A new zero-dimensional Polish group

Greg Hjorth

August 6, 1998

0. Introduction

Here we present an example of an abelian zero-dimensional Polish group G that is generated by every open neighborhood of the identity – that is to say, if O ⊂ G is open and x ∈ G then there is some finite sequence ϵ_{1}, ϵ_{2}, ..., ϵ_{n} ∈ O such that

x = ϵ_{1} + ϵ_{2} + ...ϵ_{n}.

The main point here is that any zero-dimensional locally compact Polish group has a neigborhood basis at the identity consisting of (necessarily clopen) subgroups.

1. The group

1.1 Definition For x ∈ ℝ we say that a sequence of integers (a_{i})_{i∈ℕ^{+}} is a representation of x if:

(i) at each n we have −10^{−n} < a_{n} < 10^{n};

(ii) there is some integer k with

k + ∑_{n∈ℕ^{+}} \frac{a_{n}}{10^{1} · 10^{2} · ... · 10^{n}} = x.

For a⃗ = (a_{i})_{i∈ℕ^{+}} a representation of some real we let

||a⃗|| = max_{n∈ℕ^{+}}|\frac{a_{n}}{10^{n}} |.

We let ||x|| be the infinum of the set {||a⃗|| : a is a representation of x}. We then let G be the set of x ∈ ℝ for which there is some representation a⃗ of x with

\frac{a_{n}}{10^{n}} → 0 as n → ∞.

A representation a⃗ of a real x is legitimate if we do indeed have that

\frac{a_{n}}{10^{n}} → 0

as n → ∞.

1.2 Lemma G is a subgroup of (ℝ, +).

Proof. It is immediately clear from the structure of the definitions that −x ∈ G if x ∈ G. For additive closure, suppose x, y ∈ G with legitimate representations a⃗, b⃗. Then we may find a representation c⃗ of x + y with

|c_{n}| ≤ |a_{n}| + |b_{n}| + 1

for each n ∈ ℕ^{+}. □

1.3 Lemma For all x ∈ G we have

||x|| ≤ \frac{6}{10}.

Proof. Suppose a⃗ is a legitimate representation of x and

k + ∑_{n∈ℕ^{+}} \frac{a_{n}}{10^{1} · 10^{2} · ... · 10^{n}} = x.

Then we may successively produce integers l, â_{1}, â_{2}, ... and then b_{1}, b_{2}, b_{3}, ... such that at each M

l + ∑_{n<M} \frac{b_{n}}{10^{1} · 10^{2} · ... · 10^{n}} + â_{M} + ∑_{n>M} \frac{a_{n}}{10^{1} · 10^{2} · ... · 10^{n}} = x

such that

|\frac{â_{n}}{10^{n}} | ≤ \frac{1}{2},

|b_{n}| ≤ |â_{n}| + 1.

The first step of this process is the obvious one: If |a_{1}| > 5 then we simply choose an integer l ∈ {k − 1, k + 1} and â_{1} such that

l + \frac{â_{1}}{10^{1}} = k + \frac{a_{1}}{10^{1}} ,

|â_{1}| < 5.

If on the other hand |a_{1}| ≤ 5 we have no work to do, and simply set l = k and â_{1} = a_{1}.

At the next stage it is similar: If |a_{2}| > 50 then we choose b_{1} ∈ {â_{1} − 1, â_{1} + 1} and â_{2} such that

b_{1} + \frac{â_{2}}{10^{2}} = â_{1} + \frac{a_{2}}{10^{2}} ,

|â_{2}| < 50.

Again if |a_{2}| ≤ 50 then simply set b_{1} = â_{1} and â_{2} = a_{2}.

We continue in this manner. □

In general if a⃗ and b⃗ are two legitimate representations of a real x then at each N

| ∑_{n>N} \frac{a_{n}}{10^{1} · 10^{2} · ... · 10^{n}} − ∑_{n>N} \frac{b_{n}}{10^{1} · 10^{2} · ... · 10^{n}}| ≤ \frac{1}{10^{1} · 10^{2} · ... · 10^{N}}.

Therefore for some i ∈ {0, 1, −1}

| ∑_{n≤N} \frac{a_{n}}{10^{1} · 10^{2} · ... · 10^{n}} − ∑_{n≤N} \frac{b_{n}}{10^{1} · 10^{2} · ... · 10^{n}} + 1| ≤ \frac{1}{10^{1} · 10^{2} · ... · 10^{N}},

and thus a_{N} must equal b_{N} − 1, b_{N}, or b_{N} + 1 modulo 10^{N}.

1.4 Lemma If x ∈ G then there is a legitimate representation a⃗ with

||x|| = ||a⃗||.

Proof. Let a⃗ be a legitimate representation of x. By the previous lemma we may assume

||a⃗|| < \frac{8}{10}.

Choose N ∈ ℕ^{+} with

\frac{|a_{n}| + 1}{10^{n}} < 10^{−1}

all n > N. Let b⃗ be a competing legitimate representation of x with ||b⃗|| ≤ ||a⃗||. Claim: For all n > N

a_{n} = b_{n}.

Proof of claim: Assume for a contradiction that a_{n} ≠ b_{n} and n > N. There are two possibilities.

Case(i) a_{n} = b_{n} mod 10^{n}. Then since −10^{n} < a_{n}, b_{n} < 10^{n} we must have

|a_{n}| + |b_{n}| = 10^{n}.

∴ \frac{|b_{n}|}{10^{n}} = 1 − \frac{|a_{n}|}{10^{n}} > 1 − 10^{−1},

by the assumption on n > N, with a contradiction to ||b⃗|| ≤ ||a⃗||.

Case(ii) a_{n} = b_{n} − 1 mod 10^{n} or a_{n} = b_{n} − 1 mod 10^{n}.

There is a further split.

Case(iia) a_{n} ≠ b_{n} − 1, b_{n} + 1. Then it is as case(i) with |a_{n}| + |b_{n}| = 10^{n} ± 1, and hence

\frac{|b_{n}|}{10^{n}} = 1 − \frac{|a_{n}| ± 1}{10^{n}} > 1 − \frac{2}{10} = \frac{8}{10},

violating ||b⃗|| ≤ ||a⃗||.

Case(iib) a_{n} = b_{n} − 1 or a_{n} = b_{n} + 1. Thus for some η with η ∈ {−1, 0, 1}. Then

\frac{a_{n}}{10^{n}} + \frac{a_{n+1}}{10^{n} · 10^{n+1}} = \frac{b_{n}}{10^{n}} + \frac{b_{n+1}}{10^{n} · 10^{n+1}} + \frac{η}{10^{n} · 10^{n+1}}.

Then we have a_{n+1} not equal to b_{n+1} − 1, b_{n+1}, or b_{n+1} + 1, and we are back into the contradiction of case(i). (Claim □)

Thus in determining ||x|| we are taking the infinum over a finite set. Thus this inf is attained. □

1.5 Lemma For x, y ∈ G,

||x|| = || − x||

and

||x + y|| ≤ ||x|| + ||y||.

Proof. The equality ||x|| = ||−x|| is obvious from the very form of the definitions. For the inequality, we may as well assume ||x|| + ||y|| < 1, or else we are done by

lemma 1.3. Fixing representations a⃗, b⃗ with

||a⃗|| = ||x||

and

||b⃗|| = ||y||

we have in particular at each n that

\frac{|a_{n}|}{10^{n}} + \frac{|b_{n}|}{10^{n}} < 1.

Therefore if we just let c_{n} = a_{n} + b_{n} we obtain a representation for x + y with

|c_{n}| ≤ |a_{n}| + |b_{n}|.
□

1.6 Corollary If we set d̂(x, y) = ||x − y||, then d̂ defines a pseudo-metric on G.

1.7 Definition We let d_{ℝ} be the usual euclidean metric on ℝ and define d_{G} on G by

d_{G}(x, y) = d_{ℝ}(x, y) + d̂(x, y).

Then d_{G} is the sum of a pseudo-metric and metric, and therefore is a metric.

1.8 Lemma d_{G} is a complete metric on G.

Proof. Let (x_{i})_{i∈ℕ} be a Cauchy sequence in (G, d_{G}). Fix at each i a representation a^{i} = (a_{1}^{i}, a_{2}^{i}, a_{3}^{i}, ...) of x_{i} with

||a^{i}|| = ||x_{i}||.

After perhaps thinning out the original sequence of reals we may assume that for each n the sequence (a_{n}^{0}, a_{n}^{1}, a_{n}^{2}, ...) is eventually constant. We may also assume that at each i

x_{i} = ∑_{n∈ℕ^{+}} \frac{a_{n}^{i}}{10^{1} · 10^{2} · ... · 10^{n}};

in other words, the k is always zero. Letting a_{n}^{∞} be the limit of the sequence (a_{n}^{0}, a_{n}^{1}, a_{n}^{2}, ...), we will show that a^{∞} is a legitimate representation of some x_{∞} ∈ G. Claim: For all ϵ > 0 there is some N_{ϵ} and I_{ϵ} such that for all m > N_{ϵ} and j > I_{ϵ}

\frac{a_{m}^{j}}{10^{m}} < ϵ.

Proof of claim: Fix ϵ > 0 and assume without loss of generality that ϵ < 10^{−1}. Choose I_{ϵ} so that for all i, j ≥ I_{ϵ} we have

d_{G}(x_{i}, x_{j}) < \frac{ϵ}{3}

and

\frac{ϵ}{3} > 10^{−I_{ϵ}}.

Let N_{ϵ} be such that for all m > N_{ϵ}

\frac{a_{m}^{I_{ϵ}}}{10^{m}} < \frac{ϵ}{3}.

Then for all j > I_{ϵ} we may obtain a legitimate representation c⃗ of x_{j} such that at each m > N_{ϵ}

\frac{c_{m}}{10^{m}} < \frac{ϵ}{3} + \frac{ϵ}{3} = \frac{2ϵ}{3}.

If

\frac{a_{m}^{j}}{10^{m}} ≥ ϵ,

then

|a_{m}^{j} − c_{m}| > 1,

and thus we must be in the situation that there is some η ∈ {−1, 0, 1} such that

|a_{m}^{j} − c_{m}| = η mod 10^{j}

but

|a_{m}^{j} − c_{m}| ≠ η

Thus

\frac{|a_{m}^{j}|}{10^{m}} > 1 − \frac{|c_{m}|}{10^{m}} − \frac{1}{10^{m}} ,

entailing the conclusion that

||a^{j}|| > \frac{8}{10},

which was already prohibited by lemma 1.3 and the assumption ||x_{j}|| = ||a^{j}||. (Claim□)

Thus we may indeed obtain a real x_{∞} ∈ G with representation

x_{∞} = ∑_{n∈ℕ^{+}} \frac{a_{n}^{∞}}{10^{1} · ...10^{n}}.

To see that

x_{i} → x_{∞}

in (G, d_{G}) we appeal to the claim once more: Fix δ > 0 and assume that δ < 1. Then we may find N_{δ/2} and I_{δ/2} as in claim; in particular this gives

\frac{a_{n}^{∞}}{10^{n}} < \frac{δ}{2}

all n > N_{δ/2}. By possibly enlarging I_{δ/2} we may assume that for all i > I_{δ/2} and n ≤ N_{δ/2}

a_{n}^{i} = a_{n}^{∞}.

Then we obtain that for all i > I_{δ/2} and at all n ∈ ℕ

\frac{|a_{n}^{i} − a_{n}^{∞}|}{10^{n}} < δ,

and thus ||x_{n} − x_{∞}|| has a representation c⃗ with

||c⃗|| < δ.
□

1.9 Lemma (G, d_{G}) is separable.

Proof. Let A ⊂ G be the set of all reals in G with a representation that has finite support – that is to say, if x ∈ A then there will be some k ∈ ℤ and (a_{i})_{i∈ℕ} such that at each n we have −10^{−n} < a_{n} < 10^{n}, k + ∑_{n∈ℕ^{+}} \frac{a_{n}}{10^{1}·10^{2}·...·10^{n}} = x, and a_{n} = 0 for sufficiently large n.

Then given any y ∈ G and ϵ > 0 we need to show

{x ∈ G : d_{G}(x, y) < ϵ} ∩ A ≠ 0.

Suppose that

l + ∑_{n∈ℕ^{+}} \frac{b_{n}}{10^{1} · 10^{2} · ... · 10^{n}} = y

with l ∈ Z and b⃗ a legitimate representation. Then we may find N such that for all n > N

\frac{|b_{n}|}{10^{n}} < ϵ.

Then if we let

x = l + ∑_{n≤N} \frac{b_{n}}{10^{1} · 10^{2} · ... · 10^{n}}

we have our member of A. □

1.10 Lemma The group operations in (G, d_{G}) are continuous.

Proof. Suppose that x_{0}, y_{0} ∈ G. Then for any ϵ > 0 and x, y ∈ G with

||x_{0} − x||, ||y_{0} − y|| < ϵ

then by 1.5 we have

||(−x_{0}) − (−x)|| = || − (x_{0} − x)|| = ||x_{0} − x|| < ϵ

and

||(x_{0} + y_{0}) − (x + y)|| = ||(x_{0} − x) + (y_{0} − y)|| ≤ ||x_{0} − x|| + ||y_{0} − y|| < 2ϵ.
□

1.11 Lemma (G, d_{G}) is zero-dimensional.

Proof. It suffices by 1.10 to show that there is a neighborhood basis at the identity consisting of clopen sets. So fix ϵ > 0, and we will show that there is some clopen set O such that

0 ∈ O ⊂ {x ∈ G : d_{G}(x, 0) < ϵ}.

Choose irrational δ > 0 with

δ < \frac{1}{10^{2}} , ϵ.

Let O = {x ∈ G : d_{ℝ}(x, 0) < 10δ, ||x|| < δ}.

Clearly O is open in (G, d_{G}), so we just need to show it is closed. For this purpose, fix x ∈ G \ O.

If ||x|| ≥ δ then since ||x|| rational by 1.4, and we have ||x|| > δ. Since {y ∈ G : ||y|| > δ} is an open set, we have shown that there is an open neighborhood around x disjoint from O.

Alternatively if ||x|| < δ then we must have d_{ℝ}(x, 0) ≥ 10δ. Then it follows that x has a representation a⃗ with ||a⃗|| < δ and

k + ∑_{n∈ℕ^{+}} \frac{a_{n}}{10^{1} · 10^{2} · ... · 10^{n}} = x,

and necessarily k ≠ 0. Thus in fact

d_{ℝ}(x, 0) > \frac{1}{2}

and we have an open set containing x and disjoint from ). □

1.12 Lemma Every neighborhood of the identity in (G, d_{G}) generates the group. Proof. Fix ϵ > 0 and x ∈ G with a legitimate representation a⃗ with

k + ∑_{n∈ℕ^{+}} \frac{a_{n}}{10^{1} · 10^{2} · ... · 10^{n}} = x.

Choose N ∈ ℕ with

\frac{|a_{n}|}{10^{n}}, \frac{1}{10^{n}} < ϵ

all n ≥ N.

Then we may certainly obtain

k + ∑_{n<N} \frac{a_{n}}{10^{1} · 10^{2} · ... · 10^{n}}

in the subgroup of G generated by

\frac{1}{10^{N}} .

Since

d_{G}(0, \frac{1}{10^{n}}), d_{G}(0, ∑_{n≥N} \frac{a_{n}}{10^{1} · 10^{2} · ... · 10^{n}}) < ϵ

we have shown that x is generated by {y ∈ G : d_{G}(0, y) < ϵ}. □

1.13 Theorem In the topology generated by the metric d_{G}, G is a zero-dimensional Polish group.

Proof. This is exactly the content of the last five lemmas. □

References

[1] R.D. Mauldin, The Scottish Book (Mathematics from the Scottish Café), Birkhäuser, Boston, 1981.

6363 MSB Mathematics UCLA CA90095-1555 greg@math.ucla.edu www.math.ucla.edu/˜ greg