Analysis with an introduction to proof 5th by steven lay ch03e

Open Sets and Closed SetsA set may contain all of its boundary, part of its boundary, or none of its boundary.. How do intersections and unions relate to open sets and closed sets?Theore

Chapter 3

The Real Numbers

Section 3.4 Topology of the Real Numbers

( )

Recall that the distance between two real numbers x and y is given by | x – y|

Many of the central ideas in analysis are dependent on the notion of two points being “close” to each other.

If we are given some positive measure of closeness, say ε, we may be interested in all points y that are less than ε away from x:

{ y : | x – y | < ε }.

We formalize this idea in the following definition

Definition 3.4.1

Let x ∈ and let ε > 0 A neighborhood of x is a set of the form  

N (x ; ε ) = { y ∈  : | x – y | < ε }.

The number ε is referred to as the radius of N  (x ;  ε ).

•

x

N  (x ;  ε )

Sometimes want to consider points y that are close to x but different from x

We can accomplish this by requiring | x – y | > 0.

Definition 3.4.2

Let x ∈ and let ε > 0 A deleted neighborhood of x is a set of the form  

N* (x ; ε ) = { y ∈  : 0 < | x – y | < ε }.

Clearly, N*(x  ;ε ) = N (x ;ε )\{x}.

•

x

If S ⊆ , then a point x in can be thought of as being “inside” S, on the “edge” of S,

or “outside” S.

Saying that x is “outside” S is the same as saying that x is “inside” the complement of S, \ S.

Using neighborhoods, we can make the intuitive ideas of

“inside” and “edge” more precise

5

|

3

|

0

Definition 3.4.3

Let S be a subset of A point x in is an interior point of S if there exists a

neighborhood N of x such that N If for every neighborhood N of x, ⊆ S.

N ∩ S ≠∅ and N ∩ ( \ S) ≠ ∅, then x is called a boundary point of S.

The set of all interior points of S is denoted by int S,

and the set of all boundary points of S is denoted by bd S

Example 3.4.4(c)

The point 3 is an interior point of S because there exists a neighborhood of 3 that is contained in S The same is true for any point x such that 0 < x < 5

So int S = (0, 5).

But the point 0 is different Every neighborhood of 0 (no matter how small) will

contain points of S and points that are not in S.

( )

The same is true for 5

( )

So bd S = {0, 5}.

Open Sets and Closed Sets

A set may contain all of its boundary, part of its boundary, or none of its boundary

Those sets in either the first or last category are of particular interest

Definition 3.4.6

Let S be a subset of If bd S ⊆ S, then S is said to be closed If bd S ⊆ \ S, then S is said to be open.

Theorem 3.4.7

interior point of S.

(b) A set S is closed iff its complement \ S is open.

Proof: (a) If none of the points in S are boundary points of S, then all the points in S must be interior points and S = int S The converse also applies.

(b) We have bd S = bd ( \ S) So all the boundary is in S iff none of the boundary

is in \ S ♦

How do intersections and unions relate to open sets and closed sets?

Theorem 3.4.10

(a) The union of any collection of open sets is an open set

(b) The intersection of any finite collection of open sets is an open set

Proof:

(a) Let A be an arbitrary collection of open sets and let S =  {A  : A ∈A }

But A ⊆ S, so N ⊆ S and x is an interior point of S Hence, S is open

(b) Let A1, …, An be a finite collection of open sets and let

Since each set Ai is open, there exist neighborhoods Ni  (x ;  εi) of x such that Ni  (x ;  εi) ⊆ Ai

Let ε  = min {ε1, …, εn} Then N  (x ;  ε ) ⊆ Ai for each i = 1, …,  n, so N  (x ;  ε ) ⊆ T

Thus x is an interior point of T, and T is open ♦

Note: Part (b) of Theorem 3.4.10 does not necessarily hold for infinitely many open sets

n i i

Example 3.4.12

For each n ∈ , let An = ( − 1/n,  1/n)

But what is in

Each of these sets is an open set (an open interval)

which is not open.

So the open sets have “shrunk down” (by intersecting) to a set that is not open

Our next corollary follows easily from Theorem 3.4.10 and the fact that a set is open iff its complement is closed

Corollary 3.4.11

(a) The intersection of any collection of closed sets is closed

(b) The union of any finite collection of closed sets is closed

Practice 3.4.13

To see that the union of an infinite collection of closed sets may not be closed,

consider the closed sets An = [1/n, 2] for each n ∈

?

A

∞

Accumulation Points

By using deleted neighborhoods we can define another important property of points and sets

Definition 3.4.14

Let S be a subset of A point x in is an accumulation point of S if every deleted neighborhood of x contains a point of S That is, for every ε  >  0, N*(x ;  ε ) ∩ S ≠ ∅

The set of all accumulation points of S is denoted by S  ′

(a) If S is the interval (0, 1], then S  ′ =

Example 3.4.15

[0, 1].

{0}

If x ∈ S and x ∉ S′, then x is called an isolated point of S.

Thus in either case

the neighborhood N must intersect S.

Definition 3.4.15

Let S be a subset of Then the closure of S, denoted cl S, is defined by

cl S = S ∪ S ′ , where S′ is the set of accumulation points of S.

If x ∉ S, then x ∈ S  ′ and every deleted neighborhood intersects S

Conversely, suppose that every neighborhood of x intersects S

If x ∉ S, then every neighborhood of x intersects S in a point other than x

Thus x ∈ S  ′, and so x ∈ cl S

The basic relationships between accumulation points, closure, and closed sets are presented in the following theorem

To this end,

let x ∈ \ S.

If x ∉ S, then x

is in the open set \ S.

Theorem 3.4.17

Let S be a subset of Then

(a) S is closed iff S contains all of its accumulation points, (b) cl S is a closed set,

(c) S is closed iff S = cl S, (d) cl S = S ∪ bd S.

Proof:

(a) Suppose that S is closed and let x ∈ S  ′ We must show that x ∈ S

Thus there exists a neighborhood N of x such that N ⊆ \ S.

But then N ∩ S = ∅, which contradicts x ∈ S  ′ So we must have x ∈ S

Conversely, suppose that S  ′ ⊆ S We shall show that \ S is open

Then x ∉ S  ′ , so there exists a deleted neighborhood N*(x ;  ε ) that misses S

Since x ∉ S, the whole neighborhood N (x ;  ε) misses S; that is, N  (x ;  ε ) ⊆ \ S

Thus \ S is open and S is closed by Theorem 3.4.7(b)

Theorem 3.4.17

Let S be a subset of Then

(a) S is closed iff S contains all of its accumulation points, (b) cl S is a closed set,

(c) S is closed iff S = cl S, (d) cl S = S ∪ bd S.

Proof:

(b) By part (a) it suffices to show that if x ∈ (cl S  )′, then x ∈ cl S So suppose that

x is an accumulation point of cl S.

Then every deleted neighborhood N*(x  ;  ε )

•

x

|

y

N  ( y  ;  δ )

z

Since N*(x  ;  ε) is an open set, there exists a neighborhood N  (  y  ;  δ ) contained in N*(x  ;  ε )

But y ∈ cl S, so every neighborhood of y intersects S That is, there exists a point z

in N  (  y  ;  δ ) ∩ S. But then z ∈ N  (  y  ;  δ ) ⊆ N*(x  ;  ε), so that x ∈ S  ′ and x ∈ cl S