Square-to-Cube Lindelöfness and Power Separations in Hattori Spaces

1. Introduction

For $A\subseteq \mathbb{R}$, put $B=\mathbb{R}\setminus A$. The Hattori space $H\left(A\right)$ is the real line with Euclidean local bases at points of A and Sorgenfrey local bases at points of B [7, Definition 5.1]. Every Hattori space is hereditarily Lindelöf [4, Proposition 3.1(c)]. Further properties were studied by Kulesza [10], and Hattori later surveyed this family of mixed topologies [8].

Lin and Li posed a square product question relating Lindelöfness, normality, and metrizability. Under CH they gave an example with uncountable B and Lindelöf $H{\left(A\right)}^2$ [12, Question 4.1 and Example 4.2]. Hernández-Hernández, Ramírez-Chávez, and Rojas-Hernández proved the square characterization [6, Theorem 6.7].

$$H{\left(A\right)}^2\mathrm{is}\mathrm{Lindel}\text{ö}\mathrm{f}⟺B\mathrm{contains}\mathrm{no}\mathrm{subspace}\mathrm{homeomorphic}\mathrm{to}{2}^{\omega }.$$

(1.1)

They asked for characterizations of the Hattori spaces whose cube, higher finite powers, or countable power are Lindelöf [6, Problem 7.8]. At the 2018 Joint Mathematics Meetings, Kulesza [11] announced

$$H{\left(A\right)}^{2n}\mathrm{is}\mathrm{Lindel}\text{ö}\mathrm{f}⟹H{\left(A\right)}^{2n+1}\mathrm{is}\mathrm{Lindel}\text{ö}\mathrm{f}(n\ge 1),$$

(1.2)

together with odd-to-even separations and a finite-power versus countable-power separation. The meeting abstract states these conclusions without proofs.

Theorem 2 proves the case $n=1$ of (1.2) in ZFC. Together with (1.1), it gives

$$\begin{array}{cc}\hfill H{\left(A\right)}^3\mathrm{is}\mathrm{Lindel}\text{ö}\mathrm{f}& ⟺H{\left(A\right)}^2\mathrm{is}\mathrm{Lindel}\text{ö}\mathrm{f}\hfill \\ \hfill & ⟺B\mathrm{contains}\mathrm{no}\mathrm{subspace}\mathrm{homeomorphic}\mathrm{to}{2}^{\omega }.\hfill \end{array}$$

(1.3)

The proof uses structural properties of closed discrete subsets of a Hattori cube. Under the square hypothesis, every coordinate fiber is countable, so an uncountable closed discrete set can be thinned until all three coordinate projections are one-to-one. Its Euclidean condensation set then contains a perfect subset on which the three projections remain one-to-one. Closed discreteness and the one-sided local bases force three local Pareto-maximality sets to be cocountable, while their triple intersection is countable.

The one-sided geometry also constrains structured closed discrete subsets in higher powers. A pointwise half-space criterion characterizes affine discreteness. For nonempty A, no positive-dimensional affine subspace is discrete in the first three powers. Under Lindelöfness of the square, an uncountable closed discrete monotone trace requires at least two increasing and two decreasing coordinates. Both phenomena exhibit a four-coordinate threshold.

Gruenhage and Nyikos proved the hereditary-Lindelöf characterization used for subspaces of finite products of Sorgenfrey and Euclidean lines [5, Theorem 6.9]. Hernández-Hernández, Ramírez-Chávez, and Rojas-Hernández proved the needed product theorem [6, Proposition 6.6]. These results give the reductions in Section 6. For every $n\in \mathbb{N}$, the power $H{\left(A\right)}^n$ is hereditarily Lindelöf if and only if $B^n$ is Lindelöf, and $H{\left(A\right)}^{\omega }$ is hereditarily Lindelöf if and only if every finite Sorgenfrey power of B is Lindelöf. A uniform finite-power reduction also shows that, for $n\ge 3$, Lindelöfness of $B^{n-1}$ implies Lindelöfness of $H{\left(A\right)}^n$.

Lin and Li’s example was obtained under CH [12, Example 4.2]. Section 7 gives a ZFC construction of $A\subseteq \mathbb{R}$ such that $H{\left(A\right)}^2$ and $H{\left(A\right)}^3$ are Lindelöf, whereas $H{\left(A\right)}^4$ contains a closed discrete affine line of cardinality $\mathfrak{c}$. For this set A, $B^2$ is not Lindelöf, showing that non-Lindelöfness of the Sorgenfrey square is necessary but not sufficient for non-Lindelöfness of the Hattori cube. Neither $H{\left(A\right)}^2$ nor $H{\left(A\right)}^3$ is hereditarily Lindelöf.

All finite and countable powers in what follows carry the product topology.

2. Preliminaries

Write $2^{\omega }$ for the Cantor space and $\mathfrak{c}$ for the cardinality of the continuum. The bounding number $\mathfrak{b}$ is the least cardinality of an unbounded family in ${\omega }^{\omega }$ under eventual domination. Here $\mathbb{N}=\{1,2,\dots \}$. In every finite-dimensional Euclidean space, $\parallel ·\parallel$ denotes the Euclidean norm and $B_{\mathrm{E}}(p,r)$ the open ball with center p and radius $r>0$. A point p is a Euclidean condensation point of a set S if every Euclidean neighborhood of p meets S in uncountably many points. Unless a different topology is explicitly specified, subsets of $\mathbb{R}$ carry the Euclidean subspace topology.

Definition 1 

([7, Definition 5.1]). Let $A\subseteq \mathbb{R}$, and put $B=\mathbb{R}\setminus A$. If $x\in A$, the usual Euclidean intervals

$$(x-\epsilon ,x+\epsilon ),\epsilon >0,$$

form a local base at x. If $x\in B$, the half-open intervals

$$[x,x+\epsilon ),\epsilon >0,$$

form a local base at x. The resulting space is denoted by $H\left(A\right)$.

The Hattori topology refines the Euclidean topology, so $H\left(A\right)$, all its powers, and all their subspaces are $T_1$.

Theorem 1 

([6, Theorem 6.7]). Let $A\subseteq \mathbb{R}$ and $B=\mathbb{R}\setminus A$. Then $H{\left(A\right)}^2$ is Lindelöf if and only if B contains no subspace homeomorphic to $2^{\omega }$.

Lemma 1 

([4, Proposition 2.3]). Let $A\subseteq \mathbb{R}$, and put $B=\mathbb{R}\setminus A$. Then $H\left(A\right)$ is second countable if and only if B is countable.

Proposition 1 

([3, Lemma 3.2]). Let $X\subseteq \mathbb{R}$ carry the Sorgenfrey subspace topology and let $m\in \mathbb{N}$. Then $X^m$ is Lindelöf if and only if it contains no uncountable set D such that for all distinct $x,y\in D$ there are indices $i,j\in \{1,\dots ,m\}$ with

$$x_i<y_i\mathrm{and}{x}_j>y_j.$$

Such a set is called a discrete surface.

A space is subparacompact if every open cover has a $\sigma$-locally finite closed refinement. By Burke’s characterization, this is equivalent to the usual formulation in terms of $\sigma$-discrete closed refinements [2, Theorem 1.2]. A space is perfect if every closed subset is a $G_{\delta }$-set, equivalently if every open subset is an $F_{\sigma }$-set. A space is perfectly subparacompact if it is both perfect and subparacompact [12, Definition 2.2]. This meaning of “perfect” is distinct from a perfect subset of a Euclidean space, which means a closed subset with no isolated points.

Proposition 2 

([12, Theorem 3.23]). For every $A\subseteq \mathbb{R}$, the countable power $H{\left(A\right)}^{\omega }$ is perfectly subparacompact.

Lemma 2 

([5, Lemma G]). Every perfectly subparacompact space is hereditarily subparacompact.

Lemma 3. 

For every $A\subseteq \mathbb{R}$ and every $m\in \mathbb{N}\cup \left\{\omega \right\}$, every subspace of $H{\left(A\right)}^m$ is subparacompact.

Proof. 

Proposition 2 and Lemma 2 show that every subspace of $H{\left(A\right)}^{\omega }$ is subparacompact. Every finite power is homeomorphic to a subspace of $H{\left(A\right)}^{\omega }$, which proves the finite cases. □

Lemma 4. 

A $T_1$ subparacompact space is Lindelöf if and only if every closed discrete subspace is countable.

Proof. 

If the space is Lindelöf, every closed discrete subspace is countable. For the converse, the assumption that every closed discrete subspace is countable means that the space is ${\omega }_1$-compact in the terminology of Gruenhage and Nyikos. They observe that every subparacompact ${\omega }_1$-compact space is Lindelöf [5, p. 584]. □

Every nonempty Polish space without isolated points contains a subspace homeomorphic to $2^{\omega }$ [9, Theorem 6.2], and every uncountable analytic subset of a Polish space contains a nonempty perfect subset [9, Theorem 29.1].

3. The Square-to-Cube Implication

The lemma below follows from Mycielski’s perfect-set theorem for meager relations [9, Theorem 19.1]. See also [14].

Lemma 5. 

Let $m\in \mathbb{N}$, and let $C\subseteq {\mathbb{R}}^m$ be a nonempty compact perfect set. Suppose that, for every $i\in \{1,\dots ,m\}$ and $a\in \mathbb{R}$, the fiber

$$\{x\in C:x_i=a\}$$

has empty interior in C. Then there is a nonempty compact perfect set $Q\subseteq C$ such that every coordinate projection is one-to-one on Q.

Proof. 

For each $i\in \{1,\dots ,m\}$, let

$$R_i=\{(x,y)\in C^2:x_i=y_i\}.$$

The relation $R_i$ is closed in $C^2$. It has empty interior. If nonempty relatively open sets $U,V\subseteq C$ satisfied $U\times V\subseteq R_i$, then, for any fixed $y\in V$, the set U would be contained in the coordinate fiber $\{x\in C:x_i=y_i\}$, contrary to the hypothesis. Thus every $R_i$ is nowhere dense, and $R={\bigcup }_{i=1}^m{R}_i$ is meager in $C^2$. The compact metric space C is Polish and has no isolated points. Mycielski’s theorem gives a nonempty perfect set $Q\subseteq C$. Since Q is closed in the compact set C, it is compact. Put ${\Delta }_Q=\{(x,x):x\in Q\}$. Then

$$(Q^2\setminus {\Delta }_Q)\cap R=\varnothing .$$

Hence distinct points of Q differ in every coordinate, so each coordinate projection is one-to-one on Q. □

Theorem 2. 

For every $A\subseteq \mathbb{R}$,

$$H{\left(A\right)}^2\mathit{is}\mathit{Lindel}\text{ö}\mathrm{f}⟹H{\left(A\right)}^3\mathit{is}\mathit{Lindel}\text{ö}\mathrm{f}.$$

Proof. 

Put $B=\mathbb{R}\setminus A$, and suppose that the square is Lindelöf while the cube is not. Lemmas 3 and 4 give an uncountable closed discrete set $D\subseteq H{\left(A\right)}^3$.

For $i\in \{1,2,3\}$, let ${\pi }_i$ denote the i-th coordinate projection. For $t\in \mathbb{R}$, the fiber

$$D\cap {\pi }_i^{-1}\left(\left\{t\right\}\right)$$

is a closed discrete subspace of the closed slice ${\pi }_i^{-1}\left(\left\{t\right\}\right)\cong H{\left(A\right)}^2$. Every such fiber is countable. Hence the range of each coordinate projection on any uncountable subset of D is uncountable. Successive thinning of D along the three coordinate projections, with one representative chosen from every nonempty fiber at each stage, gives an uncountable set $D^{*}\subseteq D$ whose three coordinate projections are one-to-one. Every subset of a closed discrete subspace is closed and discrete in the ambient space, so $D^{*}$ is still closed discrete. Since ${\mathbb{R}}^3$ is the union of countably many bounded boxes, after a further thinning $D^{*}$ may be assumed bounded.

Let C be the set of Euclidean condensation points of $D^{*}$. Since ${\mathbb{R}}^3$ is second countable, the points of $D^{*}$ that are not condensation points of $D^{*}$ form a countable set. Each such point lies in a member of a fixed countable base whose intersection with $D^{*}$ is countable. Therefore

$$D^{*}\setminus C$$

is countable, so C is nonempty. If $p\notin C$, there is a Euclidean open neighborhood $U\ni p$ for which $U\cap D^{*}$ is countable. Every point of U is then outside C, so ${\mathbb{R}}^3\setminus C$ is open and C is closed. Every point of C lies in ${\overline{D^{*}}}^{{\mathbb{R}}^3}$. Since $D^{*}$ is bounded, so is C. Thus C is compact. It has no isolated points. If $p\in C$ were isolated in C, a Euclidean neighborhood U with $U\cap C=\left\{p\right\}$ would meet $D^{*}$ uncountably, whereas

$$U\cap D^{*}\subseteq (D^{*}\setminus C)\cup \left\{p\right\}$$

would be countable. Thus C is a nonempty perfect subset of ${\mathbb{R}}^3$. The same argument shows that every nonempty relatively open subset of C meets $D^{*}\cap C$ uncountably.

For $i\in \{1,2,3\}$ and $a\in \mathbb{R}$, the coordinate fiber

$$F_i\left(a\right)=\{x\in C:x_i=a\}$$

has empty interior in C. Otherwise a nonempty relatively open subset of C contained in $F_i\left(a\right)$ would meet $D^{*}\cap C$ uncountably, contrary to injectivity of ${\pi }_i{|}_{D^{*}}$. Lemma 5 therefore gives a nonempty compact perfect set $Q\subseteq C$ on which all three coordinate projections are one-to-one.

For $p\in Q$, put

$$I_B\left(p\right)=\{i\in \{1,2,3\}:p_i\in B\}.$$

Since $D^{*}$ is closed discrete, choose a basic Hattori box $U_p$ about p such that $U_p\cap D^{*}\subseteq \left\{p\right\}$. Choose ${\epsilon }_p>0$ so small that, whenever $z\in {\mathbb{R}}^3$ satisfies $\parallel z-p\parallel <{\epsilon }_p$, each coordinate $z_i$ with $i\notin I_B\left(p\right)$ lies in the corresponding two-sided factor of $U_p$, and each coordinate $z_i$ with $i\in I_B\left(p\right)$ lies below the upper endpoint of the corresponding one-sided factor of $U_p$. Since $U_p\cap (D^{*}\setminus \left\{p\right\})=\varnothing$, every $d\in D^{*}\setminus \left\{p\right\}$ with $\parallel d-p\parallel <{\epsilon }_p$ satisfies

$$d_i<p_i\mathrm{for}\mathrm{some}i\in I_B\left(p\right).$$

(3.1)

Because p is a condensation point of $D^{*}$, there is $d\in D^{*}\setminus \left\{p\right\}$ with $\parallel d-p\parallel <{\epsilon }_p$. Equation (3.1) then forces $I_B\left(p\right)$ to be nonempty.

There is a Euclidean neighborhood $W_p$ of p such that no $q\in Q\cap W_p\setminus \left\{p\right\}$ satisfies

$$q_i>p_i\mathrm{for}\mathrm{every}i\in I_B\left(p\right).$$

(3.2)

If no such neighborhood existed, choose $q\in Q\setminus \left\{p\right\}$ close enough to p that $\parallel q-p\parallel <{\epsilon }_p/2$ and (3.2) holds. Let

$$\eta =min\{q_i-p_i:i\in I_B\left(p\right)\}>0.$$

Since $q\in C$, it is a condensation point of $D^{*}$. Hence there is $d\in D^{*}\setminus \left\{p\right\}$ with

$$\parallel d-q\parallel <min\{\eta /2,{\epsilon }_p/2\}.$$

Then $d_i>p_i$ for every $i\in I_B\left(p\right)$, while $\parallel d-p\parallel <{\epsilon }_p$, contradicting (3.1).

For $i=1,2,3$, put

$$A_i=\{x\in Q:x_i\in A\}.$$

Every $A_i$ meets every nonempty perfect subset of Q. If a nonempty perfect $P\subseteq Q$ were disjoint from $A_i$, then P would be compact because it is closed in the compact set Q. The continuous injection ${\pi }_i{|}_P$ would therefore be a homeomorphism onto a nonempty compact perfect subset of B. By [9, Theorem 6.2], such a set contains a subspace homeomorphic to $2^{\omega }$, contrary to Theorem 1.

For $i\in \{1,2,3\}$, let $E_i$ be the set of points $x\in Q$ for which $x_i$ has a right-hand gap in ${\pi }_i\left[Q\right]$, meaning that ${\pi }_i\left[Q\right]\cap (x_i,x_i+\epsilon )=\varnothing$ for some $\epsilon >0$. Each $E_i$ is countable. Apart from a possible maximum, every right-gap point of the compact perfect set ${\pi }_i\left[Q\right]$ is the left endpoint of a component of its complement, and there are only countably many such components.

For distinct $i,j,k\in \{1,2,3\}$, let $M_i$ consist of the points $x\in Q$ that are local Pareto maxima in coordinates $j,k$. Thus $x\in M_i$ if there is $\delta >0$ such that

$$\nexists y\in Q\setminus \left\{x\right\}\mathrm{such}\mathrm{that}\parallel y-x\parallel <\delta ,y_j\ge x_j,y_k\ge x_k.$$

The set $M_i$ is Borel. For $n\in \mathbb{N}$, let $M_{i,n}$ be defined by the same condition with $\parallel y-x\parallel \le 1/n$. For $r\in \mathbb{N}$, the set

$$R_{n,r}=\{(x,y)\in Q^2:1/r\le \parallel y-x\parallel \le 1/n,y_j\ge x_j,y_k\ge x_k\}$$

is compact. Write ${pr}_1:Q^2\to Q$ for projection onto its first factor. Then ${pr}_1\left(R_{n,r}\right)$ is compact, and

$$Q\setminus M_{i,n}=\bigcup _{r\in \mathbb{N}}{pr}_1\left(R_{n,r}\right).$$

Thus $M_{i,n}$ is a $G_{\delta }$ subset of Q, and $M_i={\bigcup }_{n\in \mathbb{N}}{M}_{i,n}$ is Borel.

The required inclusion is

$$A_i\subseteq M_i\cup E_j\cup E_k.$$

(3.3)

Take $x\in A_i\setminus (E_j\cup E_k)$. If both $x_j,x_k\in A$, then $I_B\left(x\right)=\varnothing$, contradicting $I_B\left(x\right)\ne \varnothing$. Suppose exactly one of $x_j,x_k$ lies in B. By symmetry in j and k, suppose $x_j\in B$. Then $I_B\left(x\right)=\left\{j\right\}$. Hence (3.2) excludes all sufficiently near points of Q whose j-th coordinate is larger than $x_j$. Since $x\notin E_j$ and ${\pi }_j{|}_Q$ is a homeomorphism onto its image, points of Q with larger j-th coordinate must approach x, a contradiction. Hence $x_j,x_k\in B$. Equation (3.2) now excludes every sufficiently near $y\ne x$ with both $y_j>x_j$ and $y_k>x_k$. Injectivity of ${\pi }_j{|}_Q$ and ${\pi }_k{|}_Q$ excludes equality for $y\ne x$, so $x\in M_i$. This proves (3.3).

Let $P\subseteq Q$ be nonempty and perfect. The Borel set $P\setminus (E_j\cup E_k)$ is uncountable and therefore contains a nonempty perfect subset [9]. That perfect subset meets $A_i$, so (3.3) gives $M_i\cap P\ne \varnothing$. Thus $M_i$ meets every nonempty perfect subset of Q. Since $M_i$ is Borel, the perfect set theorem implies that $Q\setminus M_i$ is countable. Each of $M_1,M_2,M_3$ is therefore cocountable in Q.

The intersection

$$M_1\cap M_2\cap M_3$$

is countable. For each x in this intersection, choose $n\left(x\right)\in \mathbb{N}$ such that all three local Pareto conditions hold in the Euclidean ball of radius $1/n\left(x\right)$ about x. For $n\in \mathbb{N}$, let

$$S_n=\{x\in M_1\cap M_2\cap M_3:n\left(x\right)=n\}.$$

If $x,y\in S_n$ are distinct, then $\parallel x-y\parallel \ge 1/n$. Otherwise their three coordinate differences are nonzero, so one of $x,y$ is larger than the other in at least two coordinates. The Pareto condition indexed by the remaining coordinate is then violated at the point that is smaller in those two coordinates. Thus $S_n$ is $1/n$-separated and hence finite in the compact metric space Q. Therefore $M_1\cap M_2\cap M_3={\bigcup }_{n\in \mathbb{N}}{S}_n$ is countable, contrary to the fact that the intersection of three cocountable subsets of the uncountable perfect set Q is uncountable. □

The separation argument applies the pigeonhole principle to the signs of three nonzero coordinate differences.

Corollary 1. 

For $A\subseteq \mathbb{R}$ and $B=\mathbb{R}\setminus A$, the following are equivalent.

1.

$H{\left(A\right)}^3$ is Lindelöf.

2.

$H{\left(A\right)}^2$ is Lindelöf.

3.

B contains no subspace homeomorphic to $2^{\omega }$.

Proof. 

The implication (1)⇒(2) follows by coordinate projection. The equivalence of (2) and (3) is Theorem 1, and Theorem 2 gives (2)⇒(1). □

If B is analytic, then Corollary 1 and the perfect set theorem [9, Theorem 29.1] show that B is countable exactly when $H{\left(A\right)}^2$ is Lindelöf, equivalently when $H{\left(A\right)}^3$ is Lindelöf. Under these equivalent conditions, $H\left(A\right)$ is second countable by Lemma 1, and hence $H{\left(A\right)}^{\omega }$ is second countable. Since every Borel subset of $\mathbb{R}$ is analytic, the conclusions also hold for Borel B.

For $A\subseteq \mathbb{R}$, if $H{\left(A\right)}^2$ is Lindelöf, then every discrete subspace of $H{\left(A\right)}^3$ that is analytic in the Euclidean topology is countable, and no discrete subspace of the cube contains a nonempty Euclidean perfect subset. A nonempty Euclidean perfect subset of ${\mathbb{R}}^3$ contains a compact subspace homeomorphic to $2^{\omega }$ [9, Theorem 6.2]. That subspace is Euclidean closed and hence closed in the finer Hattori product topology. If it lay inside a discrete subspace, it would be an uncountable closed discrete subspace of the Lindelöf cube, which is impossible. The analytic assertion then follows from the perfect set theorem [9, Theorem 29.1].

4. Affine Discreteness in Finite Powers

Lemma 6. 

Let $k\in \mathbb{N}$, let $q\in \{0,\dots ,k\}$, and let $r_1,\dots ,r_q\in {\mathbb{R}}^k$. Then

$$\{u\in {\mathbb{R}}^k:r_i·u\ge 0\mathrm{for}1\le i\le q\}$$

contains a nonzero vector.

Proof. 

If $q=0$, the asserted set is ${\mathbb{R}}^k$, which contains nonzero vectors. If the vectors $r_1,\dots ,r_q$ do not span ${\mathbb{R}}^k$, choose a nonzero vector orthogonal to their span. Suppose that $q=k$ and that the vectors are linearly independent. Let R be the matrix whose rows are the $r_i$. There is a vector u satisfying

$$Ru={(1,\dots ,1)}^{\mathsf{T}}.$$

Then $r_i·u=1$ for every i. □

Theorem 3. 

Let $A\subseteq \mathbb{R}$, put $B=\mathbb{R}\setminus A$, and let $k,m\in \mathbb{N}$ with $k\le m$. Let M be an $m\times k$ real matrix of rank k, and let $b\in {\mathbb{R}}^m$. Define $F:{\mathbb{R}}^k\to {\mathbb{R}}^m$ by

$$F\left(t\right)=b+Mt(t\in {\mathbb{R}}^k)$$

and write $F_i$ for the i-th coordinate function of F and $r_i\in {\mathbb{R}}^k$ for the i-th row of M. Put

$$L=F\left({\mathbb{R}}^k\right)\subseteq H{\left(A\right)}^m$$

and, for $t\in {\mathbb{R}}^k$,

$$I\left(t\right)=\{i\in \{1,\dots ,m\}:F_i\left(t\right)\in B\}.$$

Call an index i active at t if $i\in I\left(t\right)$. For $t\in {\mathbb{R}}^k$, put

$$C\left(t\right)=\bigcap _{i\in I\left(t\right)}\{u\in {\mathbb{R}}^k:r_i·u\ge 0\},$$

where an empty intersection is understood to be ${\mathbb{R}}^k$. Then L is closed in $H{\left(A\right)}^m$, and $F\left(t\right)$ is isolated in L if and only if

$$C\left(t\right)=\left\{0\right\}.$$

(4.1)

Consequently, L is discrete if and only if (4.1) holds for every $t\in {\mathbb{R}}^k$.

Proof. 

The set L is a Euclidean affine subspace of ${\mathbb{R}}^m$, hence it is closed in the Euclidean topology. The topology of $H{\left(A\right)}^m$ refines the Euclidean product topology, so L is also closed in $H{\left(A\right)}^m$.

Fix $t\in {\mathbb{R}}^k$ and put $x=F\left(t\right)$. Assume first that (4.1) holds. For each $i\in I\left(t\right)$, choose the right-hand basic neighborhood $[x_i,x_i+1)$. In the other coordinates take the whole space $H\left(A\right)$. Their product is an open neighborhood U of x in $H{\left(A\right)}^m$. If $F(t+u)\in U$, then

$$r_i·u\ge 0(i\in I\left(t\right)).$$

Equation (4.1) gives $u=0$. Thus $U\cap L=\left\{x\right\}$.

Conversely, suppose that $C\left(t\right)$ contains a nonzero vector u. Let $U={\prod }_{i=1}^m{U}_i$ be a basic neighborhood of x. If $x_i\in A$, choose ${\epsilon }_i>0$ such that

$$(x_i-{\epsilon }_i,x_i+{\epsilon }_i)\subseteq U_i.$$

If $x_i\in B$, choose ${\epsilon }_i>0$ such that

$$[x_i,x_i+{\epsilon }_i)\subseteq U_i.$$

Since there are only finitely many coordinates, choose $\lambda >0$ small enough that

$$|\lambda r_i·u|<{\epsilon }_i(x_i\in A)$$

and

$$0\le \lambda r_i·u<{\epsilon }_i(x_i\in B).$$

Hence $F(t+\lambda u)\in U$. Since M has rank k and $u\ne 0$, this point differs from $F\left(t\right)$. Therefore $F\left(t\right)$ is not isolated in L. □

For affine lines the criterion has a simpler form.

Corollary 2. 

Let $A\subseteq \mathbb{R}$, put $B=\mathbb{R}\setminus A$, let $m\in \mathbb{N}$, and let $b,v\in {\mathbb{R}}^m$ with $v\ne 0$. Put

$$L=\{b+tv:t\in \mathbb{R}\}\subseteq H{\left(A\right)}^m.$$

Then L is discrete if and only if, for every $t\in \mathbb{R}$, there are indices $i,j\in \{1,\dots ,m\}$ such that

$$v_i>0,b_i+t{v}_i\in B,$$

and

$$v_j<0,b_j+t{v}_j\in B.$$

Proof. 

For $k=1$, the intersection in (4.1) is $\left\{0\right\}$ if and only if the active coefficients contain both a positive and a negative number. □

Theorem 4. 

Let $A\subseteq \mathbb{R}$ be nonempty and let $m\in \{1,2,3\}$. Then $H{\left(A\right)}^m$ contains no discrete affine subspace of positive dimension.

Proof. 

Suppose that $L=F\left({\mathbb{R}}^k\right)\subseteq H{\left(A\right)}^m$ is a discrete affine subspace of positive dimension, with notation as in Theorem 3. Zero rows impose no restriction in (4.1), so they may be discarded when Lemma 6 is applied.

If $k=3$, then necessarily $m=3$. At each parameter t, Lemma 6 shows that at least four nonzero active rows would be needed for (4.1), which is impossible.

Suppose that $k=2$. The only case not already excluded by Lemma 6 is $m=3$. Then all three rows of M must be nonzero and active at every parameter, since otherwise at most two nonzero active rows would remain and Lemma 6 would contradict (4.1). Fix one row $r_i$. The affine function

$$t⟼b_i+r_i·t$$

is onto $\mathbb{R}$. Choose $a\in A$ and $t\in {\mathbb{R}}^2$ with $b_i+r_i·t=a$. At this parameter the i-th row is not active, so at most two nonzero active rows remain. Lemma 6 contradicts (4.1).

It remains to consider $k=1$. By Corollary 2, the nonzero coefficients of the direction vector must include both signs. This is impossible when $m=1$, so $m\in \{2,3\}$. One of the two sign classes therefore consists of a single coordinate. That coordinate must take values in B for every parameter. The corresponding affine coordinate function is onto $\mathbb{R}$, so this forces $B=\mathbb{R}$, contrary to $A\ne \varnothing$. □

For $A=\varnothing$, the conclusion fails already in the square. The anti-diagonal $\left\{\right(t,-t):t\in \mathbb{R}\}$ is a closed discrete affine line in the Sorgenfrey plane. Thus the hypothesis $A\ne \varnothing$ in Theorem 4 cannot be omitted.

5. Monotone Traces

The affine result is local and linear. Under Lindelöfness of the square, a perfect-kernel argument gives an analogous four-coordinate obstruction for monotone traces without assuming an interval parameter or continuity at the outset.

Theorem 5. 

Let $A\subseteq \mathbb{R}$, put $B=\mathbb{R}\setminus A$, let $m\in \mathbb{N}$, and suppose that $H{\left(A\right)}^2$ is Lindelöf. Let $T\subseteq \mathbb{R}$ be uncountable and let

$$F=(f_1,\dots ,f_m):T⟶{\mathbb{R}}^m$$

be one-to-one. Suppose that, for each $i\in \{1,\dots ,m\}$, the function $f_i$ is constant, strictly increasing, or strictly decreasing on T. If $F\left[T\right]$ is closed and discrete in $H{\left(A\right)}^m$, then at least two coordinate functions are strictly increasing and at least two are strictly decreasing. In particular, $m\ge 4$.

Proof. 

Since the graph of F in ${\mathbb{R}}^{m+1}$ is covered by countably many bounded boxes, one of them meets the graph in an uncountable set. Restrict T accordingly. Then T and all the sets $f_i\left[T\right]$ are bounded. The restricted image $F\left[T\right]$ is still closed and discrete because it is a subset of the original closed discrete image. Put $K={\overline{T}}^{\mathbb{R}}$. This set is compact because T is bounded. Choose a compact interval $J\supseteq K$. For each strictly increasing coordinate, define a nondecreasing extension on J by

$${\overline{f}}_i\left(x\right)=sup\left(\{f_i\left(t\right):t\in T,t\le x\}\cup \{inf{f}_i\left[T\right]\}\right)(x\in J).$$

This function agrees with $f_i$ on T. For a strictly decreasing coordinate, apply the same construction to $-f_i$ and then negate the resulting extension. Extend each constant coordinate as a constant function. Denote the resulting monotone extensions by ${\overline{f}}_i$.

Each monotone real-valued function has only countably many discontinuities. Only countably many points of K fail to be approached by points of T from both sides. Apart from the possible endpoints of K, any point not approached from one side is an endpoint of a component of $\mathbb{R}\setminus K$. Removing from K the discontinuity points of the ${\overline{f}}_i$, together with the points that fail this two-sided approach property, leaves an uncountable Borel subset of $\mathbb{R}$. The perfect set theorem [9, Theorem 29.1] therefore gives a nonempty perfect set $Q\subseteq K$ such that every ${\overline{f}}_i$ is continuous at every point of Q, and every point of Q is approached by points of T from both sides. Since Q is closed in the compact set K, it is compact.

If $f_i$ is strictly increasing on T, then ${\overline{f}}_i{|}_Q$ is strictly increasing. For $q<r$ in Q, choose $s,t\in T$ with $q<s<t<r$. Since ${\overline{f}}_i$ is nondecreasing and agrees with $f_i$ on T,

$${\overline{f}}_i\left(q\right)\le f_i\left(s\right)<f_i\left(t\right)\le {\overline{f}}_i\left(r\right).$$

The decreasing case is analogous. Thus

$$\overline{F}=({\overline{f}}_1,\dots ,{\overline{f}}_m){|}_Q$$

is continuous. Since F is one-to-one, at least one $f_i$ is strictly monotone on T. By the preceding argument, ${\overline{f}}_i{|}_Q$ is strictly monotone. Hence $\overline{F}$ is one-to-one.

Suppose first that there is no strictly increasing coordinate. Fix $q\in Q$ and choose a strictly increasing sequence $t_n\in T$ with $t_n<q$ and $t_n\to q$. For each n, choose $s_n\in T$ with $t_n<s_n<q$. Then, for every strictly decreasing coordinate i,

$$f_i\left(t_n\right)>f_i\left(s_n\right)\ge {\overline{f}}_i\left(q\right)\mathrm{and}{f}_i\left(t_n\right)⟶{\overline{f}}_i\left(q\right).$$

For each strictly decreasing coordinate i, this convergence is from above. If ${\overline{f}}_i\left(q\right)\in B$, the terms are eventually in every basic interval $[{\overline{f}}_i\left(q\right),{\overline{f}}_i\left(q\right)+\epsilon )$, and if ${\overline{f}}_i\left(q\right)\in A$, Euclidean convergence suffices. Thus $f_i\left(t_n\right)\to {\overline{f}}_i\left(q\right)$ in $H\left(A\right)$. Constant coordinates also converge. Therefore

$$F\left(t_n\right)⟶\overline{F}\left(q\right)$$

in $H{\left(A\right)}^m$. Since F is one-to-one, the points $F\left(t_n\right)$ are pairwise distinct. If $\overline{F}\left(q\right)\notin F\left[T\right]$, this convergence contradicts closedness of $F\left[T\right]$. If $\overline{F}\left(q\right)\in F\left[T\right]$, it contradicts discreteness. Thus there is at least one strictly increasing coordinate. The symmetric argument from the right, using a strictly decreasing sequence $u_n>q$ with $u_n\in T$ and $u_n\to q$, gives at least one strictly decreasing coordinate.

Suppose that $f_j$ is the unique strictly increasing coordinate. If ${\overline{f}}_j\left(q\right)\in A$ for some $q\in Q$, choose a strictly increasing sequence $t_n\in T$ with $t_n<q$ and $t_n\to q$. Then $F\left(t_n\right)\to \overline{F}\left(q\right)$ in $H{\left(A\right)}^m$, again a contradiction. Hence

$${\overline{f}}_j\left[Q\right]\subseteq B.$$

The continuous injection ${\overline{f}}_j{|}_Q$ is a homeomorphism onto its compact image. Since Q has no isolated points, that image is a nonempty perfect subset of B. By [9, Theorem 6.2], this contradicts Theorem 1. Thus there are at least two strictly increasing coordinates. The argument for strictly decreasing coordinates is symmetric. □

The four-coordinate threshold in Theorem 5 is attained by the affine line constructed in the proof of Theorem 8. Two of its coordinate functions are strictly increasing and two are strictly decreasing.

6. Reductions to Sorgenfrey Powers

For $A\subseteq \mathbb{R}$ and $B=\mathbb{R}\setminus A$, the subspace topology on A inherited from $H\left(A\right)$ is Euclidean, while the subspace topology on B is the Sorgenfrey subspace topology [4, Lemma 2.1]. Gruenhage and Nyikos proved that for a subspace of a finite product whose factors are Sorgenfrey or Euclidean lines, ${\omega }_1$-compactness and hereditary Lindelöfness are equivalent [5, Theorem 6.9]. Hernández-Hernández, Ramírez-Chávez, and Rojas-Hernández proved that the product of a hereditarily Lindelöf space with a cosmic space is hereditarily Lindelöf [6, Proposition 6.6].

Corollary 3. 

Let $r,m\in \mathbb{N}$, let $C\subseteq {\mathbb{R}}^r$ carry its Euclidean subspace topology, and let $X\subseteq \mathbb{R}$ carry the Sorgenfrey subspace topology. If $X^m$ is Lindelöf, then $C\times X^m$ is hereditarily Lindelöf. If $C\ne \varnothing$ and $C\times X^m$ is Lindelöf, then $X^m$ is Lindelöf.

Proof. 

Assume that $X^m$ is Lindelöf. As a subspace of a finite power of the Sorgenfrey line, it is ${\omega }_1$-compact and hence hereditarily Lindelöf by [5, Theorem 6.9]. The Euclidean subspace C is second countable and therefore cosmic. The product theorem [6, Proposition 6.6] now shows that $C\times X^m$ is hereditarily Lindelöf. If $C\ne \varnothing$ and $C\times X^m$ is Lindelöf, then the continuous projection $C\times X^m\to X^m$ shows that $X^m$ is Lindelöf. □

Corollary 4. 

Let $\varnothing \ne A\subseteq \mathbb{R}$, put $B=\mathbb{R}\setminus A$, and let $k,m\in \mathbb{N}$. Here A carries the Euclidean subspace topology, and B carries the Sorgenfrey subspace topology. The following conditions are equivalent.

1.

$B^m$ is Lindelöf.

2.

$B^m$ is hereditarily Lindelöf.

3.

$A^k\times B^m$ is Lindelöf.

4.

$A^k\times B^m$ is hereditarily Lindelöf.

Proof. 

If $B^m$ is Lindelöf, then $B^m$, viewed as a subspace of a finite power of the Sorgenfrey line, is hereditarily Lindelöf by [5, Theorem 6.9]. Applying Corollary 3 with $C=A^k$ and $X=B$ shows that $A^k\times B^m$ is hereditarily Lindelöf. This proves (1)⇒(2) and (1)⇒(4). The implications (4)⇒(3) and (2)⇒(1) follow from the definitions. Since $A\ne \varnothing$, projection onto the last m coordinates gives (3)⇒(1). □

Corollary 5. 

Let $A\subseteq \mathbb{R}$, put $B=\mathbb{R}\setminus A$, and let $n\in \mathbb{N}$. Then $H{\left(A\right)}^n$ is hereditarily Lindelöf if and only if $B^n$, with its Sorgenfrey product topology, is Lindelöf.

Proof. 

If $H{\left(A\right)}^n$ is hereditarily Lindelöf, then its subspace $B^n$ is Lindelöf. Conversely, suppose that $B^n$ is Lindelöf. If $A=\varnothing$, then $H{\left(A\right)}^n=B^n$ is ${\omega }_1$-compact and hence hereditarily Lindelöf by [5, Theorem 6.9]. Suppose that $A\ne \varnothing$. For each $r\in \{1,\dots ,n\}$, the projection of $B^n$ onto its first r coordinates shows that $B^r$ is Lindelöf. The stratum $A^n$ is second countable. The Lindelöf stratum $B^n$ is ${\omega }_1$-compact and hence hereditarily Lindelöf by [5, Theorem 6.9]. Every mixed stratum is hereditarily Lindelöf by Corollary 4. These finitely many strata cover $H{\left(A\right)}^n$. Intersecting them with an arbitrary subspace of $H{\left(A\right)}^n$ expresses that subspace as a finite union of Lindelöf spaces. Hence $H{\left(A\right)}^n$ is hereditarily Lindelöf. □

Theorem 6. 

Let $A\subseteq \mathbb{R}$, put $B=\mathbb{R}\setminus A$, and let $n\in \mathbb{N}$ with $n\ge 3$. If $B^{n-1}$, with its Sorgenfrey product topology, is Lindelöf, then $H{\left(A\right)}^n$ is Lindelöf. Consequently,

$$H{\left(A\right)}^n\mathrm{is}\mathrm{not}\mathrm{Lindel}\text{ö}\mathrm{f}⟹B^{n-1}\mathrm{is}\mathrm{not}\mathrm{Lindel}\text{ö}\mathrm{f}.$$

For $n=3$, the conclusion also follows from Corollaries 5 and 1. The proof below treats all $n\ge 3$ uniformly.

Proof. 

The hypothesis implies that $B^r$ is Lindelöf for every $r\in \{1,\dots ,n-1\}$ by coordinate projection. In particular, $B^2$ is Lindelöf, so Corollary 5 and Theorem 1 show that B contains no subspace homeomorphic to $2^{\omega }$.

If $A=\varnothing$, then $B=\mathbb{R}$ and the hypothesis would make the Sorgenfrey plane Lindelöf by projection, which is impossible. Hence $A\ne \varnothing$. Every stratum of $H{\left(A\right)}^n$ with at least one coordinate in A is, up to a coordinate permutation, either $A^n$ or $A^{n-r}\times B^r$ for some $r\in \{1,\dots ,n-1\}$. The first is second countable, and all the others are Lindelöf by Corollary 4. There are only finitely many strata, so

$$H{\left(A\right)}^n\setminus B^n$$

is Lindelöf.

Suppose that $H{\left(A\right)}^n$ is not Lindelöf. Choose an open cover with no countable subcover and refine it by basic open boxes. Let $\mathcal{U}$ denote the resulting cover. It also has no countable subcover. A countable subfamily ${\mathcal{U}}_0$ covers $H{\left(A\right)}^n\setminus B^n$. The remainder

$$C=H{\left(A\right)}^n\setminus \bigcup {\mathcal{U}}_0$$

is contained in $B^n$ and is uncountable. Otherwise, choosing one member of $\mathcal{U}$ for each point of the countable remainder would give a countable subcover of $H{\left(A\right)}^n$. Every member of ${\mathcal{U}}_0$ is a finite product of Euclidean open intervals and half-open intervals, so C is Borel in the Euclidean space ${\mathbb{R}}^n$. The perfect set theorem [9, Theorem 29.1], followed by [9, Theorem 6.2], gives a compact subspace $K\subseteq C$ homeomorphic to $2^{\omega }$. At least one coordinate projection of K has uncountable image. This image is an uncountable compact subset of B. By [9, Theorem 29.1] and [9, Theorem 6.2], it contains a subspace homeomorphic to $2^{\omega }$, contradicting the fact that B contains no subspace homeomorphic to $2^{\omega }$. □

Theorem 7. 

Let $A\subseteq \mathbb{R}$ and put $B=\mathbb{R}\setminus A$. The following conditions are equivalent.

1.

$H{\left(A\right)}^{\omega }$ is hereditarily Lindelöf.

2.

$H{\left(A\right)}^m$ is hereditarily Lindelöf for every $m\in \mathbb{N}$.

3.

$B^m$, with its Sorgenfrey product topology, is Lindelöf for every $m\in \mathbb{N}$.

Proof. 

Assume first that (3) holds. Let $Y\subseteq H{\left(A\right)}^{\omega }$, and suppose that Y is not Lindelöf. By Lemmas 3 and 4, Y contains an uncountable closed discrete set D.

For each $x={\left(x_i\right)}_{i<\omega }\in D$, choose a basic neighborhood isolating x from D. Shrink this neighborhood if necessary, and choose a finite set $F_x\subseteq \omega$ and an integer $n_x\ge 1$ with the following property. For $i<\omega$, define

$$U_i(x,F_x,n_x)=\left\{\begin{array}{cc}(x_i-1/n_x,x_i+1/n_x),\hfill & i\in F_x,x_i\in A,\hfill \\ [x_i,x_i+1/n_x),\hfill & i\in F_x,x_i\in B,\hfill \\ \mathbb{R},\hfill & i\notin F_x.\hfill \end{array}\right.$$

Then the basic set

$$Q(x,F_x,n_x)=\prod _{i<\omega }{U}_i(x,F_x,n_x)$$

meets D only at x. There are only countably many pairs $(F_x,n_x)$. Hence there exist a finite set $F\subseteq \omega$, an integer $n\ge 1$, and an uncountable set $D_0\subseteq D$ for which $F_x=F$ and $n_x=n$ for every $x\in D_0$. Pass to a further uncountable subset $D_1$ such that the set

$$I=\{i\in F:x_i\in B\}$$

does not depend on $x\in D_1$.

Let ${\pi }_F:H{\left(A\right)}^{\omega }\to {\mathbb{R}}^F$ denote projection onto the coordinates indexed by F. The map ${\pi }_F$ is one-to-one on $D_1$, since two distinct points with the same coordinates on F would belong to each other’s isolating basic sets. Thus ${\pi }_F\left[D_1\right]$ is an uncountable subset of the Euclidean space ${\mathbb{R}}^F$. Choose a Euclidean condensation point p of this image. Then choose an uncountable set $E\subseteq D_1$ such that

$${\pi }_F\left[E\right]\subseteq B_{\mathrm{E}}(p,1/\left(4n\right)).$$

For distinct $x,z\in E$, all coordinates indexed by F differ by less than $1/n$. Since $z\notin Q(x,F,n)$, some $i\in I$ satisfies $z_i<x_i$. Interchanging x and z gives some $j\in I$ with $x_j<z_j$. Necessarily $i\ne j$, so $\left|I\right|\ge 2$. Write $I=\{i_1<\dots <i_s\}$, where $s=\left|I\right|\ge 2$, and define

$${\pi }_I\left(x\right)=(x_{i_1},\dots ,x_{i_s})\in {\mathbb{R}}^s.$$

The same inequalities show that ${\pi }_I$ is one-to-one on E, and ${\pi }_I\left[E\right]$ is an uncountable discrete surface in $B^s$. This contradicts Proposition 1 and condition (3). Hence every subspace Y of $H{\left(A\right)}^{\omega }$ is Lindelöf, proving (1).

Condition (1) implies (2). Fix $m\in \mathbb{N}$ and $x_0\in \mathbb{R}$. The subspace of $H{\left(A\right)}^{\omega }$ whose first m coordinates are arbitrary and whose remaining coordinates equal $x_0$ is homeomorphic to $H{\left(A\right)}^m$, so it is hereditarily Lindelöf. Condition (2) implies (3) because $B^m$ is a subspace of $H{\left(A\right)}^m$ for every $m\in \mathbb{N}$. □

Corollary 6. 

Assume $\mathfrak{b}={\omega }_1$. There is a set $A\subseteq \mathbb{R}$ such that $B=\mathbb{R}\setminus A$ is uncountable, $H\left(A\right)$ is not second countable, and $H{\left(A\right)}^{\omega }$ is hereditarily Lindelöf.

Proof. 

Under the hypothesis $\mathfrak{b}={\omega }_1$, Bonanzinga, Giacopello, and Szewczak construct an uncountable subspace B of the Sorgenfrey line all of whose finite powers are Hurewicz and hence Lindelöf [1, Example 6.1]. Put $A=\mathbb{R}\setminus B$. Theorem 7 shows that $H{\left(A\right)}^{\omega }$ is hereditarily Lindelöf, while Lemma 1 shows that $H\left(A\right)$ is not second countable. □

Under the continuum hypothesis, Michael had already constructed an uncountable Sorgenfrey subspace B with $B^{\omega }$ hereditarily Lindelöf [13, Remark, p. 210], so every finite power $B^m$ is Lindelöf. For $A=\mathbb{R}\setminus B$, Theorem 7 and Lemma 1 give the conclusion of Corollary 6 directly. The continuum hypothesis also implies $\mathfrak{b}={\omega }_1$.

7. A ZFC Separation Between the Third and Fourth Powers

The following construction attains the four-coordinate threshold.

Lemma 7. 

There is a set $A\subseteq \mathbb{R}$ of cardinality $\mathfrak{c}$ that meets every nonempty perfect subset of $\mathbb{R}$ and satisfies

$$A\cap (A+n)=\varnothing (n\in \mathbb{Z}\setminus \{0\left\}\right).$$

Proof. 

The family of nonempty perfect subsets of $\mathbb{R}$ has cardinality $\mathfrak{c}$. There are at most $\mathfrak{c}$ closed subsets of the second-countable space $\mathbb{R}$, while the translates of the Cantor set by distinct real numbers are distinct nonempty perfect sets. Enumerate the nonempty perfect subsets as

$$\{P_{\alpha }:\alpha <\mathfrak{c}\}.$$

Every $P_{\alpha }$ has cardinality $\mathfrak{c}$. Recursively choose

$$a_{\alpha }\in P_{\alpha }\setminus \bigcup _{\beta <\alpha }(a_{\beta }+\mathbb{Z}).$$

At stage $\alpha <\mathfrak{c}$, the excluded set has cardinality at most $max\{{\aleph }_0,|\alpha \left|\right\}<\mathfrak{c}$, so the choice is possible. Put

$$A=\{a_{\alpha }:\alpha <\mathfrak{c}\}.$$

The selected points are distinct, so $\left|A\right|=\mathfrak{c}$, and A meets every nonempty perfect subset of $\mathbb{R}$. If distinct points of A differed by an integer, the one chosen later would have been excluded at its stage. Thus the displayed shift-avoidance property holds. □

The construction is a transfinite integer-shift avoidance argument. For Sierpiński’s construction technique on subsets of $\mathbb{R}$, see [15].

Theorem 8. 

There is a set $A\subseteq \mathbb{R}$ such that, with $B=\mathbb{R}\setminus A$,

1.

$\left|B\right|=\mathfrak{c}$ and B contains no subspace homeomorphic to $2^{\omega }$.

2.

$H{\left(A\right)}^2$ and $H{\left(A\right)}^3$ are Lindelöf.

3.

$B^2$, with its Sorgenfrey product topology, contains a discrete surface of cardinality $\mathfrak{c}$ and is not Lindelöf.

4.

$H{\left(A\right)}^4$ contains a closed discrete affine line of cardinality $\mathfrak{c}$. Hence $H{\left(A\right)}^n$ is not Lindelöf for every integer $n\ge 4$, and $H{\left(A\right)}^{\omega }$ is not Lindelöf.

Proof. 

Choose A as in Lemma 7. Since A meets every nonempty perfect subset of $\mathbb{R}$, the complement B contains no subspace homeomorphic to $2^{\omega }$. Also $A+1\subseteq B$, so $\left|B\right|=\mathfrak{c}$. Theorems 1 and 2 give Lindelöfness of $H{\left(A\right)}^2$ and $H{\left(A\right)}^3$.

For $i,j\in \{0,1\}$, put

$$T_{ij}=\{t\in \mathbb{R}:t+i\in B\mathrm{and}-t+j\in B\}.$$

The four sets $T_{ij}$ cover $\mathbb{R}$. By Lemma 7, neither $\{t,t+1\}$ nor $\{-t,-t+1\}$ can be contained in A. Thus at least one point of each pair belongs to B. Hence some $T_{ij}$ has cardinality $\mathfrak{c}$. For that pair, the set

$$\{(t+i,-t+j):t\in T_{ij}\}\subseteq B^2$$

is a discrete surface of cardinality $\mathfrak{c}$. If $s<t$, the first coordinate increases and the second decreases. Proposition 1 therefore shows that $B^2$ is not Lindelöf.

Consider the affine line

$$D=\{(t,t+1,-t,-t+1):t\in \mathbb{R}\}\subseteq {\mathbb{R}}^4.$$

Its direction vector is $(1,1,-1,-1)$. For every t, at least one of $t,t+1$ belongs to B, and at least one of $-t,-t+1$ belongs to B. Corollary 2 shows that D is discrete in $H{\left(A\right)}^4$, while Theorem 3 shows that it is closed. Thus D is a closed discrete subspace of cardinality $\mathfrak{c}$, so $H{\left(A\right)}^4$ is not Lindelöf. Projection onto the first four coordinates then gives the assertions for every $n\ge 4$ and for the countable power. □

Together with Theorem 4, Theorem 8 shows that 4 is the least exponent m for which there exists a nonempty $A\subseteq \mathbb{R}$ such that $H{\left(A\right)}^m$ contains a closed discrete affine line.

For the set A in Theorem 8, the Sorgenfrey square $B^2$ is non-Lindelöf while the Hattori cube $H{\left(A\right)}^3$ is Lindelöf. Theorem 6 shows that non-Lindelöfness of $B^2$ is necessary for non-Lindelöfness of $H{\left(A\right)}^3$, whereas this construction shows that it is not sufficient.

The same example also shows that $H{\left(A\right)}^2$ and $H{\left(A\right)}^3$ are Lindelöf but not hereditarily Lindelöf. The subspace $B^2\subseteq H{\left(A\right)}^2$ is not Lindelöf, and for any $x_0\in \mathbb{R}$, the subspace $B^2\times \left\{x_0\right\}\subseteq H{\left(A\right)}^3$ is homeomorphic to $B^2$.

8. Further Questions

Corollary 5 and Theorem 7 characterize hereditary Lindelöfness of the finite and countable Hattori powers. The questions below concern Lindelöfness itself.

Corollary 1 gives the cube characterization requested in [6, Problem 7.8], while Theorem 8 shows that the square obstruction no longer controls powers $n\ge 4$. Kulesza’s meeting abstract [11] announces the general even-to-odd implication (1.2). Theorem 2 proves its first case.

Question 9. 

For each fixed integer $n\ge 4$, characterize the sets $A\subseteq \mathbb{R}$ for which $H{\left(A\right)}^n$ is Lindelöf. Also characterize the sets A for which $H{\left(A\right)}^{\omega }$ is Lindelöf. In particular, with $B=\mathbb{R}\setminus A$, determine structural conditions on the Sorgenfrey stratum B that account for the general even-to-odd implication announced by Kulesza and for the possible odd-to-even losses of Lindelöfness.