Proof. We have . In addition, , so is porous.
Let be a norm on . Let be a countable dense subset of . Choose such that . Then, .
By Lemma 1, choose such that is porous and such that is periodic to all orders. Because is porous, is dense in , so fix such that .
By Lemma 1, choose such that is porous and such that is periodic to all orders. Because is porous, is dense in , so fix such that .
By Lemma 1, choose such that is porous and such that is periodic to all orders. Because is porous, is dense in , so fix such that .
Continuing yields a countable dense subset of , and a sequence in . For each integer ,
We have and . It follows both that and that . Then, and . Define by the rule: For all integers , on . Let .
Claim 1: For all , for all , we have .
Proof of Claim 1. Let and let . Assume, for a contradiction, that we have .
Since
, we see that
is compact. Then, since
and
, fix an integer
such that
. We have
on
. Then, by Lemma 6.1 of [
6],
.
Then, . However because , it follows that . Thus, we have , contradiction. □
Claim 2: For all integers , is periodic to all orders.
Proof of Claim 2. Fix an integer . We wish to show that is periodic to all orders.
Because
and
is periodic, we conclude, from Lemma 8.13 of [
6], that
. For all
, because
and
agree on
, which is an open neighborhood of
, it follows that they agree to all orders at
. Thus, for all
,
and
agree to all orders at
. Then, by Lemma 6.3 of [
6], for all
and
agree to all orders at
. Thus, as
is periodic to all orders, it follows
is periodic to all orders as well. □
Because Z is infinite cyclic, it follows that Z is isomorphic to the additive discrete group . Let be an isomorphism. Define a Z-action on by: for all , for all , . Let . Because the Z-action on is , it follows that the G-action on M is as well. By construction, M is (the total space of) a fiber bundle over with fiber , so, because is connected, M has the same number of connected components as does .
By Corollary 8.4(i) of [
6],
is an open subset of
. For all integers
,
is porous. Then,
is a dense open subset of
. Then, because
, we see, by the Baire Category Theorem, that
X is dense in
. By Claim 1, for all
, for all
, we have
, i.e., we have
. Thus, the
Z-action on
is fixpoint rare. Then, the
G-action on
M is also fixpoint rare.
Let be the canonical map. Define an injection by . Let . Because is dense in , is dense in M.
Fix an integer , and let be the kth order frame bundle of M. Let . Since A is dense in M, and since is open, it follows that D is dense in . Fix . We wish to show that is infinite.
Let . Then, . We therefore wish to show that is infinite.
Since , fix , such that . Let . Then, , i.e., . Let . Then, . As , we have . Then, , so it suffices to show that is infinite.
Recall that is the identity map defined by . Let be the identity map defined by .
Since , fix an integer such that . By Claim 2, fix such that agrees with the identity to all orders at .
Let . Then, , and, for all , we have . Then, the map is equal to , and therefore agrees with to all orders at . Then, since and since , it follows that the map agrees with at to all orders. In particular, agrees with at to order k. Then, for all , we have . Thus, since , we get . That is, . Let be the cyclic subgroup of generated by . Every nontrivial subgroup of an infinite cyclic group is infinite, so is infinite. Thus, because , it follows that is infinite, as desired. □