Let’s Make Some T-sequences

(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 T-sequences, which provide a rich method for producing topological groups with extreme behaviour. Here I will present two techniques involving T-sequences that help to answer two different questions about topological groups; one is about dynamics and the other is about combinatorial properties of $\mathbb{N}^\N$ . These results all come from “Topologies on the Integers” by Protasov and Zelenyuk.

The Words You Should Know

Definition. A sequence $(a_n \in \mathbb{Z}: n <\omega)$ is a T-sequence (in $\mathbb{Z}$ ) if there is a topology $\tau$ on $\mathbb{Z}$ such that $(\Z, \tau)$ is a $T_2$ group topology in which $(a_n)$ converges to 0. (i.e. Every open neighbourhood of $0$ contains a tail of $(a_n)$ ).

Definition. For a T-sequence $(a_n)$ we let $(\Z, (a_n))$ be the strongest $T_2$ topological group where $(a_n)$ converges to $0$ .

In general we will think of a T-sequence as being a (relatively) quickly growing sequence. How fast? Well we will be thinking about geometric series. A T-sequence needs to be at least a little sparse as the next sequence illustrates.

(Non-)example 1. Let $(a_n)$ be an ennumeration of $\mathbb{Z}$ . It cannot be a T-sequence. This is because any open set must be co-finite. So a topology in which this converges would not be $T_2$ .

Example 2. In contrast, trival T-sequences are sequences that end in a tail of zeros. These sequences converge in $\mathbb{Z}$ with the discrete topology.

(Non-)example 3. An arithmetic progression cannot be a T-sequence. We will see this later.

Example 4. The geometric progression $(2^n : n<\omega)$ is a T-sequence. We will also see this later

The Canonical Topology

Suppose that $(a_n)$ is a sequence in $\mathbb{Z}$ and let $f: \omega \rightarrow \omega$ be an non-decreasing function. When $(a_n)$ is a T-sequence, a basis for its topology in which $(a_n)$ converges to 0 is given by $\{U_f : f \in \omega^\omega\}$ , (with $U_f$ to be defined), but the notation will not require $(a_n)$ to be a T-sequence.

Define $A_N := \{a_k : k \geq N \}$ , to be the tail of the sequence (after $N$ ).

For $X \subseteq \mathbb{Z}$ define $X^* := -X \cup \{0\} \cup X$ , the symmetrization of $X$ , where $-X := \{-x : x \in X\}$ .

Finally, $U_f := \bigcup_{n < \omega} (A_{f(0)}^* + A_{f(1)}^* + ... + A_{f(n)}^*)$ .

Lemma 1. Fix a (not necessarily T-)sequence $(a_n)$ :

$\forall f, U_f \subseteq \mathbb{Z}$ ,
$\forall f, 0 \in U_f$ ,
$\forall f, U_f = -U_f$ ,
$\forall f,g \exists h$ such that $U_h \subseteq U_f \cap U_g$ ,
$\forall h, \exists f,g,$ such that $U_f + U_g \subseteq U_h$ .

Proof. The first three are obvious from staring. For 4, take $h$ to be the pointwise maximum of $f,g$ , (as Stevo said: “It is so simple!”.). For 5, let $f(n) = h(2n)$ and $g(n) = h(2n+1)$ . [QED]

This tells us that $\{U_f : f \in \omega^\omega\}$ forms a neighbourhood base of $0$ for some group topology on $\mathbb{Z}$ . (All we are missing is $T_2$ .)

Lemma 2. If $(\Z, \tau)$ is a topological group and $(a_n)$ converges to 0 in $\mathbb{Z}$ , then for every neighbourhood $U$ of 0, there is an $f: \omega \rightarrow \omega$ such that $U_f \subseteq U$ .

proof. Since $(a_n)$ converges to 0 in $(\Z, \tau)$ , then there is a tail $A_{N_0} \subseteq U := U_0$ . Let $f(0) = N_0$ . Since $(\Z, \tau)$ is a topological group, we can inductively create $U_{n+1}$ such that $U_{n+1} + U_{n+1} \subseteq U_n$ , and a tail $A_{N_{n+1}} \subseteq U_{n+1}$ . Let $f(n+1) = N_{n+1}$ . We see that $U_f \subseteq U$ , by construction. [QED]

Theorem 3. A sequence $(a_n : n<\omega)$ in $\mathbb{Z}$ is a T-sequence iff $\bigcap_{f \in \omega^\omega} U_f = \emptyset$ . Thus, $\{U_f : f \in \omega^\omega\}$ is a neighbourhood base of $0$ for any topological group $(\Z, \tau)$ where $(a_n)$ converges to $0$ .

proof. If $(a_n)$ converges to 0 in some $T_2$ topological group $(\Z, \tau)$ , then Lemma 2 and $T_2$ -ness will tell you $\bigcap_{f \in \omega^\omega} U_f = \emptyset$ . For the converse, the intersection being empty tells us that the group topology for which $\{U_f : f \in \omega^\omega\}$ forms a neighbourhood base of $0$ is in fact a $T_2$ group topology. [QED]

Recognizing T-sequences

The first thing we observe is a characterization of T-sequences, by restating Theorem 3.

Corollary 4. A sequence $(a_n)$ is a T-sequence iff $(\forall n \in \mathbb{Z}, n \neq 0)(\exists f \in \omega^\omega)[n \notin U_f]$ .

This tells us why arithmetic sequences (example 3) do not form T-sequences.

Somehow we would like to transfer this characterization to something “more finite”.

Definition. Let $(a_n)$ be any sequence in $\mathbb{Z}$ . For $k,N \in \omega$ , we define
$\displaystyle A(k,N) := \{g_0 + ... + g_{k-1} : g_0, ..., g_{k-1} \in A_N^*\}$
That is, $A(k,N)$ is the set of $k$ -sums taken from $A_N$ .

Theorem 5. A sequence $(a_n)$ is a T-sequence iff $(\forall g \in \mathbb{Z}, g \neq 0)(\forall k \in \omega)(\exists N \in \omega)[g \notin A(k,N)]$ .

That is, for each non-zero integer ( $g$ ), if the enemy fixes an amount ( $k$ ) of sums we can take, there is a tail ( $A_N$ ) with which we will not be able to make that non-zero integer as a $k$ -sum taken from $A_N$ .

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 $(a_n)$ and $(b_n)$ be T-sequences with corresponding topologies $\tau_a$ and $\tau_b$ . The following are equivalent:
$\tau_a \subseteq \tau_b$ ;
$(\exists k)(\forall m_k)(\exists N)(\forall n > N)$ we have $b_n \in A(k, m_k)$

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 $\tau_a \subseteq \tau_b$ iff $(a_n)$ almost contains $(b_n)$ , 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 $(b_n)$ isn’t almost a subset of $(a_n)$ , you can still (robustly) find tails of $(b_n)$ as small finite combinations of elements of $(a_n)$ .

proof. [To be included in a later edit.]

There are a lot of T-sequences

We now turn to the question of how many T-sequences there are in $\mathbb{Z}$ . First, an obvious observation.

Lemma 6. Every subsequence of a T-sequence is again a T-sequence.

Corollary 7. There are $\mathfrak{c}$ -many T-sequences in $\mathbb{Z}$ .

So there are lots of T-sequences, 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 $\mathfrak{c}$ -many different T-sequences.

Definition. Two (non-trivial) T-sequences $(a_n), (b_n)$ are strongly incomparable if there are neighbourhoods of the identity $A \in (\Z, (a_n))$ and $B \in (\Z, (b_n))$ such that $A \cap B = \emptyset$ .

We are going to show the following three Theorems, using two combinatorial facts. One fact will say that every T-sequence contains two strongly incomparable T-sequences, and the second will say that every T-sequence can be refined to an alternative sequence, i.e., every (infinite-coinfinite) subsequence is strongly incomparable with its complement.

Theorem 8. There is a family $(A_\alpha : \alpha < \mathfrak{c})$ of pairwise strongly incomparable T-sequences.

Theorem 9. There is a family $(A_x : x \in \mathbb{R})$ of T-sequences such that if $x < y$ then $(\Z, A_x)$ is refined by $(\Z, A_y)$ .

Theorem 10. There is a family $(A_\alpha : \alpha \in \omega_1)$ of T-sequences such that if $\alpha < \beta$ then $(\Z, A_\alpha)$ is refined by $(\Z, A_\beta)$ .

Combinatorial Fact 11. Every T-sequence contains two strongly incomparable T-sequences.

Combinatorial Fact 12. Every T-sequence $(a_n)$ contains an alternative subsequence $(b_n)$ . That is, if $(b_n) = B_1 \cup B_2$ , disjoint, both infinite, then $B_1$ and $B_2$ are strongly incomparable.

Question (Protasov and Zelenyuk, 1999): Is there always a decreasing (or equivalently, increasing) chain, of topologies given by T-sequences, of order type $\omega_2$ ?

2 thoughts on “Let’s Make Some T-sequences”

Aaron Hill says:

February 8, 2013 at 10:33 am

Question: Does there exist a topological group G so that for every non-trivial T-sequence (one that does not have a tail consisting solely of 0s) there is a nontrivial $g \in G$ (g not the identity) so that $g^{i_n}$ converges to the identity of G?

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1. Micheal Pawliuk says:
  
  February 10, 2013 at 6:16 pm
  
  Do you mean: “For each non-trivial T-sequence we can find a g, and a sequence of integers $i_n$ so that $g^{i_n}$ converges to the identity?”
  
  My initial guess is yes, with G being the infinite product of $Z_{2n}$ , with n increasing. This seems like cheating though.
  
  LikeLike

Comments are closed.

Introduction

The Words You Should Know

The Canonical Topology

Lemma 1. Fix a (not necessarily T-)sequence :

,

,

,

such that ,

such that .

Lemma 2. If is a topological group and converges to 0 in , then for every neighbourhood of 0, there is an such that .

Theorem 3. A sequence in is a T-sequence iff . Thus, is a neighbourhood base of for any topological group where converges to .

Recognizing T-sequences

Corollary 4. A sequence is a T-sequence iff .

Theorem 5. A sequence is a T-sequence iff .

There are a lot of T-sequences

Lemma 6. Every subsequence of a T-sequence is again a T-sequence.

Corollary 7. There are -many T-sequences in .

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2 thoughts on “Let’s Make Some T-sequences”

Lemma 1. Fix a (not necessarily T-)sequence $(a_n)$ :

$\forall f, U_f \subseteq \mathbb{Z}$ ,

$\forall f, 0 \in U_f$ ,

$\forall f, U_f = -U_f$ ,

$\forall f,g \exists h$ such that $U_h \subseteq U_f \cap U_g$ ,

$\forall h, \exists f,g,$ such that $U_f + U_g \subseteq U_h$ .

Lemma 2. If $(\Z, \tau)$ is a topological group and $(a_n)$ converges to 0 in $\mathbb{Z}$ , then for every neighbourhood $U$ of 0, there is an $f: \omega \rightarrow \omega$ such that $U_f \subseteq U$ .

Theorem 3. A sequence $(a_n : n<\omega)$ in $\mathbb{Z}$ is a T-sequence iff $\bigcap_{f \in \omega^\omega} U_f = \emptyset$ . Thus, $\{U_f : f \in \omega^\omega\}$ is a neighbourhood base of $0$ for any topological group $(\Z, \tau)$ where $(a_n)$ converges to $0$ .

Corollary 4. A sequence $(a_n)$ is a T-sequence iff $(\forall n \in \mathbb{Z}, n \neq 0)(\exists f \in \omega^\omega)[n \notin U_f]$ .

Theorem 5. A sequence $(a_n)$ is a T-sequence iff $(\forall g \in \mathbb{Z}, g \neq 0)(\forall k \in \omega)(\exists N \in \omega)[g \notin A(k,N)]$ .

Corollary 7. There are $\mathfrak{c}$ -many T-sequences in $\mathbb{Z}$ .