As promissed in ERT6, this post intends to prove

Theorem 1Let be a homeomorphism of the compact metric space . Then, for any and , there exist and such that

Moreover, given any dense subset , we can take .

Actually, we prove the more general version

Theorem 2Let be a compact metric space and commutative homeomorphisms of . Then some ismultiply recurrent, that is: there exists a sequence such that

**1. Homogeneity **

Consider a homeomorphism .

Definition 3is homogeneous if there is a group of homeomorphisms of , each element of commuting with , such that the action is minimal.

We remark that the notion of minimality of a group action is the same as that of a single homeomorphism:

Definition 4ishomogeneousif there is a group of homeomorphisms of , each element of commuting with , such that and the action is minimal.

In particular, is closed. Note that does not need to be -invariant. This is the good point of the above definition: we analyse the dynamics of not by the original map , but through the action of . We now proceed to the technical details of the proof.

Proposition 5Let be a homeomorphism and homogeneous. Suppose that, for each , there exist and such that

Then, for every , there exist and such that

In other words, the property assumed also holds for a pair on the diagonal .

*Proof:* Let be the group associated to the homegeneity of and . The -orbit of every is dense in and then becomes as close as we want to any other .

Let us prove that, in a scale of , the -orbits in are finite. More specifically, we prove the existence of for which

Consider an open cover of by , each of them with diameter at most . For each , the family is an open cover of (in fact, if , there exists such that , that is, ). Reduce this cover to a finite one:

The finite set satisfies the required property: if , then , for some , and , for some , implying that

Just rename the .

We now prove that, for every , there exist and such that

Let be such that

and , such that

Taking for which

and , we conclude the inequalities

Finally, we prove the proposition. Take and such that

By continuity, there is for which

Take and satisfying

Proceeding by induction, if

take such that

and , for which

Then, if ,

But may be chosen such that and then, taking , (1) implies the inequality

Proposition 6Under the same conditions of Proposition 5, there exists recurrent for .

*Proof:* Define by

From Proposition 5, the image of has values arbitrarily close to zero. In addition, is upper semicontinuous. In fact, given and , let such that

By continuity of , there is for which

Then

and so, as is arbitrary,

Then the set of continuity points of is residual (in particular, it is non-empty). Let us show that . This will conclude the proof.

By contradiction, suppose that for some . Take a neighbourhood of and such that

We now use the homogeneity of to prove that a similar inequality to (1) holds in all of .

Let be the group associated to the homogeneity of . By compactness of and minimality of , there are such that

Let us prove that , for every . If , there exists such that

and then, taking for which , (2) guarantees that

which contradicts (1).

**2. Proof of Theorem 2 **

We proceed by induction. For , it is sufficient to take any point of a minimal set (which exists by Zorn’s Lemma). Suppose the result is true for and consider commutative homeomorphisms of . Let be the group generated by . We can assume that acts minimaly on . If this is not the case, we restrict to a -invariant closed subset for which is minimal (such minimal set exists again by Zorn’s Lemma).

Let be the diagonal and consider the product transformation . We wish to show that there exists recurrent for . To this matter, it suffices to check the hypotheses of Proposition 5:

I. is homogeneous for .

II. For each , there exist and such that

The action of can be induced in in a natural way, associating to the map . In this setting, if , the pairs and are isomorphic and then acts minimaly on . In particular, is homogeneous for , establishing I.

To prove II, define , . By the induction hypothesis, there are and such that

Taking

we get

concluding the proof of the theorem.

**Previous posts:** ERT0, ERT1, ERT2, ERT3, ERT4, ERT5, ERT6.

## Leave a Reply