# Bounded Set, Compact Set and Connected Subsets of R

## Bounded Set

A set A β R is bounded from above if there exists a number K such that

x β€ K β x β A

K is called the upper bound of A. Every number greater than K is also an upper bound of A.

A set A β R is bounded from below if there exists a number k such that

k β€ x β x β A

k is called the lower bound of A. Every number smaller then k is also a lower bound of A.

A set is called a bounded set if it is bounded from above as well as from below. Thus the set A is a bounded set if there exist k and K such that

k β€ x β€ K β x β A or A β [k,K]

The bounds of a set may or may not belong to the set.

The upper bound of set which belongs to the set is called its greatest member. A lower bound of a set which belongs to the set is called its smallest member. A set may or may not have greatest and smallest members.

The set of upper bounds of a set which is bounded above has a smallest members called the least upper bound or the Supremum of set denoted by Sup A or M.

The set of lower bounds of a set which is bounded below has a greatest member called the greatest lower bound of Infimum of set denotes by Inf A or m.

M and m may or may not belong to the set.

• [a,b] is bounded set with m = a and M = b.
• a is also the smallest member and b is the greatest member of [a,b]
• (a,b] is bounded set with m = a and M = b
• a is not the smallest member and b is the greatest member of (a,b]
• [a,b) is bounded set with m = a and M = b
• a is the smallest member and b is not greatest member of [a,b).
• (a,b) is bounded set with m = a and M = b
• a is not the smallest member and b is not the greatest member of (a,b)
Subset of R Bounded or not? Infimum of set (m) Supremum of set (M) Whether βaβ is the smallest member of set? Whether βbβ is the greatest member of set?
[a,b] Bounded A b a is the smallest member of [a,b] b is the greatest member of [a,b]
(a,b] Bounded A b NO b is the greatest member of (a,b]
[a,b) Bounded A b a is the smallest member of [a,b] NO
(a,b) Bounded A b NO NO

The set of natural numbers βNβ is bounded from below but is not bounded from above. 1 is the infimum of set of natural numbers as well as the smallest member of the set of Natural numbers.

The set of whole numbers βWβ is bounded from below but is not bounded from above. 0 is the infimum of the set of whole numbers as well as the smallest member of the set of Whole numbers.

The set of Integers βZβ is not bounded set.

Every finite set is bounded set.

The set of rational numbers Q is not bounded set.

The set of real numbers R is not bounded set.

### Open Cover of a set

Let A β R. An open cover of A is collection of {A_Ξ»: Ξ»βΞ} of open sets in R whose union contains A, i.e.,

A sub collection of {AΞ»: Ξ» β Ξ} which also covers A is called subcover of {AΞ»: Ξ» β Ξ}

A subcover which contains finite number of open sets is called finite subcover of {AΞ»: Ξ» β Ξ}

e.g. Let A = [2, β)

Consider A1 = {(1, β)}

A2 = {(n-1, n+1) : n β N} = {(0,2), (1,3), (2,4), (3,5), β¦}

A3 = {(0, n) : n β N} = {(0, 1), (0, 2), (0, 3), (0, 4), β¦}

A4 = {(0, n): n β N, 10 β€ n} = {(0, 10), (0, 11), (0, 12), β¦}

A1, A2, A3, A4 are all open covers of A. A4 is the subcover of A3. Also A1 having only one member is finite cover of A.

### Compact Set

A subset K β R is said to be a compact set if every open cover of K has a finite subcover.

Thus, K is compact set if whenever it is contained in the union of a collection A of open set in R, then it is contained in the union of some finite number of sets in A. K is not compact set if there exists even one open cover of K which has no finite subcover.

##### The necessary and sufficient condition for a subset of R to be compact set is defined by the Heine-Borel Theorem:

Heine - Borel Theorem: A subset K of R is compact set if and only if it is closed set and bounded set.

This theorem describes all the compact subsets of R.

#### Every finite set being closed set and bounded set is compact set.

Consider K = {a1 ,a2,β¦,an } a finite subset of R.

Let A = {AΞ»: Ξ» β Ξ} be an open cover of K

i.e. , K β βΞ»βΞAΞ»

Now each ai β AΞ»i for some AΞ»i β A

Now union of {AΞ»1 ,AΞ»2,β¦,AΞ»n } contains K and hence it is a finite subcover of A

So every open cover of K has a finite subcover so by definition K is compact.

[a,b] is compact set but (a,b] , [a,b), (a,b) are not compact set

The set of all rational numbers Q is neither closed set nor bounded set therefore Q is not compact set.

Since the set of all real numbers R is not bounded set therefore R is not compact set.

Since the set of all natural numbers N is not bounded set therefore N is not compact set.

{1/n βΆn β N } is not compact set as it is not closed set.

If K β R is compact set and x β K then we can find two open sets A and B such that

1. x β A
2. K β B
3. A β© B = β

### Connected Sets

A set S β R is disconnected set if there exists open sets A and B such that

1. S β A βͺ B
2. S β© A β  β and S β© B β  β
3. A β© B = β

In other words, S is disconnected set if it can be broken onto two non-empty sets such that the two parts are respectively contained in two disjoint sets,

If S is not disconnected set, then we can say that S is connected set.

#### Theorem: A subset S of R is connected if and only if whenever a,b β S and a< c < b then c β S

All the intervals [a,b], (a,b), [a,b), (a,b], (-β,a) and (a, β) are connected set.

The set of all real numbers R = (-β,β) is also connected set.

A finite subset of R is disconnected set.

The set of rational numbers Q is disconnected set as :

1. Q β(-β,β2) βͺ (β2,β) = R -{β2}
3. (-β,β2) β© (β2,β) = β

The set of all integers Z is not connected set.

For more information on bounded sets in real analysis, connected sets proofs and help with compact sets in topology and other real analysis topics, schedule a Math live tutoring session with our MathHelp service.

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