## Proof Exercise #3: Introduction:

proving the Heine-Borel Covering Theorem

(jump to further reading option)

Definition 1. "fif" means "if, and only if,"

Definition 2. If I is a set of real numbers and G is a collection of sets of real numbers, then G is said to cover I fif every member of I is a member of some member of G.

Definition 3. The statement that I is an open interval means that there exist real numbers a and b such that

I = {x | a < x < b}. I can be denoted by "(a,b)".

Definition 4. The statement that I is a closed interval means that there exist real numbers a and b such that

I = {x | a is less than or equal to x is less than or equal to b}. I can be denoted by "[a,b]".

Definition 5. If M is a set of real numbers, then the statement that h is an upper bound of M means that no member of M is greater than h.

Definition 6. If M is a set of real numbers, then the statement that h is the least upper bound of M means that h is an upper bound of M and no upper bound of M is less than h.

Axiom 1. (The Least-Upper-Bound Axiom). Every non-empty set of real numbers that has an upper bound has a least upper bound.

Theorem. (Heine-Borel Covering Theorem). If G is a collection of open intervals covering a closed interval I, then there exists a finite subcollection G' of G such that G' covers the closed interval I.

Proof: Since I is a closed interval, there exist real numbers a and b such that I = [a,b].

Case 1. b < a. Then I is the null set, and is obviously covered by any finite subcollection of G.

Case 2. a = b. Then I is finite, and is covered by G', where G' consists of any one interval to which a belongs.

Case 3. a < b. This is the non-trivial case to which we must apply our wits.

What, from the following list, do you think, will be our strategy for proving that there exists a finite subcollection G' of G such that G' covers [a,b]?

choice #1: Notice that the Pythagorean Theorem can be quoted here.

choice #2: Notice that the distance from a to be might be less than 1/2.

choice #3: Notice that some of the members of G might overlap one another.

choice #4: Notice that the singleton {a} covered by a finite subcollection of G.

choice #5: Notice that some of the members of G might be disjoint from one another.

(end of page)

(jump to further reading option)

(return to list of proof exercises)

Definition 1. "fif" means "if, and only if,"

Definition 2. If I is a set of real numbers and G is a collection of sets of real numbers, then G is said to cover I fif every member of I is a member of some member of G.

Definition 3. The statement that I is an open interval means that there exist real numbers a and b such that

I = {x | a < x < b}. I can be denoted by "(a,b)".

Definition 4. The statement that I is a closed interval means that there exist real numbers a and b such that

I = {x | a is less than or equal to x is less than or equal to b}. I can be denoted by "[a,b]".

Definition 5. If M is a set of real numbers, then the statement that h is an upper bound of M means that no member of M is greater than h.

Definition 6. If M is a set of real numbers, then the statement that h is the least upper bound of M means that h is an upper bound of M and no upper bound of M is less than h.

Axiom 1. (The Least-Upper-Bound Axiom). Every non-empty set of real numbers that has an upper bound has a least upper bound.

Theorem. (Heine-Borel Covering Theorem). If G is a collection of open intervals covering a closed interval I, then there exists a finite subcollection G' of G such that G' covers the closed interval I.

Proof: Since I is a closed interval, there exist real numbers a and b such that I = [a,b].

Case 1. b < a. Then I is the null set, and is obviously covered by any finite subcollection of G.

Case 2. a = b. Then I is finite, and is covered by G', where G' consists of any one interval to which a belongs.

Case 3. a < b. This is the non-trivial case to which we must apply our wits.

What, from the following list, do you think, will be our strategy for proving that there exists a finite subcollection G' of G such that G' covers [a,b]?

choice #1: Notice that the Pythagorean Theorem can be quoted here.

choice #2: Notice that the distance from a to be might be less than 1/2.

choice #3: Notice that some of the members of G might overlap one another.

choice #4: Notice that the singleton {a} covered by a finite subcollection of G.

choice #5: Notice that some of the members of G might be disjoint from one another.

(end of page)

(jump to further reading option)

(return to list of proof exercises)