As promissed in ERT6, this post intends to prove
Theorem 1 Let 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
Consider a homeomorphism .
Definition 3 is 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 4 is homogeneous if 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.
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 6 Under the same conditions of Proposition 5, there exists recurrent for .
Proof: Define by
By continuity of , there is for which
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.
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
concluding the proof of the theorem.