Continuous Image of Compact Set is Closed

Lecture 2: Compact Sets and Continuous Functions

2.1 Topological Preliminaries

What does it mean for a function to be continuous? An elementary calculus course would define:

Definition 1: Let and be a function. Let and . The function has limit as x approaches a if for every , there is a such that for every with , one has . This is expressed as

Definition 2:The function f is said to be continuous at if

On the other hand, in a first topology course, one might define:

Definition 3: A topological space is a pair (X, ) where X is a set and is a collection of subsets of X (called the open sets of the topological space) such that

The Union of any number of open sets is an open set.
The Intersection of a finite number of open set is an open set and.
Both X and the empty set are open.

As an abbreviation, we speak of the topological space X when we don't need to refer to . A set is closed if its complement is open.

Exercise 1: If (X, ) is a topological space and , then (A, ) is also a topological space. We say that this is the topology induced on A by the topology on X.

Definition 4: A function of topological spaces is continuous if for every open subset of , is an open subset of X.

Definition 5:An open rectangle is defined to be a subset of of the form , i.e. There are real numbers and where for all i and

In the case of , we can define a topology by saying that a subset of is open if every point is contained in an open rectangle which is entirely contained in . We call this the usual topology of and for any subset of , the induced topology is called the usual topology of .

Exercise 2: Make a reasonable definition of closed rectangle. Prove that open rectangles are open sets and closed rectangles are closed sets.

Exercise 3: Instead of rectangles, we could have used open balls. Show that the resulting topology would be the same.

We are now ready for:

Proposition 1: With the usual topology on and , the two notions of continuous function are equivalent.

Proof: This is easy if you did Exercise 3.

2.2 Compact Sets

The most important theorem in one variable calculus is:

Theorem 1: (Mean Value Theorem) If (where ) is continuous on and differentiable on , then there is at least one satisfying

The principal ingredient of the proof of this theorem is:

Theorem 2: (Extreme Value Theorem) Every continuous function (where has at least one absolute minimum and at least one absolute maximum.

Usually, in an elementary calculus course, one proves the Mean Value Theorem, but not the Extreme Value Theorem -- the second result is usually taken to be intuitively obvious. Actually, this is a rather subtle and difficult result. To begin with, let us note that in fact, the Extreme Value Theorem holds when the interval is replaced with a closed and bounded subset of the real numbers, where bounded is defined as:

Definition 6: A subset of is said to be bounded if it is contained ins ome open rectangle.

Exercise 4: Find counterexamples to Theorem 2 if you either just assume that the interval is not closed or that it is not bounded.

Definition 7: A cover for a subset of a topological space X is simply any collection of subsets of X whose union contains . The cover is called open (respectively finite) if the subsets are all open (respectively are finite in number).

Definition 8: A subset A of a topological space X is said to be compact if every open cover of A contains a finite subcover (i.e. a finite subset of the cover is itself a cover).

Proposition 2: If is continuous and is compact, then so is .

Proof: If is an open cover of , then is an open cover of A (Why?). Since is compact, has a finite subcover, say , where the are all in . Now, one can verify that form a finite cover of .

Exercise 5: With the usual topology on , if is compact, then is both closed and bounded.

Theorem 5: (Heine-Borel Theorem) With the usual topology on , a subset of is compact if and only if it both closed and bounded.

Note: The Extreme Value Theorem follows: If is continuous, then is the image of a compact set and so is compact by Proposition 2. So, it is both closed and bounded by Exercise 5. In particular, the least upper bound of (which exists by the completeness property of the real numbers) is in (Why?). So, has an absolute maximum.

The Heine-Borel Theorem can be proved in at least two ways. The first method proceeds by building up larger and larger sets which are known to be compact. One starts with Lemma 1 below and then uses Lemma 2 to inductively conclude that any closed rectangle is compact. Finally, Lemma 3 completes the proof.

Lemma 1: A closed interval is compact.

Proof: Let be an open cover of (assumed non-empty). The set of all such that contains a finite subcover of is bounded. So it has a least upper bound, say . Let contain and hence an open interval containing . Choose in this open interval. Then and so there is a finite subcover of . But then is a finite subcover of for any contrary to the choice of .

Lemma 2: If and are compact, then so is .

Proof: Suppose is an open cover of . If , is compact. Further, is also an open cover of and so this set has a finite subcover . For every , there is an open rectangle containing and contained in some one of the ; the set of these is rectangles is an open cover of and so admits of an open subcover of the same set. Each rectangle where and are open rectangles. Letting , it follows that is actually a finite open cover of .

For each , we can do the construction of the last paragraph. The set of all the for all the is an open cover of and so it admits of a finite subcover . But then the set of all the 's associated with all the 's is a finite subcover of .

Lemma 3: If every rectangle is compact, then every closed and bounded subset of is compact.

Proof: If is closed and bounded, then it is contained in a closed rectangle . If is an open cover of , then adding the complement of in to the cover gives an open cover of . Since is compact, there is a finite subcover. But then after possibly removing the complement of in , we get a finite subcover of .

The second proof of the Heine-Borel Theorem proceeds by binary search. We will prove it in the case an interval and leave the generalization to a rectangle as an exercise.

Proof:(Heine-Borel, case where ) Suppose the result is false. Then there is an open cover of without a finite subcover. Let's look for smaller intervals with the same property with respect to this fixed cover. Clearly, either the left or the right half of must not have a finite subcover (Otherwise, the union of the subcovers for each half would give a subcover for the whole interval.) We can now apply the same reasoning to a half that does not have a finite subcover, etc. This gives a sequence of intervals

where for every and such that each subinterval does not admit of a finite subcover of . Clearly, one has . Let be the least upper bound of the set of these . Then is in and so there is a set in with . It is now easy to see that contains almost all the intervals , which is a contradiction.

Exercise 6: Generalize the above proof to show that every closed rectangle in is compact.

2.3 The Mean Value Theorem

Definition 9: The derivative of the function is whenever it exists.

Proposition 3: With as in the definition, suppose that the derivative of at exists. If is a local maximum or local minimum of , then the derivative of at is zero.

Proof:Suppose, for example, that is a local maximum of , i.e. for all in an open interval containing . But then for in this interval with , one has and so the limit as approaches from the right must be non-positive. By taking in the interval with , one has and so the limit as approaches from the left must be non-negative. If the limit exists, then these two must be equal, and so the derivative is zero. The proof in case of a local minimum is analogous.

Proposition 4: (Rolle's Theorem) If is continuous on and differentiable on and , then for some , one has .

Proof: If the result is false, then Proposition 3 says that there is no local maximum or local minimum in . On the other hand, the Extreme Value Theorem says that there is at least one absolute maximum and at least one absolute minimum; so both must be at the endpoints. But, then the function must be constant in the entire interval. This is a contradiction.

We can now prove:

Corollary 1: (Mean Value Theorem) If and are both real valued functions continuous on and differentiable on and if the graphs of and intersect at and , then there is at least one satisfying .