Proof. We show that $X^{\mathbb N}\setminus \Sigma _E$ is open. Let $\beta \in X^{\mathbb N}\setminus \Sigma _E$ . Then either $\beta _1=1$ or there exists $k\ge 1$ with $\beta _{k}>\beta _{k+1}$ .

If $\beta _1=1$ , the cylinder $\{\alpha \in X^{\mathbb N}:\alpha _1=1\}$ is an open neighbourhood of $\beta $ disjoint from $\Sigma _E$ (since Engel sequences satisfy $\alpha _1\ge 2$ ).

Otherwise pick k with $\beta _{k}>\beta _{k+1}$ . There are two subcases. (Note that $\beta _{k+1}\neq \infty $ , since $\beta _k>\beta _{k+1}$ would be impossible.)

Case I: Both coordinates finite. If $\beta _k,\beta _{k+1}\in \mathbb N$ , the cylinder

$$ \begin{align*} U {:=}q \{\alpha\in X^{\mathbb N}: \alpha_k=\beta_k,\ \alpha_{k+1}=\beta_{k+1}\} \end{align*} $$

is open and every $\alpha \in U$ still violates monotonicity at k. Hence U is disjoint from $\Sigma _E$ .

Case II: Left coordinate $\infty $ . If $\beta _k=\infty $ and $\beta _{k+1}\in \mathbb N$ , let

$$ \begin{align*} U_k {:=}q \{\infty\}\cup\{m\in\mathbb N: m\ge \beta_{k+1}+1\} \quad\text{and}\quad U_{k+1} {:=}q \{\beta_{k+1}\}. \end{align*} $$

Then $U_k$ is an open neighbourhood of $\infty $ in $\mathbb N_\infty $ (its complement is finite) and $U_{k+1}$ is open (singleton in the discrete part $\mathbb {N}$ ). The basic open set

$$ \begin{align*} V {:=}q \{\alpha\in X^{\mathbb N}: \alpha_k\in U_k, \, \alpha_{k+1}\in U_{k+1}\} \end{align*} $$

contains $\beta $ and is disjoint from $\Sigma _E$ , because for any $\alpha \in V$ one has either $\alpha _k=\infty>\alpha _{k+1}$ or $\alpha _k\ge \beta _{k+1}+1>\alpha _{k+1}$ .

In all cases we have found an open neighbourhood of $\beta $ disjoint from $\Sigma _E$ , so $X^{\mathbb N}\setminus \Sigma _E$ is open, that is, $\Sigma _E$ is closed.

Proof. Since X is compact (one-point compactification), Tychonoff’s theorem implies that $X^{\mathbb N}$ is compact in the product topology, which is metrised by the metric d in (2.1). Then Lemma 2.1 yields compactness of $\Sigma _E$ .

Proof. By construction, the Engel expansion of $x\in (0,1]$ is unique, so $\varphi (f(x))=x$ . If $\eta \in \Sigma _E^{\mathrm {irr}}$ , then $\eta _n\in \mathbb N$ with $\eta _n\to \infty $ , and $\varphi (\eta )$ has Engel digits $\eta _n$ ; hence, $f(\varphi (\eta ))=\eta $ .

Proof. Let $\eta ,\eta '\in \Sigma _E$ and let n be the first index at which they differ. Put ${A{:=}q 1/(\eta _1\cdots \eta _{n-1})=1/(\eta _1'\cdots \eta _{n-1}')}$ . Then

$$ \begin{align*} \varphi(\eta)-\varphi(\eta') = A\bigg(\!\sum_{j=0}^{\infty}\frac{1}{\eta_n\cdots \eta_{n+j}} -\sum_{j=0}^{\infty}\frac{1}{\eta_n'\cdots \eta_{n+j}'}\bigg). \end{align*} $$

Since the digits are nondecreasing, $\eta _{n+\ell }\ge \eta _n$ and $\eta _{n+\ell }'\ge \eta _n'$ for all $\ell \ge 0$ . Hence,

$$ \begin{align*} \sum_{j=0}^\infty \frac{1}{\eta_n\cdots\eta_{n+j}} \leq \sum_{j=0}^\infty \frac{1}{\eta_n^{j+1}}=\frac{1}{\eta_n-1}\le\frac{2}{\eta_n}, \end{align*} $$

and similarly with $\eta $ replaced by $\eta '$ . Therefore,

$$ \begin{align*} |\varphi(\eta)-\varphi(\eta')| \le A\bigg(\frac{2}{\eta_n}+\frac{2}{\eta_n'}\bigg) \le 2^{2-n}\bigg(\frac{1}{\eta_n}+\frac{1}{\eta_n'}\bigg), \end{align*} $$

since $\eta _k,\eta _k'\ge 2$ imply $A\le 2^{-(n-1)}$ . On the other hand,

$$ \begin{align*} d(\eta,\eta')\ge 2^{-n}\rho(\eta_n,\eta_n')=2^{-n}\bigg(\frac{1}{\eta_n}+\frac{1}{\eta_n'}\bigg). \end{align*} $$

Combining gives $|\varphi (\eta )-\varphi (\eta ')|\le 4\,d(\eta ,\eta ')$ .

Proof. The description of $J_n(\sigma )=(\ell (\sigma ),r(\sigma )]$ in Proposition 2.3 shows that x is the (right) endpoint of a unique fundamental interval, giving the stated $\sigma $ and $\alpha $ . For the terminating code, one checks directly that replacing the rightmost digit greater than or equal to $3$ by $1$ less and appending an $\infty $ -tail preserves admissibility and the value under $\varphi $ (the usual carry rule); see also [Reference Galambos6]. Uniqueness of the two codes follows from monotonicity and the endpoint structure of fundamental intervals.

Proof. Suppose $x\in (0,1]\setminus \mathbb Q$ and f is not continuous at x. Then there are $\varepsilon>0$ and $x_m\to x$ with $d(f(x_m),f(x))\ge \varepsilon $ for all m. Since $(\Sigma _E,d)$ is compact (Lemma 2.2), the sequence $(f(x_m))_{m\ge 1}$ has a convergent subsequence, say $f(x_{m_k})\to \beta \in \Sigma _E$ . Since $\varphi $ is continuous (Theorem 3.1),

$$ \begin{align*} x=\lim_{k\to\infty} x_{m_k}=\lim_{k\to\infty} \varphi(f(x_{m_k}))=\varphi(\beta). \end{align*} $$

As x is irrational, $\beta \in \Sigma _E^{\mathrm {irr}}$ and, hence, $\beta =f(\varphi (\beta ))=f(x)$ (Proposition 2.5). Thus $f(x_{m_k})\to f(x)$ , contradicting $d(f(x_{m_k}),f(x))\ge \varepsilon $ .

Proof. If $x=1$ , the proof is similar to that of Theorem 3.3, using the fact that

$$ \begin{align*} \varphi^{-1}(\{1\})=\{(2,2,2,\dotsc)\}. \end{align*} $$

Thus f is left-continuous at $1$ . We may therefore assume that $x\in(0,1)\cap\mathbb Q$ .

Let $\sigma =(\sigma _1,\ldots ,\sigma _n)$ be as in Lemma 3.2, so $x=r(\sigma )$ and $J_n(\sigma )=(\ell (\sigma ),r(\sigma )]$ with right endpoint x. Write $\alpha {:=}q f(x)=(\sigma _1,\ldots ,\sigma _{n-1},\sigma _n,\sigma _n,\ldots )$ and $\alpha '{:=}q (\sigma _1,\ldots ,\sigma _{j-1},\sigma _j-1,\infty ,\infty ,\ldots )$ for the terminating code (Lemma 3.2). Recall from Proposition 2.3 that the cylinder

$$ \begin{align*} C_n(\sigma){:=}q\{\eta\in\Sigma_E: \eta_1=\sigma_1,\ldots,\eta_n=\sigma_n\} \end{align*} $$

is clopen in $\Sigma _E$ .

Left-continuity. Take any sequence $y_m\uparrow x$ with $y_m<x$ . For m large, $y_m\in J_n(\sigma )$ , hence $f(y_m)\in C_n(\sigma )$ . Because $C_n(\sigma )$ is closed, any subsequential limit $\gamma $ of $(f(y_m))_{m\ge 1}$ lies in $C_n(\sigma )$ . On the other hand, $\varphi $ is continuous (Theorem 3.1), so $\varphi (\gamma )=\lim _m \varphi (f(y_m))=\lim _m y_m=x$ . Thus, $\gamma \in \varphi ^{-1}(\{x\})=\{\alpha ,\alpha '\}$ (Lemma 3.2). Since $\alpha \in C_n(\sigma )$ but ${\alpha '\notin C_n(\sigma )}$ (as its first n digits do not match $\sigma $ ), we must have $\gamma =\alpha =f(x)$ . Therefore, $f(y_m)\to f(x)$ as $m\to \infty $ .

Failure of right-continuity. Take any sequence $z_m\downarrow x$ with $z_m>x$ . For each m, $z_m\notin J_n(\sigma )$ , hence $f(z_m)\notin C_n(\sigma )$ . Since $\Sigma _E\setminus C_n(\sigma )$ is also closed (as $C_n(\sigma )$ is clopen), any subsequential limit $\delta $ of $(f(z_m))_{m\ge 1}$ must lie in $\Sigma _E\setminus C_n(\sigma )$ . Again by continuity of $\varphi $ , $\varphi (\delta )=\lim _m \varphi (f(z_m))=\lim _m z_m=x$ , so $\delta \in \{\alpha ,\alpha '\}$ . But $\alpha \in C_n(\sigma )$ while $\alpha '\in \Sigma _E\setminus C_n(\sigma )$ , hence $\delta =\alpha '\neq f(x)$ . Therefore $\lim _{t\downarrow x} f(t)=\alpha '\ne f(x)$ , and f is not right-continuous at x.

Proof. The bijectivity of $\varphi \colon \Sigma ^{\mathrm {irr}}_E \to (0,1] \setminus \mathbb {Q}$ with inverse f is established by Proposition 2.5. The continuity of $\varphi $ is a consequence of Theorem 3.1 (Lipschitz continuity). The continuity of the inverse map f on $(0,1]\setminus \mathbb {Q}$ is established by Theorem 3.3. Since $\varphi $ is a continuous bijection with a continuous inverse, it is a homeomorphism.

Proof. For $m\in \mathbb N$ , define

$$ \begin{align*} \Upsilon_m{:=}q \bigcup_{\substack{\sigma=(\sigma_1,\ldots,\sigma_n) \text{ admissible}\\ \sum_{k=1}^n \phi(\sigma_k)>m}} C_n(\sigma) \subseteq \Sigma_E. \end{align*} $$

Each $\Upsilon _m$ is open (union of clopen cylinders). We claim that $\Upsilon _m\cap \Sigma _E^{\mathrm {irr}}$ is dense in $\Sigma _E^{\mathrm {irr}}$ . Indeed, given a nonempty cylinder $C_N(\tau )$ with $\tau _N\in \mathbb N$ (equivalently, $C_N(\tau )\cap \Sigma _E^{\mathrm {irr}}\neq \varnothing $ ), choose $t\ge \max \{2,\tau _N\}$ with $\phi (t)>0$ (possible since $\{t: \phi (t)>0\}$ is unbounded), and append a long enough t-tail to $\tau $ so that $\sum _{k=1}^{|\sigma |}\phi (\sigma _k)>m$ . Then $C_{|\sigma |}(\sigma )\subseteq C_N(\tau )\cap \Upsilon _m$ , proving density in $\Sigma _E^{\mathrm {irr}}$ .

Hence,

$$ \begin{align*} \Gamma_\phi{:=}q\bigcap_{m=1}^\infty \Upsilon_m \end{align*} $$

is such that $\Gamma _\phi \cap \Sigma _E^{\mathrm {irr}}$ is a dense $G_\delta $ in $\Sigma _E^{\mathrm {irr}}$ ; moreover, if $\eta \in \Gamma _\phi \cap \Sigma _E^{\mathrm {irr}}$ , then the partial sums $\sum _{k=1}^n\phi (\eta _k)$ are unbounded in n, so $\sum _{k\ge 1}\phi (\eta _k)=\infty $ .

Set

$$ \begin{align*} \mathcal G_\phi{:=}q \varphi(\Gamma_\phi\cap\Sigma_E^{\mathrm{irr}}) \subseteq (0,1]\setminus\mathbb Q. \end{align*} $$

By Corollary 3.5, $\varphi :\Sigma _E^{\mathrm {irr}}\to (0,1]\setminus \mathbb Q$ is a homeomorphism; hence $\mathcal G_\phi $ is a dense $G_\delta $ in $(0,1]\setminus \mathbb Q$ , and for $x\in \mathcal G_\phi $ with $\eta {:=}q f(x)$ ,

$$ \begin{align*} \sum_{n=1}^\infty\phi(d_n(x))=\sum_{n=1}^\infty\phi(\eta_n)=\infty. \end{align*} $$

Since $(0,1]\setminus \mathbb Q$ is dense $G_\delta $ in $(0,1]$ , it follows that $\mathcal G_\phi $ is a dense $G_\delta $ set in $(0,1]$ . As $\mathcal G_\phi \subseteq \mathcal D_\phi $ , we conclude that $\mathcal D_\phi $ is comeagre (residual) in $(0,1]$ .

Proof. Apply Theorem 4.1 with $\phi (t){:=}q t^{-s}$ .

Proof. For each $k\in \mathbb N$ , Corollary 4.2 shows that $\mathcal D (k)$ is comeagre. Since

$$ \begin{align*} \{x\in(0,1]:\lambda(x)=\infty\}=\bigcap_{k=1}^\infty\mathcal D (k), \end{align*} $$

this set is a countable intersection of comeagre sets and is therefore comeagre. For the equivalence, note that if $\lambda (x)> \alpha $ , there exists a rational $q> \alpha $ such that ${\sum _{n=1}^\infty d_n(x)^{-q} = \infty }$ . Thus,

$$ \begin{align*} \{x\in(0,1]:\lambda(x)> \alpha\} = \bigcup_{q\in\mathbb Q,\ q>\alpha} \mathcal D(q). \end{align*} $$

As a countable union of comeagre sets, this set is comeagre, which implies that its complement $\{x\in (0,1]:\lambda (x)\le \alpha \}$ is meagre.