(This was the basis for a talk I gave at the Toronto Student Set Theory and Topology seminar on January 15, 2013. The assumed knowledge is an undergraduate course in general topology. This is only a draft, and will be updated soon.)
Introduction
There are many questions in mathematics and sciences in general whose main object of study is the topological group. These objects are very versatile and can represent many of the structures we encounter. One question that I’ve been working on examines the (extreme) dynamical properties of topologies on the integers. On the recommendation of Vladimir Pestov (one of my advisors) I have been learning about Tsequences, which provide a rich method for producing topological groups with extreme behaviour. Here I will present two techniques involving Tsequences that help to answer two different questions about topological groups; one is about dynamics and the other is about combinatorial properties of . These results all come from “Topologies on the Integers” by Protasov and Zelenyuk.
The Words You Should Know
Definition. A sequence is a Tsequence (in ) if there is a topology on such that is a group topology in which converges to 0. (i.e. Every open neighbourhood of contains a tail of ).
Definition. For a Tsequence we let be the strongest topological group where converges to .
In general we will think of a Tsequence as being a (relatively) quickly growing sequence. How fast? Well we will be thinking about geometric series. A Tsequence needs to be at least a little sparse as the next sequence illustrates.
(Non)example 1. Let be an ennumeration of . It cannot be a Tsequence. This is because any open set must be cofinite. So a topology in which this converges would not be .
Example 2. In contrast, trival Tsequences are sequences that end in a tail of zeros. These sequences converge in with the discrete topology.
(Non)example 3. An arithmetic progression cannot be a Tsequence. We will see this later.
Example 4. The geometric progression is a Tsequence. We will also see this later
The Canonical Topology
Suppose that is a sequence in and let be an nondecreasing function. When is a Tsequence, a basis for its topology in which converges to 0 is given by , (with to be defined), but the notation will not require to be a Tsequence.
Define , to be the tail of the sequence (after ).
For define , the symmetrization of , where .
Finally, .
Lemma 1. Fix a (not necessarily T)sequence :

,

,

,

such that ,

such that .
Proof. The first three are obvious from staring. For 4, take to be the pointwise maximum of , (as Stevo said: “It is so simple!”.). For 5, let and . [QED]
This tells us that forms a neighbourhood base of for some group topology on . (All we are missing is .)
Lemma 2. If is a topological group and converges to 0 in , then for every neighbourhood of 0, there is an such that .
proof. Since converges to 0 in , then there is a tail . Let . Since is a topological group, we can inductively create such that , and a tail . Let . We see that , by construction. [QED]
Theorem 3. A sequence in is a Tsequence iff . Thus, is a neighbourhood base of for any topological group where converges to .
proof. If converges to 0 in some topological group , then Lemma 2 and ness will tell you . For the converse, the intersection being empty tells us that the group topology for which forms a neighbourhood base of is in fact a group topology. [QED]
Recognizing Tsequences
The first thing we observe is a characterization of Tsequences, by restating Theorem 3.
Corollary 4. A sequence is a Tsequence iff .
This tells us why arithmetic sequences (example 3) do not form Tsequences.
Somehow we would like to transfer this characterization to something “more finite”.
Definition. Let be any sequence in . For , we define
That is, is the set of sums taken from .
Theorem 5. A sequence is a Tsequence iff .
That is, for each nonzero integer (), if the enemy fixes an amount () of sums we can take, there is a tail () with which we will not be able to make that nonzero integer as a sum taken from .
Proof. The picture is very helpful here, and there is a neat method involved. The next proof will show it off.
Theorem 6 (Exercise 2.1.2). Let and be Tsequences with corresponding topologies and . The following are equivalent:
;
we have
Before we go on the the proof, what is this saying? The big picture is that it is giving us a way to compare two topologies if we know something about how the sequences relate. We would like to say that iff almost contains , this just isn’t true. (If we know that one sequence almost contains the other, then we do get that the topologies are comparable, but we don’t get the other direction.)
The second alternative of the theorem should be interpreted as a way of saying that even though isn’t almost a subset of , you can still (robustly) find tails of as small finite combinations of elements of .
proof. [To be included in a later edit.]
There are a lot of Tsequences
We now turn to the question of how many Tsequences there are in . First, an obvious observation.
Lemma 6. Every subsequence of a Tsequence is again a Tsequence.
Corollary 7. There are many Tsequences in .
So there are lots of Tsequences, in fact there are as many as there could be, but many of these can generate the same topologies. Defining what it means to be “different”, we see that there are many different Tsequences.
Definition. Two (nontrivial) Tsequences are strongly incomparable if there are neighbourhoods of the identity and such that .
We are going to show the following three Theorems, using two combinatorial facts. One fact will say that every Tsequence contains two strongly incomparable Tsequences, and the second will say that every Tsequence can be refined to an alternative sequence, i.e., every (infinitecoinfinite) subsequence is strongly incomparable with its complement.
Theorem 8. There is a family of pairwise strongly incomparable Tsequences.
Theorem 9. There is a family of Tsequences such that if then is refined by .
Theorem 10. There is a family of Tsequences such that if then is refined by .
Combinatorial Fact 11. Every Tsequence contains two strongly incomparable Tsequences.
Combinatorial Fact 12. Every Tsequence contains an alternative subsequence . That is, if , disjoint, both infinite, then and are strongly incomparable.
Question (Protasov and Zelenyuk, 1999): Is there always a decreasing (or equivalently, increasing) chain, of topologies given by Tsequences, of order type ?
Question: Does there exist a topological group G so that for every nontrivial Tsequence (one that does not have a tail consisting solely of 0s) there is a nontrivial (g not the identity) so that converges to the identity of G?
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Do you mean: “For each nontrivial Tsequence we can find a g, and a sequence of integers so that converges to the identity?”
My initial guess is yes, with G being the infinite product of , with n increasing. This seems like cheating though.
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