Lindelöf Powers of Hattori Spaces and the Perfect Set Property

1. Introduction

For a set $A\subseteq \mathbb{R}$, put $B=\mathbb{R}\setminus A$. The Hattori space $H\left(A\right)$ is $\mathbb{R}$ with the topology for which points of A have their usual Euclidean local bases, while points of B have their local bases inherited from the Sorgenfrey line [17]. Then $H\left(\mathbb{R}\right)$ is the usual real line, and $H(\varnothing )$ is the Sorgenfrey line. Every Hattori space is hereditarily Lindelöf ([7], Proposition 3.1). The question considered below concerns the finite and countable products, not the Lindelöfness of $H\left(A\right)$ itself.

Several closely related lines of work concern Hattori spaces and neighboring hybrid topologies. Bouziad and Sukhacheva studied completeness-type and weak-separation properties of Hattori spaces, while Kulesza considered spaces lying between the Sorgenfrey topology and the usual topology on $\mathbb{R}$ [1,13]. Later work treats compactness-like properties and Hattori subspaces [3,4], countable dense homogeneity and $\sigma$-compactness [6,8], and function-space questions involving Hattori spaces [9,12,18]. Related hybrid topologies on the real line, and Hattori-type versions on almost topological groups, are studied in [2,5,14,16].

In their work on Lindelöf properties of spaces of continuous functions and of Hattori spaces, Hernández-Hernández, Ramírez-Chávez and Rojas-Hernández proved the square characterization ([9], Theorem 6.7)

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

They then asked for characterizations of when $H{\left(A\right)}^3$, all finite powers $H{\left(A\right)}^n$, or $H{\left(A\right)}^{\omega }$ are Lindelöf ([9], Problem 7.8). With the square theorem in hand, the question becomes whether this same condition suffices for the cube, for the higher finite powers, or for the countable power. Related product questions for Hattori spaces have also been studied by Lin and Li. They asked whether the Lindelöfness of $H{\left(A\right)}^2$, the normality of $H{\left(A\right)}^2$, and the metrizability of $H\left(A\right)$ are equivalent, and they gave, under CH, an example with uncountable complement and Lindelöf square ([15], Question 4.1 and Example 4.2). The example below is obtained in ZFC rather than under CH, and it shows that the Lindelöf square need not force the fourth power to be Lindelöf.

We first treat the case of ([9], Problem 7.8) in which B has the perfect set property, deriving it from the square theorem, the perfect set property dichotomy, and the second-countability criterion for Hattori spaces. In that setting, the absence of a subspace homeomorphic to $2^{\omega }$ in B is equivalent to the countability of B. Together with the square characterization, this gives the corresponding finite-power and countable-power characterization for this class of Hattori spaces. No descriptive set theory beyond this dichotomy is needed. The conclusion applies when B is analytic, and in particular when B is Borel.

We then show that the perfect set property assumption is essential for the higher-power part of the conclusion. We construct, in ZFC, a set $A\subseteq \mathbb{R}$ for which B has cardinality $\mathfrak{c}$ and contains no subspace homeomorphic to $2^{\omega }$. By the square theorem, $H{\left(A\right)}^2$ is Lindelöf, but $H{\left(A\right)}^n$ is not Lindelöf for every $n\ge 4$, and $H{\left(A\right)}^{\omega }$ is not Lindelöf. The construction makes A meet every nonempty perfect set while $\left\{1\right\}\cup A$ remains linearly independent over $\mathbb{Q}$. This gives the one-sided coordinates for an uncountable closed discrete subspace of $H{\left(A\right)}^4$. Thus the reduction of the higher-power problem to the cardinality of B that the perfect set property provides cannot come from the square theorem alone. The corresponding cube implication remains undecided.

All finite and countable powers in this note are endowed with the product topology.

2. Preliminaries

We write $2^{\omega }$ for the Cantor space and $\mathfrak{c}$ for the cardinality of the continuum. We use the following definition.

Definition 1

([10], Definition 5.1). Let $A\subseteq \mathbb{R}$, and put $B=\mathbb{R}\setminus A$. The Hattori topology associated with A is the topology on $\mathbb{R}$ defined as follows. If $x\in A$, then the usual Euclidean intervals

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

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

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

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

We use the square theorem in the following form.

Theorem 1

([9], 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 $2^{\omega }$.

We also use the following countable-base characterization due to Chatyrko and Hattori, stated in the language of Hattori spaces. The same characterization is noted in later Hattori-space literature ([15], p. 1919). One direction is already contained in the network-weight computation $nw\left(H\left(A\right)\right)=|\mathbb{R}\setminus A|+{\aleph }_0$ of ([18], Proposition 7): if $H\left(A\right)$ is second countable, then $nw\left(H\left(A\right)\right)={\aleph }_0$, and B is countable.

Lemma 1

([7], 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.

For the perfect-set-property classification, we need only the following form of the perfect set property.

Definition 2

([11], Section 29.A). A set $B\subseteq \mathbb{R}$ has the perfect set property if either B is countable or B contains a nonempty perfect subset.

For subsets of the real line, this is equivalent to the form used below. Either B is countable or B contains a subspace homeomorphic to $2^{\omega }$. A subspace of $\mathbb{R}$ homeomorphic to $2^{\omega }$ is compact, closed in $\mathbb{R}$, and has no isolated points, so it is a nonempty perfect subset of $\mathbb{R}$. Conversely, every nonempty perfect subset of $\mathbb{R}$ contains a subspace homeomorphic to $2^{\omega }$ (see [11], Theorem 6.2). Analytic subsets of $\mathbb{R}$ have the perfect set property (see [11], Theorem 29.1).

Consequently, when $B\subseteq \mathbb{R}$ has the perfect set property, B is countable exactly when it contains no subspace homeomorphic to $2^{\omega }$: a countable set contains no such subspace, while an uncountable B contains a nonempty perfect subset. By [11, Theorem 6.2], this subset contains a subspace homeomorphic to $2^{\omega }$.

3. Main Results

The next proposition follows directly from Lemma 1 and Theorem 1.

Proposition 1.

Let $A\subseteq \mathbb{R}$, and put $B=\mathbb{R}\setminus A$.

1.

If B is countable, then $H{\left(A\right)}^{\omega }$ is Lindelöf, and $H{\left(A\right)}^n$ is Lindelöf for every integer $n\ge 2$.

2.

If $H{\left(A\right)}^{\omega }$ is Lindelöf, or if $H{\left(A\right)}^n$ is Lindelöf for some integer $n\ge 2$, then B contains no subspace homeomorphic to $2^{\omega }$.

Proof. 

If B is countable, then $H\left(A\right)$ is second countable by Lemma 1. It follows that $H{\left(A\right)}^{\omega }$ is second countable and Lindelöf. Each finite power is a continuous image of $H{\left(A\right)}^{\omega }$ under a coordinate projection, so $H{\left(A\right)}^n$ is Lindelöf for every $n\ge 2$.

For the second assertion, suppose first that $H{\left(A\right)}^{\omega }$ is Lindelöf. The projection onto the first two coordinates maps $H{\left(A\right)}^{\omega }$ continuously onto $H{\left(A\right)}^2$, so $H{\left(A\right)}^2$ is Lindelöf. The same conclusion holds if $H{\left(A\right)}^n$ is Lindelöf for some integer $n\ge 2$, again by projecting onto the first two coordinates. By Theorem 1, the set B contains no subspace homeomorphic to $2^{\omega }$. □

The perfect set property is used only to convert the square obstruction into countability.

Corollary 1.

Let $A\subseteq \mathbb{R}$, put $B=\mathbb{R}\setminus A$, and suppose that B has the perfect set property. The following assertions are equivalent.

1.

B is countable.

2.

$H\left(A\right)$ is second countable.

3.

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

4.

$H{\left(A\right)}^n$ is Lindelöf for every integer $n\ge 2$.

5.

$H{\left(A\right)}^3$ is Lindelöf.

6.

$H{\left(A\right)}^n$ is Lindelöf for some integer $n\ge 3$.

7.

$H{\left(A\right)}^2$ is Lindelöf.

Proof. 

The equivalence (1)⇔(2) is Lemma 1. If (1) holds, then Proposition 1 gives (3) and (4). Since (4) includes (5) and (7), and also implies (6), condition (1) implies all remaining conditions. It remains to prove that each of (3), (4), (5), (6), and (7) implies (1).

If (3) holds, or if (4), (5), or (6) holds, then a projection onto the first two coordinates shows that $H{\left(A\right)}^2$ is Lindelöf. It is enough to consider (7). By Theorem 1, (7) implies that B contains no subspace homeomorphic to $2^{\omega }$. Since B has the perfect set property, the preceding observation implies that B is countable. □

Corollary 2.

Let $A\subseteq \mathbb{R}$, and put $B=\mathbb{R}\setminus A$. If B is analytic, then the seven conditions in Corollary 1 are equivalent. In particular, this holds when B is Borel.

Proof. 

Analytic subsets of $\mathbb{R}$ have the perfect set property by ([11], Theorem 29.1). Corollary 1 applies. The Borel case follows because every Borel subset of $\mathbb{R}$ is analytic. □

Corollary 1 gives the part of ([9], Problem 7.8) in which B has the perfect set property. In this case, the square characterization of Hernández-Hernández, Ramírez-Chávez and Rojas-Hernández reduces the problem to the cardinality of B, because a set with the perfect set property and with no subspace homeomorphic to $2^{\omega }$ must be countable. Thus the perfect set property enters only in passing from the absence of a subspace homeomorphic to $2^{\omega }$ in B to the countability of B.

Another result of Hernández-Hernández, Ramírez-Chávez and Rojas-Hernández characterizes the countability of B through the Lindelöf $\Sigma$ property of $H\left(A\right)$ ([9], Theorem 6.2). Here the countable-power assertion is obtained instead from the second-countability conclusion of Lemma 1.

The perfect set property in Corollary 1 cannot be omitted from the higher-power conclusion. We prove this in ZFC by choosing A so that B contains no subspace homeomorphic to $2^{\omega }$, but still contains enough real numbers to isolate a large closed discrete subset of a higher power. The recursive choice is a transfinite construction of Sierpiński type (compare [19]). Rational linear independence supplies the one-sided coordinates.

Lemma 2.

There is a set $A\subseteq \mathbb{R}$ such that, putting $B=\mathbb{R}\setminus A$, the set B has cardinality $\mathfrak{c}$, contains no subspace homeomorphic to $2^{\omega }$, and $H{\left(A\right)}^4$ contains a closed discrete subspace of cardinality $\mathfrak{c}$. Consequently, $H{\left(A\right)}^2$ is Lindelöf and $H{\left(A\right)}^4$ is not Lindelöf.

Proof. 

By ([11], Theorem 6.2), every nonempty perfect subset of $\mathbb{R}$ contains a subspace homeomorphic to $2^{\omega }$. Hence, as a subset of $\mathbb{R}$, it has cardinality $\mathfrak{c}$. The nonempty perfect subsets of $\mathbb{R}$ are also $\mathfrak{c}$ in number. There are at most $\mathfrak{c}$ closed subsets of the second-countable space $\mathbb{R}$, while the translates of the Cantor set already give $\mathfrak{c}$ examples. Enumerate them as $\mathcal{P}=\{P_{\alpha }:\alpha <\mathfrak{c}\}$, and choose $a_{\alpha }\in P_{\alpha }$ recursively so that

$$a_{\alpha }\notin {span}_{\mathbb{Q}}\left(\left\{1\right\}\cup \{a_{\beta }:\beta <\alpha \}\right).$$

This is possible because, at stage $\alpha <\mathfrak{c}$, the set of previously chosen points has cardinality less than $\mathfrak{c}$. Its rational linear span has cardinality at most $max\{{\aleph }_0,|\alpha \left|\right\}<\mathfrak{c}$, whereas $P_{\alpha }$ has cardinality $\mathfrak{c}$. Put

$$A=\{a_{\alpha }:\alpha <\mathfrak{c}\}\mathrm{and}B=\mathbb{R}\setminus A.$$

The set $\left\{1\right\}\cup A$ is linearly independent over $\mathbb{Q}$. A nonzero rational dependence among its members would involve a largest-indexed $a_{\alpha }$, which would then lie in ${span}_{\mathbb{Q}}(\left\{1\right\}\cup \{a_{\beta }:\beta <\alpha \})$, contradicting its choice. No rational is ever chosen, since each rational already lies in ${span}_{\mathbb{Q}}\left(\left\{1\right\}\right)$. It follows that $A\cap \mathbb{Q}=\varnothing$, and in particular $1\notin A$. The set A meets each coset of $\mathbb{Q}$ in at most one point. If $x,y\in A$ are distinct and $x-y=q\in \mathbb{Q}\setminus \left\{0\right\}$, the relation $1·x-1·y-q·1=0$ is a nonzero rational dependence among $x,y,1$, which are distinct because $x\ne y$ and $1\notin A$. This cannot happen. Each $P_{\alpha }$ meets A at $a_{\alpha }$, so A meets every nonempty perfect subset of $\mathbb{R}$.

To bound $\left|B\right|$ from below, fix $t\in [0,1)$. The pair $\{t,t+1\}$ lies in a single $\mathbb{Q}$-coset, so at most one of its points lies in A, and the pair meets B. These pairs are disjoint as t ranges over $[0,1)$, giving $\left|B\right|\ge \mathfrak{c}$. The reverse inequality holds because $B\subseteq \mathbb{R}$, so $\left|B\right|=\mathfrak{c}$. Since A meets every nonempty perfect set, B contains no nonempty perfect subset of $\mathbb{R}$. Hence B contains no subspace homeomorphic to $2^{\omega }$. By Theorem 1, $H{\left(A\right)}^2$ is Lindelöf.

It remains to construct a closed discrete subspace of $H{\left(A\right)}^4$ of cardinality $\mathfrak{c}$. For $t\in \mathbb{R}$, put

$$d_t=(t,t+1,-t,-t+1)$$

and set

$$D=\{d_t:t\in \mathbb{R}\}\subseteq {\mathbb{R}}^4.$$

The map $t\mapsto d_t$ is injective, so $\left|D\right|=\mathfrak{c}$. The coordinates are chosen so that among $t,t+1$ at least one coordinate lies in B, and likewise among $-t,-t+1$. The two selected B-coordinates force opposite inequalities on a candidate parameter s.

The set D is closed in the Euclidean topology of ${\mathbb{R}}^4$, being the solution set of

$$x_2=x_1+1,x_3=-x_1,x_4=-x_1+1.$$

Because the topology of $H\left(A\right)$ refines that of $\mathbb{R}$, the product topology on $H{\left(A\right)}^4$ refines the Euclidean one. The Euclidean-open set ${\mathbb{R}}^4\setminus D$ is open in $H{\left(A\right)}^4$, and D is closed there as well.

Fix $t\in \mathbb{R}$. Since A meets each $\mathbb{Q}$-coset in at most one point, neither $\{t,t+1\}$ nor $\{-t,-t+1\}$ lies wholly in A. Each pair contains a point of B.

Choose a coordinate among the first two entries of $d_t$ which belongs to B, and choose a coordinate among the last two entries of $d_t$ which belongs to B. In the two selected coordinates, take right-hand basic neighbourhoods of length one. In the remaining coordinates take the whole space $H\left(A\right)$. This gives an open neighbourhood $U_t$ of $d_t$ in $H{\left(A\right)}^4$.

Suppose $d_s\in U_t$. The selected first-pair coordinate gives $s\ge t$: from the first coordinate this is direct, and from the second coordinate it is $s+1\ge t+1$. The selected last-pair coordinate gives $s\le t$, since either $-s\ge -t$ or $-s+1\ge -t+1$. Together these force $s=t$. It follows that $U_t\cap D=\left\{d_t\right\}$, and $d_t$ is isolated in D.

The set D is a closed discrete subspace of $H{\left(A\right)}^4$ of cardinality $\mathfrak{c}$. If $H{\left(A\right)}^4$ were Lindelöf, its closed subspace D would be Lindelöf too, which is impossible for an uncountable discrete space. This proves that $H{\left(A\right)}^4$ is not Lindelöf. □

Theorem 2.

There is a set $A\subseteq \mathbb{R}$ such that, putting $B=\mathbb{R}\setminus A$, the set B has cardinality $\mathfrak{c}$ and contains no subspace homeomorphic to $2^{\omega }$, while $H{\left(A\right)}^n$ is not Lindelöf for every integer $n\ge 4$. Consequently, $H{\left(A\right)}^2$ is Lindelöf, but $H{\left(A\right)}^{\omega }$ is not Lindelöf.

Proof. 

Choose A as in Lemma 2. Then B has cardinality $\mathfrak{c}$ and contains no subspace homeomorphic to $2^{\omega }$, so $H{\left(A\right)}^2$ is Lindelöf by Theorem 1. The same lemma gives that $H{\left(A\right)}^4$ contains a closed discrete subspace of cardinality $\mathfrak{c}$, so $H{\left(A\right)}^4$ is not Lindelöf.

If $H{\left(A\right)}^n$ were Lindelöf for some $n\ge 4$, its projection onto the first four coordinates would make $H{\left(A\right)}^4$ Lindelöf, a contradiction. It follows that $H{\left(A\right)}^n$ is not Lindelöf for every $n\ge 4$. The same projection rules out Lindelöfness of $H{\left(A\right)}^{\omega }$. □

The preceding theorem gives an obstruction above the square, but it leaves open whether such an obstruction can already appear in the cube. The remaining implication is the following question.

Question 3.

Let $A\subseteq \mathbb{R}$, and put $B=\mathbb{R}\setminus A$. Suppose that B contains no subspace homeomorphic to $2^{\omega }$, or equivalently that $H{\left(A\right)}^2$ is Lindelöf. Must $H{\left(A\right)}^3$ be Lindelöf? Equivalently, can $H{\left(A\right)}^2$ be Lindelöf while $H{\left(A\right)}^3$ is not Lindelöf?

Theorem 2 shows that Lindelöfness of $H{\left(A\right)}^2$ need not pass to the fourth power, to every finite power $n\ge 4$, or to the countable power. The cube in Question 3 is the only finite-power implication left undecided by this example.