exercise 3.2.14 (Complement, Closure, and Interior)

1 (E)^c= (E^c)^◦

2 (E^◦)^c= E^c.

First Proof of 1.

We’ll first show that (E)^c= (E^c)^◦. First, note that E is closed and hence (E)^cis open.

x ∈ (E)^c ⇐⇒ ∃ϵ > 0, Vϵ(x) ⊆ (E)^c

⇐⇒ ∃ϵ > 0, Vϵ(x) ∩ E = ∅

⇐⇒ ∃ϵ > 0, Vϵ(x) ∩ E = ∅

⇐⇒ ∃ϵ > 0, V_ϵ(x) ⊆ E^c

⇐⇒ x ∈ (E^c)^◦. This shows that (E)^c= (E^c)^◦.

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Second Proof of 1.

We’ll show that (E)^c⊆ (E^c)^◦and (E^c)^◦⊆ (E)^c.

Since E is closed, E^cis open. Let x ∈ E^c. There exists ϵ > 0, such that Vϵ(x) ⊆ E^c. But since E ⊆ E, we have E^c⊆ E^c, and so Vϵ(x) ⊆ E^c, which implies x ∈ (E^c)^◦. Hence, E^c⊆ (E^c)^◦. Conversely, let x ∈ (E^c)^◦. There exists some ϵ > 0 such that Vϵ(x) ⊆ (E^c)^◦⊆ E^c. This implies Vϵ(x) ∩ E = ∅. Hence V_ϵ/2(x) ∩ E = ∅ which implies x ∈ (E)^cand so (E^c)^◦⊆ (E)^c.

This shows that (E)^c= (E^c)^◦.

Proof of 2.

In part 1, replace E with E^cto get

(E^c)^c= E^◦ Then take the complement to get

E^c= (E^◦)^c.

End of §3.2

Next: §3.3 – Compact Sets

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§3.3 – Compact Sets The Three Most Important Kinds of Sets in Analysis

The three most important kinds of subsets of R in elementary real analysis are:

1 Open sets

2 Closed sets

3 Compact sets.

(In more advanced real analysis, the measurable sets are of supreme importance.)

Definition of Compactness

A set K ⊆ R is compact if every sequence in K contains a convergent subsequence whose limit belongs to K.

Example

Let (an) be an arbitrary sequence in the closed interval [0, 1]. Since this sequence is bounded, the Bolzano-Weierstrass Theorem tells us there is a convergent subsequence. Its limit belongs to [0, 1]. So, [0, 1] is compact.

Definition of Bounded Set

A set A of real numbers is bounded if there exists some M > 0 such that |x| < M for all x ∈ A.

First Theorem about Compact Sets

A set K ⊆ R is compact if and only if K is both closed and bounded.

Proof.

(⇒) Assuming K is compact, we will show that K is closed and bounded. First, we’ll show K is bounded. Suppose, by way of contradiction, that K is not bounded. Then there exists x1∈ K with |x1| > 1, there exists x2∈ K with |x2| > 2, and in general, there exists xn∈ K with

|xn| > n. If (xn_k) is any subsequence of (xn) whatsoever, then |xn_k| > n_k≥ k which is unbounded and hence does not converge. This contradicts the fact that every sequence in a compact set contains some convergent subsequence that converges to a limit in K. Therefore K is bound.

Next, we’ll show that K is closed. Let (xn) be a convergent sequence in K with x = lim xn. We must show that x ∈ K. By the definition of compactness, there is some convergent subsequence (xn_k) that converges to a point in K. But every subsequence of the convergent sequence converges to the same limit as the limit of (xn). So, x ∈ K. This shows that K is closed.

We have shown that if K is compact, then K is both bounded and closed.

(⇐) Conversely, assume K is closed and bounded. Since K is bounded, the Bolzano-Weierstrass Theorem tells us that any sequence in K has a convergent subsequence. This subsequence converges to a point in K since K is closed. Hence, K is compact.

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Although compact sets are not necessarily intervals or the union of intervals, many theorems about closed intervals are true if the phrase “closed interval” is replace with “compact set”.

Theorem (Nested Compact Set Property)

If

K1⊇ K2⊇ K3⊇ K4⊇ · · · is a nested sequence of nonempty compact sets, then

∞

\

n=1

Kn̸= ∅.

Proof.

For every n ∈ N, choose a point xn∈ Kn. Since the sets are nested the sequence (xn) is contained in K1and so there exists a convergent subsequence (xn_k) that converges to a point x ∈ K1. But for any n ∈ N and for all sufficiently large k, the terms of the subsequence also belong to Knand so converge to x ∈ Kn. Since x ∈ Knfor every n ∈ N, ∩^∞_n=1Kn̸= ∅.

Definition of Open Cover

1 Let A ⊆ R. An open cover for A is a (possibly infinite) collection of open sets {Oλ: λ ∈ Λ}

such that

A ⊆ [

λ∈Λ

O_λ.

2 Given an open cover for A, a finite subcover is a finite subcollection {Oλ₁, Oλ₂, . . . , Oλ_n}

of sets from the original cover such that

A ⊆

n

[

k=1

O_λk.

See Example 3.3.7 on page 98 of the textbook for a nice example of an open set that has an open covering but does not have a finite subcovering.

Heine-Borel Theorem

Let K be a subset of R. The following three statements are logically equivalent:

1 K is compact.

2 K is closed and bounded.

3 Every open cover for K has a finite subcover.

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Proof.

The equivalence of the first two statement has already been established. Please study the proof on page 99 that the third statement is equivalent to the first two.