# Compact Sets and Continuous Functions on Compact Sets

by | Feb 23, 2022 | Math Learning, Math Lecture

# INTRODUTION

In advance analysis, the notion of ‘Compact set’ is of paramount importance. In , Heine-Borel theorem provides a very simple characterization of compact sets. The definition and techniques used in connection with compactness of sets in are extremely important. In fact, the real line sets the platform to initiate the idea of compactness for the first time and the notion of compactness plays its important role in topological spaces.

The definition of compactness of sets in uses the notation of open cover of sets in . For this propose we need some definitions and illustrative examples to clear the meaning of cover of a set in .

# COVER, OPEN COVER, SUB COVER

Definition (Cover): Let be subset of and be a collection of sub sets of . is said to cover or, in other words, is said to be a covering of if

i.e. if for some .

If , for each , is an open set and then is said to be an open cover of .

For example

• (i) The family is an open cover of , since and the sets in family are open sets.
• (ii) The family is a collection of open sets in , but is not a cover of , since and do not belong to .
• (iii) The family is a collection of open sets in and for some . Hence is an open cover of .

Note: If be a collection of open intervals in such that then is also an open cover of .

Definition (Sub-Cover): Let and be a collection of sets in which covers . If   be a sub-collection of such that   itself is a cover of then   is said to be a sub cover of . If   is a finite sub collection of such that   is a cover of then   is said to be a finite sub cover of .

For example if then is an open cover of .

If then and ; this implies is an open sub cover of .

Definition (Countable set): set in is said to be a countable set if either it is finite or if it is infinite, it is enumerable i.e. there exists a bijective mapping from to .

For example

• (i) every finite set is countable,
• (ii) are all countable sets,
• (iii) the sets are not countable sets.

Definition (Countable Sub cover): Let and be a collection of sets in such that covers . If be a countable sub-collection of such that covers then is said to be a countable sub cover of .

For example, if then is an open cover of and is a countable sub-collection of , since if , the set has one-to-one correspondence with . also covers . Hence is a countable sub cover of . Note that there are infinitely many countable sub covers of , since is a countable set and is the set of open intervals with rational end points and hence itself is a countable family of open sets so that every sub cover of is countable.

We now give some examples of open cover of a set in which has no finite sub cover.

Example 1. Let and where . Show that is an open cover of but it has no finite sub cover.

Solution: Let . Then and by the Archimedean property of , there exists a natural number such that for some  . Hence  . This shows that is a collection of open sets in which covers i.e is an open cover of .

If possible, let where are natural numbers such that i.e. covers .

Let then and for

Since but  , so we have a contradiction. Hence is not a cover of .

Thus there exists no finite subcollection of that will cover .

Example 2. Let  and . Show that is an open cover of , but it has no finite sub cover.

Solution: Let . Then . By the Archimedean property of there exists a natural number such that for some  , which shows that is an open cover of ( is also a countable cover of ).

If possible, let where are natural numbers such that i.e. is a sub cover of .

Let  . Then and for all  .

Thus it is proved that no finite subcollection of can cover .

Example 3. Let and  Let . Show that is an open cover of but no finite subcollection of can cover .

Solution: Let . Then . For . Let . Then . By the Archimedean property of , there exists a natural number such that . Since , for for some natural number .

Hence is an open cover of .

If possible, let be a finite subcollection of , where are natural numbers , such that .

Let and then for natural numbers .

Thus . Since . Hence but , which is a contradiction. Hence no finite subcollection of covers .

Example 4. Let and . Let where . Show that is an cover of but it has no finite sub cover.

Solution: Let . Then .

Hence which implies is an open cover of .

If possible, let where for and .

Let

And .

Then

. Since but they do not belong to , we have a contradiction. Hence has no finite sub-collection that can cover , i.e. has no finite sub cover.

Example 5. The collection of open intervals is an uncountable cover of but  where is the set of integers, is countable sub cover of .

Solution: Let . Since is both unbounded above and unbounded below, thee always exist two real numbers and such that such that in such that . Hence .

Thus . As open interval is an uncountable subset of , so is an uncountable cover of .

Thus . Then by the Archimendeon property of real numbers, there exists an integer such that . This implies for some . Hence . Thus is a countable sub cover of .

# COMPACT SETS IN

Definition (Compact set):  set () is said to be a compact set in if every open cover of has a finite sub cover. More explicitly, is said to be compact if for any open cover of , there is a finite sub collection of such that i.e. is a finite sub-cover of .

NOTE: To prove that a set is compact in , we must examine an arbitrary collection of open sets whose union contains , and show that is contained in the union of some finite number of sets in the given collection, i.e. we must have to show that any open cover of has a finite sub-cover. But to prove that a set is not compact, it is sufficient to choose one particular open cover has no finite sub-cover, i.e. union of any finite number of sets in fails to contain .

## Theorem 1 (Heine-Borel Theorem) :

Statement: close and bounded subset of is a compact set in , or in other words every open cover of a closed and bounded subset of has a finite sub cover.

Proof, Let be a closed and bounded subset of .

Let be an open cover of . We assume that has no finite sub-cover. Then is not a subject of the union of finite number of open sets in .

Since H is a bounded subject of , there exist real number such that .

Let . If then at least one of the two subsets and are subset of the union of finite number of open sets in , for otherwise both and are subsets of the union of finite number of open sets in contains , contradicting our assumption that has no finite sub-cover.

We call or according as and it is not a subset of the union of finite number of open sets in or and it is not a subset of the union of finite number of open sets in .

Let and . The at least one of the subsets and is non-empty and it is not a subset of the union of finite number of open sets in . If the first subset is non-empty and it is not a subset of the union of finite number of open sets in , we call , otherwise we call .

Let , and .

Continuing this process of bisection of intervals, we have a family of close and bounded intervals   such that

, for all ,

For all is non-empty and it is not a subset of the union of finite number of open sets in .

such That as .

Then by Nested Interval Theorem, , a singleton set. We shall show that .

Since , for any positive , there exists a natural number such that i.e. and . Hence . Since and it is not a subset of the union of finite number of open sets in , contains infinite number of elements of is a limit point of . Since is closed, .

Now for some . is an open set, hence there exists a positive and hence . Since is an open cover of , for some , which goes against the construction of (.

Hence our assumption that is not a subset of the union of finite number of sets in is wrong and it is established that if is closed and bounded, any open cover of has a finite sub cover so that is a compact set in .

Remark:  In the Heine-Borel theorem neither of the two conditions (i) is closed (ii) is bounded can be dropped. The theorem fails if one of the two conditions is withdrawn – this is evident if we go through the example 1.2.3 and the example 1.2.1. In example 1.2.3, is closed but bounded and in example 1.2.1, is closed but no bounded.

Thus both the conditions (i) and (ii) are necessary for a set in to be compact. Next we shall show that these two conditions are also sufficient for a set to be compact in .

## Theorem 2 (Converse of Heine-Borel Theorem):

Statement: compact subset of is closed and bounded in .

Proof. Let be a compact in . First we shall prove that is a closed set in .

Let and . Then exist two positive numbers and such that .

Let .

Then is an open cover of is compact, has a finite sub cover i.e. there exist elements of and positive numbers such that . For each there exists positive numbers such that .

Let .

Then . Therefore is an interior point of .

Since is arbitrary point of , is open. Hence is closed.

Nest we shall prove that is bounded.

Let be a fixed positive number. Then is an open cover of . Since is compact, has a finite sub-cover. Then there exist points of such that is a finite cover of . If and then is bounded.

Hence it is proved that if is a compact set in , it is closed and bounded in . This completes the proof.

Combining the theorems 1 and 2 we have the following theorem which gives a complete characterization of compact sets in .

Note: set in is compact if and only if is closed and bounded in .

Definition (Heine-Borel Property): set () is said to possess Heine-Borel property if every open cover of has a finite sub cover.

set is said to be compacted if it has the Heine-Borel property.

Example 6. Using the definition of compact set, prove that the set is not compact although it is a closed set in .

Solution: In example 1.2.1, it is shown that , where , is an open cover of and has no finite sub cover. Hence from definition is not compact.

is a closed set in , since is open.

Note: In the example 1.3.1, does not satisfy Heine-Borel property, since is not bounded in .

Example 7. Using definition of compact set show that a finite subset of is a compact set in .

Solution: Let be a finite subset of . Let be an open cover of . Then each is contained in some open set of for some . Let . Then . Thus also covers .

Hence is a finite sub cover of . Therefore, by definition, is a compact set in .

## Theorem 3.

Statement:– If be a compact subset of , then every infinite subset of has a limit point belonging to .

Let be an infinite subset of the compact subset of of such that has no limit point belonging to .

Let . Then is not a limit point of . There exists a positive such that where , called deleted .

Let , which is a collection of open sets in . Since so is an open cover of .

Since is compact, there exists a finite sub collection of where and such that covers i.e.

i.e.

[Since and for ]

which shows that is a finite subset of a compact set in has a limit point in .

## Theorem 4.

Statement: – If be such that every infinite subset of has a limit point in then is closed and bounded in .

Proof. First we shall prove that is bounded. If possible, let be unbounded above. Let be any point of . Since is unbounded above, there exists a point in such that . By similar argument there exists a point in such that and so on. Continuing this process indefinitely we ultimately have an infinite subject of , which being a discrete set, has no limit point in is a bounded above subset of . Similarly, If is unbounded below we can construct an infinite subset of which has no limit point. Hence is also bounded below so that is a bounded subset of .

Next we shall prove that is closed in .

Since S is an infinite and bounded subset of , by the Bolzano-Weierstrass theorem on set, has a limit point in .

Let be a limit point of . Then for any is infinite.

For is infinite. Let us take a point .

For is infinite. Let us take a point ,

such that , Continuing this process, we have an infinite subset

of such that

for . We shall show that has a unique limit point which is .

Let be any positive number. Then by the Archimedean property of , there exists a natural number such that and contains infinite subset of . Thus for every positive , is infinite which proves that is a limit point of .

To prove uniqueness, let be a limit point of . Let . Then the neighborhoods and are disjoint (since either or ). By the Archimedean property of , there exists a natural number such that . Since each of belongs to , so contains all elements of expect some finite number of elements and hence can contain almost finite number of elements of . This implies is not a limit point of . Hence is the only limit point of . By the condition of the theorem . Hence is closed.

Thus it is proved that is closed and bounded in .

Note: subset of a compact subset of is compact if and only if every infinite subset of has a limit point belonging to .

## Theorem 5.

Statement: – A Subset of is compact if and only if every sequence in has a subsequence that converges to a point in .

Proof. Let be compact. Then is closed and bounded.

Let be a sequence of points in . Since is bounded, is bounded. By the Bolzano-Weierstrass theorem on sequence, there exists a subsequence of that converges to a point, say . Since is closed, if , and is open. Then there exists a neighborhood of which contains no point of . This implies contains no element of the sequence which contradicts that . Thus . Hence every sequence in has a subsequence converging to a point of .

Suppose is not closed. Then has a limit point, say which is not in . Since is a limit point of , there is a sequence in , where for all , such that . Then every subsequence of converges to . Since , there is no subsequence of that converges to a point of .

Suppose is not bounded. Then there exists a sequence in such that for all . Then every subsequence of unbounded sequence is unbounded and hence no subsequence of converges to a point in .

Hence, by contrapositive argument, it is proved that if every sequence in has a subsequence that converges to a point of the is closed and bounded and hence by Heine-Borel theorem is compact.

Note:  Following theorem 5., an alternative definition of compact set can be given in the from:

“A set in is called a compact set in if every sequence in has a subsequence that converges to a point of .”

Theorem 5 and the Heine-Borel theorem together prove the equivalence of the two definitions.

Example 8. If is a closed subset of a compact set in then using definition of compact set, prove that is compact.

Solution: is open, since is closed.

Let be an open cover of . Suppose is not an open cover of . Let . Then i.e., is an open cover of . Since , is also an open cover of . being compact, has a finite sub collection such that , where . must contain , otherwise which implies is an open cover of , which is contrary to our assumption.

Since we have .

Which shows that,

is a finite sub collection of and covers , which implies is a finite sub cover of . Therefore, by definition, is compact.

Example 9. Every compact set in has greatest as well as least element.

Solution: Let be any compact set in . If possible, let have no greatest element. Then for each element . let . Then is an open set. Let , a family of open sets in . Let . Since has no greatest element, there exists an element in such that . Thus . Hence is an open cover of . being compact, has a finite sub-cover, say .

Let Then for and . Let .

Then and . This leads to a contradiction, since . Hence our assumption is not tenable and has greatest element.

To prove the next part, let, if possible, has no least element. Then for each element , let is an open set. Let , a family of open sets in . Let . Since has no least element there exists an element in such that . Then and . Thus is an open cover of . Since is compact, has a finite sub collection that covers .

Let where and . If  then .

Now but which contradicts our assumption. Hence has least element.

Example 10.  If and are component sets in , show that is also compact. Give an example to show that union of an infinite number of compact sets in is not necessarily a compact set in .

Solution: Let be a family of open sets in such that i.e. is an open cover of .

Since and ,

is an open cover of both and . Since and are both compact sets in , then there exist two finite sub collections

and of such that

and where and .

Let . Then is a finite sub collection of such that

or

for some

Or for some