Proof. By [Reference Chalebgwa and Morris2, remark after Proposition 2], every nonzero integer multiple of a Liouville number is a Liouville number. (See also Remark 2.8.) Thus, the group $\operatorname {\mathrm {{gp}}}\{g\}$ generated by each Liouville number is a discrete cyclic subgroup of $\mathbb {R}$ and is a Liouville number group. Further, $\operatorname {\mathrm {{gp}}}\{g\}$ is isomorphic as a topological group to $\mathbb {Z}$ . Since each cyclic subgroup of $\mathbb {R}$ is countable, it contains only countably many Liouville numbers. As there are $\frak c$ Liouville numbers, the family $\{\operatorname {\mathrm {{gp}}}\{g\}:g\in \mathcal {L}\}$ contains $\frak c$ distinct groups.

Proof. We use Petruska’s criterion [Reference Petruska9]: if $a\ge 2$ is an integer and $ \xi =\sum _{n=1}^{\infty } a^{-v_n}, $ where $v_n$ is a strictly increasing sequence of positive integers satisfying $ v_{n+1}\ge 2v_n+2 $ for all sufficiently large n, then $\xi $ is not a strong Liouville number.

For the Liouville constant, take $a=10$ and $v_n=n!$ . Then, $ \ell =\sum _{n=1}^{\infty }10^{-v_n}. $ Moreover, $ v_{n+1}=(n+1)!=(n+1)n!. $ For every $n\ge 2$ , we have $ (n+1)n! \ge 2n!+2, $ because $ (n-1)n!\ge 2. $ Thus, $ v_{n+1}\ge 2v_n+2 $ for all $n\ge 2$ . Therefore, Petruska’s criterion applies and $\ell $ is not a strong Liouville number.

Proof. The determinant identity $ p_{k+1}q_k-p_kq_{k+1}=(-1)^k $ is standard in continued fraction theory and immediately yields the first formula. Since $p_k/q_k\to \mu $ , summing from $k=0$ to N and letting $N\to \infty $ gives

$$ \begin{align*} \mu-\frac{p_0}{q_0} = \sum_{k=0}^{\infty} \bigg(\frac{p_{k+1}}{q_{k+1}}-\frac{p_k}{q_k}\bigg). \end{align*} $$

As $p_0/q_0=a_0$ , the result follows.

Proof. By definition, a strong Liouville number is in particular a Liouville number. Hence, x is Liouville.

Write $ r = {a}/{b} $ with $a \in \mathbb {Z} \setminus \{0\}$ and $b \in \mathbb {N}$ . We prove that $rx$ is Liouville. Let $N \ge 1$ be arbitrary. Since x is Liouville, there exist infinitely many rational numbers ${p}/{q}$ with $q \ge 1$ such that $ 0 < | x - {p}{/q} | < {1}/{q^{N+1}}. $ Then, $ rx - {ap}/{bq} = r(x - {p}/{q}) $ and, therefore, $ 0 < | rx - {ap}/{bq} | = |r| | x - {p}/{q} | < {|r|}/{q^{N+1}}. $

Given p and q, set $P = ap$ and $Q = bq$ . Then, $Q \ge q$ and

$$ \begin{align*} 0 < \bigg| rx - \frac{P}{Q} \bigg| < \frac{|r|}{q^{N+1}} = \frac{|r| b^{N+1}}{Q^{N+1}}. \end{align*} $$

Since there are infinitely many such q, we may choose q sufficiently large so that $ |r| b^{N+1} \le q. $ For such q, we have $ {|r| b^{N+1}}/{Q^{N+1}} \le {1}/{Q^N}. $ Hence, for every $N \ge 1$ , there exist integers $P, Q$ with $Q \ge 1$ such that

$$ \begin{align*} 0 < \bigg| rx - \frac{P}{Q} \bigg| < \frac{1}{Q^N}. \end{align*} $$

Since there are infinitely many choices for $p/q$ , there are infinitely many such $P/Q$ . This shows that $rx$ is a Liouville number.

Proof. Let ${p_n}/{q_n}$ be the sequence of convergents of x. Then, the convergents of $-x$ are given by $ -{p_n}/{q_n}. $ In particular, the sequence of denominators $(q_n)$ is unchanged. Since x is a strong Liouville number, its convergents satisfy the defining growth condition required in Definition 2.1 (or the equivalent characterisation in Remark 2.2). Because the denominators and approximation quality are preserved under negation, the same condition holds for $-x$ . Hence, $-x$ is a strong Liouville number.

Proof. Set $ S := m_1x_1 + \cdots + m_nx_n$ and $M := |m_1| + \cdots + |m_n|. $ We argue by induction on M.

If $M=0$ , then all $m_i=0$ , so $S=0$ , which is rational.

Assume the result holds for all integer coefficients with total weight strictly less than M and suppose $M \ge 1$ . Choose an index j such that $m_j \neq 0$ . Write $ m_j = \varepsilon + m_j', $ where

$$ \begin{align*} \varepsilon = \begin{cases} 1 & m_j> 0, \\ -1 & m_j < 0, \end{cases} \qquad m_j' = m_j - \varepsilon. \end{align*} $$

Then, $|m_j'| = |m_j| - 1$ and, hence, $ |m_1| + \cdots + |m_j'| + \cdots + |m_n| = M - 1. $ Thus, we can write $ S = \varepsilon x_j + T, $ where

$$ \begin{align*} T := m_1x_1 + \cdots + m_{j-1}x_{j-1} + m_j' x_j + m_{j+1}x_{j+1} + \cdots + m_n x_n. \end{align*} $$

By the induction hypothesis, T is either rational or Liouville. By Lemma 2.9, $\varepsilon x_j$ is a strong Liouville number. We now consider two cases.

Case 1: T is Liouville. Theorem 2.6 implies that the sum of a strong Liouville number and a Liouville number is either rational or Liouville. Hence, $S = \varepsilon x_j + T$ is rational or Liouville.

Case 2: T is rational. Then, $S = \varepsilon x_j + T$ is irrational, since $\varepsilon x_j$ is irrational. Moreover, S is Liouville because the sum of a Liouville number and a rational number is Liouville. Indeed, if x is Liouville, then for every $N \ge 1$ , there exist infinitely many ${p}/{q}$ such that $ 0 < | x - {p}/{q} | < {1}/{q^N}. $ Thus, for $r \in \mathbb {Q}$ ,

$$ \begin{align*} 0 < \bigg| (x + r) - \bigg( \frac{p}{q} + r \bigg) \bigg| < \frac{1}{q^N} \end{align*} $$

and ${p}/{q} + r \in \mathbb {Q}$ , proving that $x + r$ is Liouville.

In both cases, S is either rational or Liouville. This completes the induction.

Proof. Since $\mathcal {L}_{\mathrm {strong}} \subseteq \mathbb {R}$ , we immediately have $ |\mathcal {L}_{\mathrm {strong}}| \le |\mathbb {R}| = \mathfrak {c}. $ It therefore suffices to prove that $ |\mathcal {L}_{\mathrm {strong}}| \ge \mathfrak {c}. $

We construct an injective map from $\{0,1\}^{\mathbb {N}}$ into $\mathcal {L}_{\mathrm {strong}}$ .

Fix $\varepsilon =(\varepsilon _n)_{n\ge 1}\in \{0,1\}^{\mathbb {N}}$ . Define a continued fraction $ x_\varepsilon =[0;a_1,a_2,a_3,\ldots ] $ recursively as follows. Set $ a_1:=1. $ For each $n\ge 1$ , let $q_n$ denote the denominator of the nth convergent of $ [0;a_1,\ldots ,a_n] $ and define $ a_{n+1}:=q_n^{n}+1+\varepsilon _n. $ Since $q_n\ge 1$ and $\varepsilon _n\in \{0,1\}$ , each $a_{n+1}$ is a positive integer, so this indeed defines an irrational real number $x_\varepsilon $ .

Let $p_n/q_n$ be the nth convergent of $x_\varepsilon $ . By the standard recurrence relation for continued fractions, $ q_{n+1}=a_{n+1}q_n+q_{n-1} \ (n\ge 1), $ where $q_0=1$ and $q_{-1}=0$ . Hence, $ q_{n+1}\ge a_{n+1}q_n = (q_n^{n}+1+\varepsilon _n)q_n> q_n^{n+1}. $ Therefore, for every $n\ge 1$ , $ q_{n+1}>q_n^{n+1}. $

We now show that $x_\varepsilon $ is a strong Liouville number. Let $N\ge 1$ be arbitrary. If $n\ge N$ , then $ q_{n+1}>q_n^{n+1}\ge q_n^{N}. $ Thus, the convergents of $x_\varepsilon $ satisfy $ q_{n+1}>q_n^{N} $ for all sufficiently large n. By the definition of a strong Liouville number, $x_\varepsilon \in \mathcal {L}_{\mathrm {strong}}$ .

It remains to prove injectivity. Assume $\varepsilon ,\delta \in \{0,1\}^{\mathbb {N}}$ with $\varepsilon \neq \delta $ . Let m be the least index such that $\varepsilon _m\neq \delta _m$ . Then, the recursive construction gives

$$ \begin{align*} a_1(\varepsilon)=a_1(\delta),\ldots,a_m(\varepsilon)=a_m(\delta), \end{align*} $$

so the first m convergents, and in particular the denominator $q_m$ , are the same for both sequences. Hence, $ a_{m+1}(\varepsilon )=q_m^{m}+1+\varepsilon _m \neq q_m^{m}+1+\delta _m=a_{m+1}(\delta ). $ Therefore, the continued fraction expansions of $x_\varepsilon $ and $x_\delta $ differ, so $ x_\varepsilon \neq x_\delta. $ Thus, the map

$$ \begin{align*} \{0,1\}^{\mathbb{N}}\to \mathcal{L}_{\mathrm{strong}}, \quad \varepsilon\mapsto x_\varepsilon, \end{align*} $$

is injective. Since $ |\{0,1\}^{\mathbb {N}}|=\mathfrak {c}, $ we obtain $ |\mathcal {L}_{\mathrm {strong}}|\ge \mathfrak {c}. $ Combining this with the upper bound $ |\mathcal {L}_{\mathrm {strong}}| \le \mathfrak {c}, $ we conclude that $ |\mathcal {L}_{\mathrm {strong}}|=\mathfrak {c}. $

Proof. We first establish a cardinality bound.

Proof of the claim

Every element of $\operatorname {span}_F(S)$ can be written as a finite linear combination $ \sum _{i=1}^n a_i s_i \ (a_i \in F,\ s_i \in S,\ n \in \mathbb {N}). $ Hence, $ \operatorname {span}_F(S) \subseteq \bigcup _{n=0}^\infty F^n \times S^n $ , where we interpret $F^0 \times S^0$ as a singleton corresponding to $0$ . Thus,

$$ \begin{align*}|\!\operatorname{span}_F(S)| \le \sum_{n=0}^\infty |F|^n \cdot |S|^n. \end{align*} $$

Since F is countable, $|F|=\aleph _0$ . Hence, for each $n \ge 1$ , $ |F|^n \cdot |S|^n = \aleph _0 \cdot \kappa $ , which is equal to $\kappa $ when $\kappa $ is infinite and to $\aleph _0$ when $\kappa $ is finite. Therefore, $ |\operatorname {span}_F(S)| \le \aleph _0 \cdot \kappa < \mathfrak {c}. $ This proves the claim.

We now construct the desired set B by transfinite recursion. Since $|\mathcal {L}_{\mathrm {strong}}|=\mathfrak {c}$ by Lemma 2.11, we may well-order $\mathcal {L}_{\mathrm {strong}}$ and construct a family $ B = \{b_\alpha : \alpha < \mathfrak {c}\} \subset \mathcal {L}_{\mathrm {strong}} $ as follows.

Assume that for some ordinal $\alpha < \mathfrak {c}$ , we have already chosen $ \{b_\beta : \beta < \alpha \} \subset \mathcal {L}_{\mathrm {strong}} $ to be linearly independent over $\mathbb {A}$ . Let $ S_\alpha := \{b_\beta : \beta < \alpha \}. $ Then, $|S_\alpha | < \mathfrak {c}$ . Since $\mathbb {A}$ is countable, the claim above shows that $ |\!\operatorname {span}_{\mathbb {A}}(S_\alpha )| < \mathfrak {c}. $ However, $|\mathcal {L}_{\mathrm {strong}}|=\mathfrak {c}$ , so $ \mathcal {L}_{\mathrm {strong}} \setminus \operatorname {span}_{\mathbb {A}}(S_\alpha ) \neq \varnothing. $ Choose $ b_\alpha \in \mathcal {L}_{\mathrm {strong}} \setminus \operatorname {span}_{\mathbb {A}}(S_\alpha ). $ By construction, $b_\alpha $ is not in the $\mathbb {A}$ -linear span of the previously chosen elements, so the set $\{b_\beta : \beta \le \alpha \}$ remains linearly independent over $\mathbb {A}$ .

Continuing this process for all $\alpha < \mathfrak {c}$ , we obtain a set $ B = \{b_\alpha : \alpha < \mathfrak {c}\} \subset \mathcal {L}_{\mathrm {strong}} $ that is linearly independent over $\mathbb {A}$ . Finally, since the index set has cardinality $\mathfrak {c}$ , we have $|B|=\mathfrak {c}$ , as required.

Proof. By Lemma 2.11, the set $\mathcal {L}_{\mathrm {strong}}$ of strong Liouville numbers has cardinality $\mathfrak {c}$ . We first construct a subset $ B \subset \mathcal {L}_{\mathrm {strong}} $ such that $ |B|=\mathfrak {c} $ and $ \{1\}\cup B $ is linearly independent over the field $\mathbb {A}$ of algebraic numbers.

We do this by transfinite recursion. Assume that for some ordinal $\alpha <\mathfrak {c}$ , we have already chosen $ \{b_\beta :\beta <\alpha \}\subset \mathcal {L}_{\mathrm {strong}} $ so that $ \{1\}\cup \{b_\beta :\beta <\alpha \} $ is linearly independent over $\mathbb {A}$ . Set $ S_\alpha :=\{1\}\cup \{b_\beta :\beta <\alpha \}. $ Then, $|S_\alpha |<\mathfrak {c}$ . Since $\mathbb {A}$ is countable, its $\mathbb {A}$ -linear span also has cardinality $<\mathfrak {c}$ . Indeed, every element of $\operatorname {span}_{\mathbb {A}}(S_\alpha )$ is a finite $\mathbb {A}$ -linear combination of elements of $S_\alpha $ , so $ \operatorname {span}_{\mathbb {A}}(S_\alpha ) \subseteq \bigcup _{n=0}^{\infty }\mathbb {A}^n\times S_\alpha ^n. $

Hence, $ |\!\operatorname {span}_{\mathbb {A}}(S_\alpha )| \le \sum _{n=0}^{\infty } |\mathbb {A}|^n |S_\alpha |^n < \mathfrak {c}. $ Because $|\mathcal {L}_{\mathrm {strong}}|=\mathfrak {c}$ , there exists $ b_\alpha \in \mathcal {L}_{\mathrm {strong}}\setminus \operatorname {span}_{\mathbb {A}}(S_\alpha ). $ Then, $ \{1\}\cup \{b_\beta :\beta \le \alpha \} $ remains linearly independent over $\mathbb {A}$ .

Continuing this construction for all $\alpha <\mathfrak {c}$ yields a set $ B=\{b_\alpha :\alpha <\mathfrak {c}\}\subset \mathcal {L}_{\mathrm {strong}} $ such that $|B|=\mathfrak {c}$ and $\{1\}\cup B$ is linearly independent over $\mathbb {A}$ .

Now, let $ \mathcal {P}_{\infty }(B):=\{\Delta \subseteq B:\Delta \text { is infinite}\}. $ Since $|B|=\mathfrak {c}$ , we have $ |\mathcal {P}_{\infty }(B)|=2^{\mathfrak {c}}. $ For each $\Delta \in \mathcal {P}_{\infty }(B)$ , define

$$ \begin{align*} G_\Delta:=\langle \Delta\rangle =\bigg\{\sum_{i=1}^n m_i x_i : n\ge 1,\ m_i\in\mathbb{Z},\ x_i\in\Delta\bigg\}\cup\{0\}. \end{align*} $$

We claim that each $G_\Delta $ is a Liouville number group.

First, we show that $ G_\Delta \cap \mathbb {Q}=\{0\}. $ Let $ g=m_1x_1+\cdots +m_nx_n\in G_\Delta $ with $m_i\in \mathbb {Z}$ and $x_i\in \Delta $ , and suppose that $g\in \mathbb {Q}\setminus \{0\}$ . Since $\mathbb {Q}\subseteq \mathbb {A}$ ,

$$ \begin{align*} m_1x_1+\cdots+m_nx_n-g\cdot 1=0. \end{align*} $$

This is a nontrivial $\mathbb {A}$ -linear relation among elements of $ \{1,x_1,\ldots ,x_n\}\subseteq \{1\}\cup B, $ because the coefficient of $1$ is $-g\neq 0$ . This contradicts the $\mathbb {A}$ -linear independence of $\{1\}\cup B$ . Hence, $ G_\Delta \cap \mathbb {Q}=\{0\}. $

Next, let $g\in G_\Delta $ be nonzero. Then, g is a finite integer linear combination of elements of $\Delta $ , and every element of $\Delta $ is a strong Liouville number. By Corollary 2.10, it follows that g is either rational or Liouville. Since $g\neq 0$ and $G_\Delta \cap \mathbb {Q}=\{0\}$ , the rational alternative is impossible. Therefore, g is a Liouville number. Thus, every nonzero element of $G_\Delta $ is Liouville, so $G_\Delta $ is a Liouville number group.

Finally, we show that distinct infinite subsets yield distinct groups. Let $\Delta _1, \Delta _2\in \mathcal {P}_{\infty }(B)$ with $\Delta _1\neq \Delta _2$ . Choose $ b\in \Delta _2\setminus \Delta _1 $ without loss of generality. Then, $b\in G_{\Delta _2}$ . We claim that $b\notin G_{\Delta _1}$ . Otherwise, there would exist distinct elements $x_1,\ldots ,x_n\in \Delta _1$ and integers $m_1,\ldots ,m_n$ such that $ b=m_1x_1+\cdots +m_nx_n. $ Rearranging gives

$$ \begin{align*} b-m_1x_1-\cdots-m_nx_n=0, \end{align*} $$

which is a nontrivial $\mathbb {A}$ -linear relation among elements of B, which contradicts the $\mathbb {A}$ -linear independence of B. Hence, $ b\in G_{\Delta _2}\setminus G_{\Delta _1}, $ so $G_{\Delta _1}\neq G_{\Delta _2}$ .

Therefore, the family $ \{G_\Delta :\Delta \in \mathcal {P}_{\infty }(B)\} $ consists of $2^{\mathfrak {c}}$ pairwise distinct Liouville number groups in $\mathbb {R}$ .

Proof. By Theorem 2.13, there exists a family $\mathcal {G}$ of $2^{\mathfrak c}$ pairwise distinct Liouville number groups, each of which is a subspace of $\mathbb {R} \subseteq \mathbb {C}$ .

We now use a consequence of Lavrentiev’s theorem (see, for example, [Reference van Mill10, Theorem A8.5]): if $X \subseteq \mathbb {C}$ is a subspace, then the collection of all subspaces of $\mathbb {C}$ that are homeomorphic to X has cardinality at most $\mathfrak c$ .

Partition $\mathcal {G}$ into homeomorphism classes. Suppose, towards a contradiction, that fewer than $2^{\mathfrak c}$ homeomorphism classes are represented. Then, there exists a cardinal $\lambda < 2^{\mathfrak c}$ such that $\mathcal {G}$ is the union of $\lambda $ homeomorphism classes.

Since each homeomorphism class has cardinality at most $\mathfrak c$ , it follows that

$$ \begin{align*} |\mathcal{G}| \leq \lambda \cdot \mathfrak c = \max(\lambda, \mathfrak c). \end{align*} $$

Because $\lambda < 2^{\mathfrak c}$ and $\mathfrak c < 2^{\mathfrak c}$ , we have $ \max (\lambda , \mathfrak c) < 2^{\mathfrak c}, $ and, hence, $ |\mathcal {G}| < 2^{\mathfrak c}, $ which contradicts $|\mathcal {G}| = 2^{\mathfrak c}$ . Therefore, there must be at least $2^{\mathfrak c}$ distinct homeomorphism classes among the groups in $\mathcal {G}$ . In particular, there exist $2^{\mathfrak c}$ pairwise nonhomeomorphic Liouville number groups.

Proof. Let $(u,v)$ be a nonempty open interval in $\mathbb {R}$ . Since irrational numbers are dense in $\mathbb {R}$ and continued-fraction cylinders form a base for the topology on $\mathbb {R}\setminus \mathbb {Q}$ , there exists a finite continued-fraction block $ [a_0;a_1,\ldots ,a_N] $ such that every irrational number whose continued fraction begins with this block lies in $(u,v)$ .

We now extend this block recursively. Assume that $a_0,\ldots ,a_k$ have been chosen and let $q_k$ be the denominator of the kth convergent determined by these partial quotients. Define $ a_{k+1}:=q_k^{k}+1 \ (k\ge N). $ This produces an irrational number $ x=[a_0;a_1,\ldots ,a_N,a_{N+1},a_{N+2},\ldots ] $ whose continued fraction begins with the prescribed block; hence, $x\in (u,v)$ .

Let $p_k/q_k$ denote the convergents of x. For $k\ge N$ , the standard recurrence gives

$$ \begin{align*} q_{k+1}=a_{k+1}q_k+q_{k-1} \ge (q_k^{k}+1)q_k> q_k^{k+1}. \end{align*} $$

Now, let $n\in \mathbb {N}$ be arbitrary. If $k\ge \max \{N,n\}$ , then $ q_{k+1}>q_k^{k+1}\ge q_k^{n}. $ Therefore, there exists $N(n)=\max \{N,n\}$ such that $ q_{k+1}>q_k^{n} $ for all $k\ge N(n)$ . By Definition 2.1, x is a strong Liouville number.

Since $(u,v)$ was arbitrary, every nonempty open interval contains a strong Liouville number. Hence, ${\mathcal L}_{\mathrm {strong}}$ is dense in $\mathbb {R}$ .

Proof. We first note that every nonempty open interval contains $\frak c$ many strong Liouville numbers. Indeed, fix a nonempty open interval $(u,v)$ . Choose a finite continued-fraction block $[a_0;a_1,\ldots ,a_N]$ whose cylinder is contained in $(u,v)$ . For each binary sequence $\varepsilon =(\varepsilon _k)_{k\ge N}\in \{0,1\}^{\mathbb {N}}$ , extend this block recursively by

$$ \begin{align*} a_{k+1}:=q_k^{k}+1+\varepsilon_k \qquad (k\ge N), \end{align*} $$

where $q_k$ is the denominator of the kth convergent determined by the initial segment. Exactly as in Lemma 2.11, this defines a strong Liouville number in $(u,v)$ , and different binary sequences give different continued fractions. Hence, $|{\mathcal L}_{\mathrm {strong}}\cap (u,v)|=\frak c$ .

We construct $(x_n)$ recursively. Suppose that $x_1,\ldots ,x_{n-1}$ have already been chosen so that $\{1,x_1,\ldots ,x_{n-1}\}$ is linearly independent over $\mathbb {A}$ . Let

$$ \begin{align*} V_{n-1}:=\operatorname{span}_{\mathbb{A}}(\{1,x_1,\ldots,x_{n-1}\}). \end{align*} $$

Since $\mathbb {A}$ is countable and the generating set is finite, $V_{n-1}$ is countable. However, $ |{\mathcal L}_{\mathrm {strong}}\cap (0,2^{-n})|=\frak c, $ so $ ({\mathcal L}_{\mathrm {strong}}\cap (0,2^{-n}))\setminus V_{n-1}\neq \varnothing. $ Choose

$$ \begin{align*} x_n\in ({\mathcal L}_{\mathrm{strong}}\cap (0,2^{-n}))\setminus V_{n-1}. \end{align*} $$

Then, $0<x_n<2^{-n}$ , so $x_n\to 0$ and, by construction, $\{1,x_1,x_2,\ldots \}$ is linearly independent over $\mathbb {A}$ . Let $ W:=\operatorname {span}_{\mathbb {A}}(\{1\}\cup \{x_n:n\in \mathbb {N}\}). $ Since $\mathbb {A}$ is countable and the set $\{1\}\cup \{x_n:n\in \mathbb {N}\}$ is countable, the space W is countable.

We now construct T by transfinite recursion. Since $|{\mathcal L}_{\mathrm {strong}}|=\frak c$ by Lemma 2.11, we may choose $ T=\{t_\alpha :\alpha <\frak c\}\subset {\mathcal L}_{\mathrm {strong}} $ so that for each $\alpha <\frak c$ ,

$$ \begin{align*} t_\alpha\notin \operatorname{span}_{\mathbb{A}}(W\cup\{t_\beta:\beta<\alpha\}). \end{align*} $$

This is possible because for each $\alpha <\frak c$ , the set $ \operatorname {span}_{\mathbb {A}}(W\cup \{t_\beta :\beta <\alpha \}) $ has cardinality $<\frak c$ , whereas ${\mathcal L}_{\mathrm {strong}}$ has cardinality $\frak c$ . Then, it follows that $|T|=\frak c$ and, by construction, $ \{1\}\cup \{x_n:n\in \mathbb {N}\}\cup T $ is linearly independent over $\mathbb {A}$ .

Proof. Choose $(x_n)$ and T as in Lemma 2.16. For each $\tau \in T$ , define the group $ G_\tau :=\operatorname {\mathrm {{gp}}}(\{x_n:n\in \mathbb {N}\}\cup \{\tau \}). $ Each $G_\tau $ is countable, being generated by a countable set. Moreover, $G_\tau $ is infinite because it contains the infinitely many distinct nonzero elements $x_n$ .

We claim that $G_\tau $ is a Liouville number group. Let $g\in G_\tau \setminus \{0\}$ . Then, $ g=m\tau + \sum _{j=1}^n m_jx_j $ for some integers $m,m_1,\ldots ,m_n$ . By Corollary 2.10, g is either rational or Liouville. If g were rational, then $ m\tau +\sum _{j=1}^n m_jx_j-g\cdot 1=0 $ would be a nontrivial $\mathbb {A}$ -linear relation among elements of $ \{1,\tau ,x_1,\ldots ,x_n\}{\kern-1pt}\subseteq{\kern-1pt} \{1\}{\kern-1pt}\cup{\kern-1pt} \{x_n:n{\kern-1pt}\in{\kern-1pt} \mathbb {N}\}{\kern-1pt}\cup{\kern-1pt} T, $ contrary to Lemma 2.16. Hence, g is Liouville. Therefore, every nonzero element of $G_\tau $ is Liouville, so $G_\tau $ is a Liouville number group.

Since $x_n\in G_\tau $ for all n and $x_n\to 0$ , the point $0$ is not isolated in $G_\tau $ . Because $G_\tau $ is a topological group, translation by any element is a homeomorphism, so no point of $G_\tau $ is isolated. Thus, $G_\tau $ is a countable metrisable space with no isolated points. By the Sierpiński–Alexandroff–Urysohn theorem, $G_\tau $ is homeomorphic to $\mathbb {Q}$ .

Finally, if $\tau \neq \sigma $ in T and $G_\tau =G_\sigma $ , then $ \tau \in G_\sigma =\operatorname {\mathrm {{gp}}}(\{x_n:n\in \mathbb {N}\}\cup \{\sigma \}), $ so $ \tau =m\sigma +\sum _{j=1}^n m_jx_j $ for some integers $m,m_1,\ldots ,m_n$ . Hence, $ \tau -m\sigma -\sum _{j=1}^n m_jx_j=0, $ which is a nontrivial $\mathbb {A}$ -linear relation among elements of $ \{1\}\cup \{x_n:n\in \mathbb {N}\}\cup T, $ contrary to Lemma 2.16. Therefore, $G_\tau \neq G_\sigma $ whenever $\tau \neq \sigma $ .

Since $|T|=\frak c$ , the family $\{G_\tau :\tau \in T\}$ consists of $\frak c$ pairwise distinct countably infinite Liouville number groups, each generated by countably infinitely many strong Liouville numbers and each homeomorphic to $\mathbb {Q}$ .

Proof. Let G be a subgroup of $\mathbb {R}$ . By [Reference Chalebgwa and Morris2], the space $\mathcal L$ of Liouville numbers is homeomorphic to the space $\mathbb {P}$ of irrational real numbers. Thus, it suffices to rule out subgroups of $\mathbb {R}$ homeomorphic to $\mathcal L$ .

If $G=\mathbb {R}$ , then G is connected, whereas $\mathcal L$ is totally disconnected, so G is not homeomorphic to $\mathcal L$ .

Assume therefore that G is a proper subgroup of $\mathbb {R}$ . By [Reference Morris7, Proposition 19], every subgroup of $\mathbb {R}$ is either dense or closed. If G is closed, then by [Reference Morris7, Proposition 20], it is discrete; hence, not homeomorphic to $\mathcal L$ .

Assume finally that G is dense and homeomorphic to $\mathcal L$ . Since $\mathcal L$ is homeomorphic to $\mathbb {P}=\mathbb {R}\setminus \mathbb {Q}$ and $\mathbb {P}$ is a $G_\delta $ subset of $\mathbb {R}$ , the space $\mathbb {P}$ is completely metrisable. Hence, $\mathcal L$ is completely metrisable and, therefore, so is G. Now, G is a completely metrisable subspace of the complete metric space $\mathbb {R}$ . By the classical Alexandrov theorem, a subspace of a complete metric space is completely metrisable if and only if it is a $G_\delta $ subset. Thus, G is a dense $G_\delta $ subset of $\mathbb {R}$ . By Erdős [Reference Erdős3], every dense $G_\delta $ subset of $\mathbb {R}$ has the Erdős property: every real number is the sum of two of its elements. However, no proper subgroup of $\mathbb {R}$ has this property, because if $x\notin G$ , then x cannot be a sum of two elements of G. This contradiction shows that no proper subgroup of $\mathbb {R}$ is homeomorphic to $\mathcal L$ .

Therefore, there is no subgroup of $\mathbb {R}$ homeomorphic to $\mathcal L$ and, hence, none homeomorphic to $\mathbb {P}$ . In particular, no Liouville number group is homeomorphic to $\mathcal L$ .