Now we begin to prove the theorem.

**4.Attractor-repeller pairs**

**Definition:** A compact set is an **attractor** for if there exists open, and . is called a **basin** of attraction.

For any attractor , be a basin for , let is open. By definition, and . We also have

**Definition:** A **repeller** for is an attractor for . A basin of repelling for is a basin of attracting for .

Hence is a repeller for with basin .

It’s easy to see that is defined independent of the choice of basin for . (Exercise)

We call such a pair an attracting-repelling pair.

The following two properties of attracting-repelling pairs are going to be important for our proof of the theorem.

**Proposition 1:** There are at most countably many different attractors for .

**Proof:** Since is compact metric, there exists countable basis that generates the topology.

For any attractor , any attracting basin of is a union of sets in , i.e. latex for some subsequence of .

Since is compact, is an open cover of , we have some s.t. covers .

Let hence . We have

Since is an attracting basin for , all three sets are equal. Hence . i.e. any attractor is intersection of foreward interates of come finite union of sets in . Since is countable, the set of all finite subset of it is coubtable.

Hence there are at most countably many different attractors. This establishes the proposition.

By proposition 1, we let be a list of all attractors for . Now we are going to relate the arrtactor-repeller pairs to the chain recurrent set and chain transitive components.

**Proposition 2:**

**Proof:** i)

This is same as saying for any attractor , .

For all , let be a basin of , then there is for which (recall that is the dual basin of for ). Since and we conclude

Hence . Let be the smallest integer for which . Hence . Let is also a basin for .

Now we show such cannot be chain recurrent: Since and are compact and disjoint, we may let

Since , there exists some s.t.

so there exists s.t.

(Here denotes the ball of radius around and denotes the -neignbourhood of compact set )

Now set , for any -chain , we have: Since and , and . Hence the third term of any such chain must be in . Since , no -chain starting at can reach , in particular, the chain does not come back to . Hence we conclude that is not chain recurrent. ii) Suppose not, there is and . i.e. for some there is no -chain from to itself. Let be the open set consisting all points that can be connected from by an -chain.

We wish to generate an attractor by , to do this all we need to check is :

For any there exists with . Since , there is -chain which gives rise to -chain . Therefore .

Hence is an attractor with as a basin.

By assumption, , since and there is no -chain from to itself, cannot be in . i.e. Take any limit point of , since is compact -invariant we have . But since we can find where , gives an -chain from to , hence .

Recall that is a basin of hence is empty. Contradiction. Establishes proposition 2.

This proposition says that to study the chain recurrent set is the same as studying each attractor-repeller pair of the system. But the dynamics is very simple for each such pair as all points not in the pair will move towards the attractor under foreward iterate. We can see that such property is goint to be of importantce for our purpose since the dynamical for each attractor-repeller pair is like the sourse-sink map.

**5. Main ingredient**

Here we are going to prove a lemma that’s going to produce for us the ‘building blocks’ of our final construction. Namely a function for each attracting-repelling pair that strickly decreases along the orbits not in the pair. In light of proposition , we should expect to put those functions together to get our complete Lyapunov function.

**Lemma1:** For each attractor-repeller pair there exists continuous function s.t. and for all .

**Proof:** First we define s.t.

Note that takes value only on and only on . However, can’t care less about orbits of .

Define . Hence automatically for all , . Since no points accumulates to under positive iterations, we still have the and .

We now show that is continuous:

For and , and hence i.e. is continuous on .

For we use the fact that is attracting. Let be a basin of . For all , for any , there is s.t. . Therefore for some , all are in i.e. hence . But for some and all , . Therefore . is continuous on .

Let , for any , since , there exists s.t. for all . i.e. which is countinous. Since those ‘bands’ partitions the whole (by taking to be ), hence we have proven is continuous on the whole .

Finally, we define

We check that is continuous since is. takes values and only on and , respectively. For any ,

therefore iff for all i.e. is constant on the orbit of . But this cannot be since there is a subsequence of converging to some point in , continuity of tells us this constant has to be hence .

Therefore is strictly decreasing along orbits of not in .

Establishes lemma 1.

**6.Proof of the main theorem**

The proof of the main theorem now follows easily from what we have established so far.

First we restate the fundamental theorem of dynamical systems:

**Theorem:** Complete Lyapunov function exists for any homeomorphisms on compact metric spaces.

**Proof:** First we enumerate the countably many attractors as . For each , we have function where is on , on and is strictly decreasing on .

Define by

Since each is bounded between and , the sequence of partial sums converge uniformly. Hence the limit function is continuous.

For points , we have for all . i.e. $latex \ \forall n \in \N, \ g_n(p) = 0$ or . Hence we have

where each is in . This is same as saying the base- expansion of only contains digits and . We conclude where is the standard middle-third Cantor set in . i.e. is compact and nowhere dense in .

For , there exists such that , hence . This implies since for all . i.e. is strictly decreasing along orbits that are not chain recurrent.

To show is constant only on the chain-transitive components, we need the following lemma:

**Claim:** are in the same chain-transitive component iff there is no attracting-repelling pair where one of is in while the other in .

**Proof (of claim):** ”” Suppose and , for any attractor , if and , then are both in and we are done. Hence suppose at least one of is in . W.L.O.G. suppose . Let be a basin of . Since are closed and disjoint, we may choose . By the same arguement as in proposition , there are no -chain (with length ) from any point in to any point in . Hence there is also no -chain from any point in to any point in . Hence i.e. .

”” Suppose for any attractor , iff . For any , let be the set of all points for which there is an -chain from to , as defined in proposition . We have showed in proposition that is a basin of some attractor . Since and , hence . Hence by our assumption, must be also in . Hence i.e. there is an -chain from to . Since the construction is symetric, we may also show there is an -chain from to . i.e. .

Establishes the claim.

Finally, for , means and has the same base- expansion in the Cantor set. This is same as saying , which is to say there is no for which one of is in while the other in . Hence by Lemma, we conclude that iff are in the same chain transitive component.

This establishes our theorem.