Abstract.

Semitopological groups $G$ and $H$ are said to be $\mathrm{𝐻𝑀}$-equivalent if the Hartman-Mycielski extensions $G^{\bullet }$ and $H^{\bullet }$ of $G$ and $H$, respectively, are topologically isomorphic. It is shown that $G$ and $G^k$ are $\mathrm{𝐻𝑀}$-equivalent, for every semitopological group $G$ and an integer $k\ge 1$. We also show that if $G$ and $H$ are semitopological groups and $G$ is topologically isomorphic to a subgroup of a finite power of $H$, then $G^{\bullet }$ admits a topological monomorphism to $H^{\bullet }$.

It is established that the $\mathrm{𝐻𝑀}$-equivalence relation preserves a variety of properties, especially those expressed in terms of cardinal functions. On the other hand, we show under some extra set-theoretic assumptions that the cellularity, Lindelöf property, countable compactness, tightness, Fréchet-Urysohn property, etc., are not preserved by the $\mathrm{𝐻𝑀}$-equivalence in the class of topological groups.

1. Introduction

In [8], S. Hartman and J. Mycielski proved that every topological group $G$ is topologically isomorphic to a closed subgroup of a pathwise connected, locally pathwise connected topological group, denoted by $G^{\bullet }$. A detailed analysis of the Hartman-Mycielski construction shows that the topological groups $G$ and $G^{\bullet }$ share several topological properties, such as metrizability, separability, $\sigma$-compactness, first and second countability (see [1, Section 3.8]). Moreover, if $G$ is abelian, divisible, torsion, or torsion-free, then $G^{\bullet }$ inherits the same property. On the other hand, $G^{\bullet }$ is never compact, locally compact, countably compact, or even precompact, except in the trivial case where $|G|=1$.

Assuming that $G$ is a metrizable topological group, M. Bruguera et al. describe the Raĭkov completion of $G^{\bullet }$ in [5]. Furthermore, in [12], M. López and I. Sánchez characterize the topological groups $G$ such that $G^{\bullet }$ is Lindelöf, and show that $G^{\bullet }$ is Čech-complete if and only if $|G|=1$.

For additional information on the topological groups of the form $G^{\bullet }$, the reader may consult [1, 8]. We only mention that the Hartman-Mycielski construction has a functorial nature [1, Section 3.8].

In this paper, we consider the cases of a semitopological or topological group $G$, so the corresponding Hartman-Mycielski extension $G^{\bullet }$ of $G$ is also a semitopological or topological group. As usual, a semitopological group is a group equipped with a topology in which the left and right translations are continuous. If multiplication in a semitopological group is jointly continuous, we say that it is a paratopological group. Adding the continuity of inversion we obtain the definition of a topological group. A detailed overview of these classes of groups is available in [1].

This article’s aim is different from those in [5, 12]. Let us say that semitopological groups $G$ and $H$ are $\mathrm{𝐻𝑀}$-equivalent if the groups $G^{\bullet }$ and $H^{\bullet }$ are topologically isomorphic. We are interested in identifying (topological) properties that are preserved by the $\mathrm{𝐻𝑀}$-equivalence relation in the classes of topological, paratopological or semitopological groups (see Problem 3.2).

We start in Section 2 by reviewing certain relevant aspects of the Hartman-Mycielski construction. It is established in the instrumental Proposition 2.5 that $G$ and $G^k$ are $\mathrm{𝐻𝑀}$-equivalent for any semitopological group $G$ and an integer $k\ge 1$. Therefore, if the $\mathrm{𝐻𝑀}$-equivalence preserves a property $𝒫$, this property has to be finitely productive. Actually, we consider a more general situation, where $G^{\bullet }$ admits a topological monomorphism to $H^{\bullet }$. We complement Proposition 2.5 in Corollary 2.8 by proving that if $G$ is a subgroup of a finite power of a semitopological group $H$, then $G^{\bullet }$ admits a topological monomorphism to $H^{\bullet }$.

In Theorem 2.10, we provide a list of properties that $G^{\bullet }$ directly inherits from $G$, showing that $G^{\bullet }$ is a special extension of the group $G$. In fact, we consider the more general case of the Hartman-Mycielski extension $X^{\bullet }$ of a space $X$ in the theorem. In doing so, we follow [4, 2].

It is shown in Corollary 2.14 that the $\mathrm{𝐻𝑀}$-equivalence preserves metrizability, submetrizability, weight, $\pi$-weight, character, pseudocharacter, $\pi$-character, network weight, density and the index of narrowness in the class of semitopological groups. It is not surprising that all of the properties in question are finitely (in fact, countably) productive.

Conversely, we prove in Corollary 2.18 that the $\mathrm{𝐻𝑀}$-equivalence does not preserve countable cellularity, countable compactness, Lindelöfness, countable tightness, hereditary separability or the Fréchet-Urysohn property in the class of Hausdorff topological groups. These conclusions depend on additional set-theoretic assumptions, except for the Lindelöf property.

Several open problems with brief comments are collected in Section 3.

When using general topology and topological algebra terminology, we adhere to [6] and [1], respectively.

2. $\mathrm{𝐻𝑀}$-equivalence in semitopological groups

In [8], S. Hartman and J. Mycielski presented a universal construction that assigns to every topological group $G$ a pathwise connected, locally pathwise connected topological group $G^{\bullet }$ containing $G$ as a closed subgroup. Afterwards, R. Brown and S. Morris [4] and K. Bicknell [2] applied this construction to an arbitrary (Hausdorff) space $X$. In fact, if $X$ has an additional algebraic structure compatible with its topology (such as topological semigroup, monoid, group, or groupoid), then so does $X^{\bullet }$ (see [4]).

Let $X$ be a space. Consider the set $X^{\bullet }$ of all functions $f$ from $[0,1)=J$ to $X$ such that for some finite sequence , the function $f$ is constant on the half-open interval $[a_k,a_{k+1})$, for $k=0,1,\mathrm{\dots },n$. We endow $X^{\bullet }$ with a topology as follows. Given real numbers $a,b$ with , a nonempty open $V$ in $X$, and a real number $ϵ>0$, define a subbasic open set $O(a,b,V,ϵ)$ in $X^{\bullet }$ by

(1)

where $\mu$ is Lebesgue measure on the real line. A direct verification shows that the sets $O(a,b,V,ϵ)$ form a subbase for a topology on $X^{\bullet }$, and if $X$ is Hausdorff or Tychonoff, then so is $X^{\bullet }$ (see Propositions 3 and 6 in [4]).

Assume that $G$ is a semitopological (paratopological, topological) group with identity element $e$. Define a binary operation $\ast$ on $G^{\bullet }$ by $(f\ast g)(x)=f(x)g(x)$, for all $f,g\in G^{\bullet }$ and $x\in J$. Then every element $f\in G^{\bullet }$ has a unique inverse $f^{-1}\in G^{\bullet }$ given by $(f^{-1})(x)={(f(x))}^{-1}$, for each $x\in J$. It is easy to see that $(G^{\bullet },\ast )$ is a group with identity $e^{\bullet }$, where $e^{\bullet }(x)=e$ for each $x\in J$. It can be shown that the sets

(2)

form a neighborhood base at the identity $e^{\bullet }$ for a semitopological (paratopological, topological) group topology on $G^{\bullet }$. In what follows we omit the symbol $\ast$ for multiplication in $G^{\bullet }$. A routine verification demonstrates that the two topologies on $G^{\bullet }$ presented in (1) and (2) coincide for a semitopological group $G$ when treated as a topological space.

In this and subsequent sections, $G^{\bullet }$ (resp., $X^{\bullet }$) always refers to the Hartman–Mycielski extension of a given group $G$ (resp., space $X$).

Important properties of $X^{\bullet }$ can be found in the following two results.

Proposition 2.1 (See Theorem 1 in [4]).

The space $X^{\bullet }$ is pathwise connected and locally pathwise connected, for every space $X$.

Actually, Theorem 1 in [4] is more general than Proposition 2.1. It states that $X^{\bullet }$ is contractible and locally contractible.

For each $x\in X$, let $x^{\bullet }$ be the element of $X^{\bullet }$ defined by $x^{\bullet }(r)=x$, for all $r\in J$. Let also $i_X:X\to X^{\bullet }$ be the mapping defined by $i_X(x)=x^{\bullet }$.

Theorem 2.2 (See Theorem 1.1 in [16]).

The mapping $i_X$ is a topological embedding of $X$ to $X^{\bullet }$. If $X$ is Hausdorff, then $i_X$ is a closed embedding. Furthermore, if $G$ is a (Hausdorff) semitopological group, then $i_G:G\to G^{\bullet }$ is a topological isomorphism of $G$ onto a (closed) subgroup of the group $G^{\bullet }$.

For the special case of a topological group $G$, the conclusion of Theorem 2.2 was established by Hartman and Mycielski in [8] (see also [1, Theorem 3.8.2]). Actually, the arguments in [8], [1] and [16] are nearly identical.

For a deeper study of the spaces $X^{\bullet }$, we need to introduce some notation. Each element $f\in X^{\bullet }$ is a step function from $J=[0,1)$ to $X$, so there exist real numbers such that $f$ is constant on each subinterval $[r_k,r_{k+1})$ for every $k$ with $0\le k\le n$ and $f(r_k)\ne f(r_{k+1})$ if . We consider the sets

$$P(f)=\{r_0,r_1,\mathrm{\dots },r_n\}\subset J,V(f)=\{f(r_0),\mathrm{\dots },f(r_n)\}\subset X$$ (3)

and put $\mathrm{\Delta }(f)=\min \{r_{k+1}-r_k:0\le k\le n\}$. Clearly, for every $x\in X$, the element $x^{\bullet }\in X^{\bullet }$ satisfies $P(x^{\bullet })=\{0\}$, $V(x^{\bullet })=\{x\}$ and $\mathrm{\Delta }(x^{\bullet })=1$.

Another important fact regarding the spaces of the form $X^{\bullet }$ is established in [4, Proposition 4]. It states that the natural mapping, say, $\phi$ of ${(X\times Y)}^{\bullet }$ to $X^{\bullet }\times Y^{\bullet }$ is a homeomorphism, for arbitrary spaces $X$ and $Y$. If $X$ and $Y$ are semitopological groups (topological monoids, etc.), then the mapping $\phi$ from [4] is a topological isomorphism. Given the importance of this fact for our objective in the article, we present a concise argument closely aligned with the one outlined in [4]. First, we require a simple result extending [1, Proposition 3.8.6] to semitopological groups.

Lemma 2.3.

Let $\phi :X\to Y$ be a continuous mapping of topological spaces. Then there exists a continuous mapping ${\phi }^{\bullet }$ of $X^{\bullet }$ to $Y^{\bullet }$ satisfying ${\phi }^{\bullet }\circ i_X=i_Y\circ \phi$, where $i_X$ and $i_Y$ are the respective canonical embeddings of $X$ to $X^{\bullet }$ and $Y$ to $Y^{\bullet }$. If $\phi$ is a topological embedding, then so is ${\phi }^{\bullet }$. Furthermore, if $X$ and $Y$ are semitopological groups and $\phi$ is a continuous homomorphism, then ${\phi }^{\bullet }$ is also a continuous homomorphism. Hence, if $\phi$ is a topological monomorphism, then so is ${\phi }^{\bullet }$.

Proof 2.4.

The statements of the lemma in the case of topological spaces $X$ and $Y$ follow directly from [4, Proposition 2]. So we assume that $X$ and $Y$ are semitopological groups and $\phi$ is a continuous homomorphism.

Define ${\phi }^{\bullet }:X^{\bullet }\to Y^{\bullet }$ by letting ${\phi }^{\bullet }(f)=\phi \circ f$, where $f\in X^{\bullet }$. For $f,g$ and $r\in J$, we have

$$({\phi }^{\bullet }(fg))(r)=(\phi \circ (fg))(r)=\phi (f(r)g(r))=\phi (f(r))\phi (g(r))=({\phi }^{\bullet }(f){\phi }^{\bullet }(g))(r).$$

Since the above equalities hold for each $r\in J$, it follows that ${\phi }^{\bullet }(fg)={\phi }^{\bullet }(f){\phi }^{\bullet }(g)$. It is also clear that ${\phi }^{\bullet }(e_X^{\bullet })=e_Y^{\bullet }$, where $e_X$ and $e_Y$ are identity elements of $X$ and $Y$, respectively. Hence, ${\phi }^{\bullet }$ is a homomorphism. The continuity of ${\phi }^{\bullet }$ follows from [4, Proposition 2]. It is easy to see that the equality ${\phi }^{\bullet }\circ i_X=i_Y\circ \phi$ holds as well.

If $\phi$ is a topological monomorphism, hence a topological embedding, we apply [4, Proposition 2] once more to conclude that ${\phi }^{\bullet }$ is also a topological embedding. Consequently, ${\phi }^{\bullet }$ is a topological monomorphism.

Identifying semitopological groups $X$ and $Y$ with their isomorphic images $i_X(X)\subset X^{\bullet }$ and $i_Y(Y)\subset Y^{\bullet }$, respectively, one can reformulate Lemma 2.3 by saying that $\phi$ extends to a continuous homomorphism of $X^{\bullet }$ to $Y^{\bullet }$.

Proposition 2.5.

Let $G$ and $H$ be arbitrary semitopological groups. Then there exists a natural topological isomorphism $\mathrm{\Phi }$ of ${(G\times H)}^{\bullet }$ onto $G^{\bullet }\times H^{\bullet }$. Therefore, for each integer $k\ge 1$, the groups $G^{\bullet }$ and ${(G^k)}^{\bullet }$ are topologically isomorphic.

Proof 2.6.

Let $p:G\times H\to G$ and $q:G\times H\to H$ be the projections. According to Lemma 2.3, the homomorphisms $p$ and $q$ admit extensions to continuous homomorphisms $p^{\bullet }:{(G\times H)}^{\bullet }\to G^{\bullet }$ and $q^{\bullet }:{(G\times H)}^{\bullet }\to H^{\bullet }$. Let $\mathrm{\Phi }$ be the diagonal of $p^{\bullet }$ and $q^{\bullet }$, $\mathrm{\Phi }:{(G\times H)}^{\bullet }\to G^{\bullet }\times H^{\bullet }$. Clearly, $\mathrm{\Phi }$ is a continuous homomorphism. It also follows from [4, Proposition 4] that $\mathrm{\Phi }$ is a homeomorphism. Therefore, $\mathrm{\Phi }$ is a topological isomorphism.

Taking $G=H$, we see that the groups ${(G^2)}^{\bullet }$ and ${(G^{\bullet })}^2$ are topologically isomorphic. Induction on $k$ shows that the groups ${(G^k)}^{\bullet }$ and ${(G^{\bullet })}^k$ are also topologically isomorphic for any integer $k\ge 1$. It remains to refer to [2, Theorem 2] implying that the groups ${(G^{\bullet })}^k$ and $G^{\bullet }$ are topologically isomorphic as well. [To be precise, the result in [2] states that ${(X^{\bullet })}^2$ and $X^{\bullet }$ are homeomorphic, for any space $X$. However, the homeomorphism constructed in [2] results in an isomorphism if $X$ is a semitopological group, and the square of $X^{\bullet }$ can be replaced with ${(X^{\bullet })}^k$, for any integer $k\ge 2$.]

The last statement of Proposition 2.5 can be given the following equivalent form.

Corollary 2.7.

The groups $G$ and $G^k$ are $\mathrm{𝐻𝑀}$-equivalent for every semitopological group $G$ and every integer $k\ge 1$.

Corollary 2.8.

Let $H$ be a semitopological group, $k\ge 1$ an integer, and $G$ be a subgroup of $H^k$. Then $G^{\bullet }$ admits a topological monomorphism to $H^{\bullet }$.

Proof 2.9.

It follows from Proposition 2.5 that the groups $H^{\bullet }$ and ${(H^k)}^{\bullet }$ are topologically isomorphic. Denote by $j$ the identity embedding of $G$ to $H^k$. By Lemma 2.3, $j$ extends to a topological monomorphism $j^{\bullet }:G^{\bullet }\to {(H^k)}^{\bullet }$. The groups ${(H^k)}^{\bullet }$ and $H^{\bullet }$ are topologically isomorphic according to Proposition 2.5. Consequently, $G^{\bullet }$ admits a topological monomorphism to $H^{\bullet }$.

The subsequent two theorems show that the spaces $X$ and $X^{\bullet }$ share many properties. We recall that a space is submetrizable if it admits a coarser metrizable topology. Every submetrizable space is Hausdorff and has countable pseudocharacter. Also, we say that a semitopological group $G$ is $\kappa$-narrow, where $\kappa \ge \omega$, if for every neighborhood $U$ of the identity in $G$, there exists a set $A\subset G$ with $|A|\le \kappa$ such that $AU=G=UA$ (see [1, Section 3.4]).

Theorem 2.10.

Let $\kappa$ be an infinite cardinal and $X$ be a space. If either space $X$ or $X^{\bullet }$ possesses any of the properties listed below, then the other space does as well:

1. metrizability and submetrizabilty;

2. having a base of cardinality $\le \kappa ;$

3. having a local base at every point of cardinality $\le \kappa ;$

4. having a local pseudo-base at every point of cardinality $\le \kappa ;$

5. having a network of cardinality $\le \kappa ;$

6. having a dense subset of cardinality $\le \kappa ;$

7. $\kappa$-narrowness, provided $X$ is a semitopological group.

Proof 2.11.

Part I. Assuming that $X$ has one of the properties in (a)–(g), we show that $X^{\bullet }$ retains the same property.

In the special case where $X$ is a topological group, the arguments regarding items (c) and (e)–(f) presented in [1, Theorem 3.8.8] are applicable to the space $X$ without any substantial modifications. The same is valid regarding item (g), since the arguments in [1] use only the continuity of translations in the group. Hence, it suffices to consider the properties in items (a), (b) and (d) of the theorem.

(a) Let $\varrho$ be a compatible metric on $X$. We can assume that $\varrho$ is bounded by $1$. We define a distance $d(f,g)$ between elements $f,g\in X^{\bullet }$ by the formula

$$d(f,g)={\int }_0^1\varrho (f(r),g(r))𝑑r.$$

According to [4, Proposition 5], $d$ is a compatible metric on $X^{\bullet }$ and the canonical embedding $i_X:X\to X^{\bullet }$ is an isometry. In particular, the space $X^{\bullet }$ is metrizable.

Assume that the space $X$ is submetrizable, and let $\tau$ be a coarser metrizable topology on $X$. Then the space $Y=(X,\tau )$ is metrizable, and so is $Y^{\bullet }$. Denote by $j$ the identity mapping of $X$ onto $Y$. By [8, Proposition 2], $j$ extends to a continuous mapping $j^{\bullet }$ of $X^{\bullet }$ onto $Y^{\bullet }$. Clearly, $j^{\bullet }$ is a bijection. As the space $Y^{\bullet }$ is metrizable, we conclude that $X^{\bullet }$ is submetrizable.

(b) We modify the argument presented in the proof of [1, Theorem 3.8.8 (d)]. Let $ℬ$ be a base for $X$ with $|ℬ|=\kappa$, where $\kappa =w(X)$. For every $m\in \mathbb{N}$, denote by $J(m)$ the set of all $m$-tuples $\overrightarrow{c}=(c_1,\mathrm{\dots },c_m)$ of rationals such that . If $\overrightarrow{c}=(c_1,\mathrm{\dots },c_m)\in J(m)$ and $\overrightarrow{d}=(d_1,\mathrm{\dots },d_m)\in J(m)$, we write if for each $k=1,\mathrm{\dots },m$. Given $m,n\in \mathbb{N}$, elements $\overrightarrow{c},\overrightarrow{d}\in J(m)$ with and $\overrightarrow{V}=(V_1,\mathrm{\dots },V_m)\in {ℬ}^m$, we define a subset $Q(m,n,\overrightarrow{c},\overrightarrow{d},\overrightarrow{V})$ of $X^{\bullet }$ as the set of all $g\in X^{\bullet }$ such that Lebesgue measure of the set of all $r\in J$ satisfying and $g(r)\notin V_k$ is less than $1/n$, for each $k=1,\mathrm{\dots },m$.

Claim. The sets $Q(m,n,\overrightarrow{c},\overrightarrow{d},\overrightarrow{V})$ are open in $X^{\bullet }$.

The required conclusion regarding the sets $Q(m,n,\overrightarrow{c},\overrightarrow{d},\overrightarrow{V})$ follows from the equality

$$Q(m,n,\overrightarrow{c},\overrightarrow{d},\overrightarrow{V})=\bigcap _{k=1}^{m}O(c_k,d_k,V_k,1/n),$$

where the sets $O(c_k,d_k,V_k,1/n)$ are defined in (1).

The cardinality of the family $𝒬$ of all sets $Q(m,n,\overrightarrow{c},\overrightarrow{d},\overrightarrow{V})$ with $m,n\in \mathbb{N}$, $\overrightarrow{c},\overrightarrow{d}\in J(m)$ and $\overrightarrow{V}\in {ℬ}^m$ is less than or equal to $\kappa$, and we claim that $𝒬$ is a base for $X^{\bullet }$.

Indeed, take an arbitrary element $f\in X^{\bullet }$ and a basic open neighborhood $O={\bigcap }_{k=1}^{m}O(a_k,b_k,U_k,{ϵ}_k)$ of $f$ in $X^{\bullet }$, where , $U_k$ is open in $X$, and ${ϵ}_k>0$ for each $k=1,\mathrm{\dots },m$. Diminishing the set $O$ and increasing the number $m$, if necessary, we can assume without loss of generality that $f$ is constant on $[a_k,b_k)$ and that for $k=1,\mathrm{\dots },m$. Then $f(a_k)\in U_k$. For every $k\le m$, choose an element $V_k\in ℬ$ such that $f(a_k)\in V_k\subset U_k$.

Take $n\in \mathbb{N}$ with and choose $\overrightarrow{c}=(c_1,\mathrm{\dots },c_m)\in J(m)$ and $\overrightarrow{d}=(d_1,\mathrm{\dots },d_m)\in J(m)$ such that and for each $k=1,\mathrm{\dots },m$. Since $f(r)=f(a_k)\in V_k\subset U_k$ for all $r\in [c_k,d_{k+1})$ and $k=1,\mathrm{\dots },m$, we see that $f\in Q(m,2n,\overrightarrow{c},\overrightarrow{d},\overrightarrow{V})$, where $\overrightarrow{V}=(V_1,\mathrm{\dots },V_m)\in {ℬ}^m$. Therefore, all we need to verify is that $Q(m,2n,\overrightarrow{c},\overrightarrow{d},\overrightarrow{V})\subset O$.

Take an arbitrary element $h\in Q(m,2n,\overrightarrow{c},\overrightarrow{d},\overrightarrow{V})$. It follows from the definition of $Q(m,2n,\overrightarrow{b},\overrightarrow{V})$ that for each $k\le m$, the set

satisfies . Also, our choice of $\overrightarrow{c}$ and $\overrightarrow{d}$ implies that for each $k\le m$, the measure of the complement $[a_k,b_k)\setminus [c_k,d_k)$ is less than $1/(2n)$. We see therefore that $h\in O(a_k,b_k,V_k,1/n)$ for each $k\le m$, whence it follows that $h\in {\bigcap }_{k=1}^{m}O(a_k,b_k,V_k,1/n)\subset {\bigcap }_{k=1}^{m}O(a_k,b_k,U_k,1/n)=O$. This proves the inclusion $Q(m,2n,\overrightarrow{c},\overrightarrow{d},\overrightarrow{V})\subset O$. Hence, the family $𝒬$ is a base for $X^{\bullet }$ and $w(X^{\bullet })\le \kappa$.

(d) Let $\kappa =\psi (X)$. Take any element $f\in X^{\bullet }$ and choose such that $f$ is constant on each half-open interval $[a_k,a_{k+1})$, $0\le k\le m$. For every $i\le m$, denote by ${ℬ}_i$ a pseudobase of $X$ at the point $x_i=f(a_i)$ satisfying $|{ℬ}_i|\le \kappa$. The family

$$\lambda =\{O(a_i,a_{i+1},V_i,1/n):0\le i\le m,n\in \mathbb{N},V_i\in {ℬ}_i\}$$

of open neighborhoods of $f$ in $X^{\bullet }$ satisfies $|\lambda |\le \kappa$. Let us verify that $\{f\}=\bigcap \lambda$.

Assume that $h\in \bigcap \lambda$. It follows from the definition of the sets $O(a_i,a_{i+1},V_i,1/n)$ that for all $i\le m$ and $V_i$, every element $g\in {\bigcap }_{n\in \mathbb{N}}O(a_i,a_{i+1},V_i,1/n)$ satisfies $g(r)\in V_i$ for each $r\in [a_i,a_{i+1})$. Therefore, our choice of $h$ implies that $h(r)=f(a_k)$ for each $r\in [a_i,a_{i+1})$, where $i=0,1,\mathrm{\dots },m$. Hence, $h=f$. We conclude that $\psi (X^{\bullet })\le |\lambda |\le |\gamma |=\psi (X)$.

Part II. Conversely, assuming that $X^{\bullet }$ possesses one of the properties in (a)–(g), we verify that $X$ has the same property. We only verify (f) and (g) because $X$ admits a topological embedding into $X^{\bullet }$ and the properties in items (a)–(e) are hereditary with regard to taking subspaces.

(f) Let $D$ be a dense subset of $X^{\bullet }$ with $|D|=\kappa$, where $\kappa =d(X^{\bullet })$. Then the subset $S=\bigcup \{V(f):f\in D\}$ of $X$ satisfies $|S|\le \kappa$ (see (3) for the definition of the set $V(f)$). Let us verify that $S$ is dense in $X$. Suppose that $U$ is a nonempty open set in $X$. Take an arbitrary element $x\in U$. Then $O=O(0,1,U,1)$ is an open neighborhood of $x^{\bullet }$ in $X^{\bullet }$. Take an element $f\in D\cap O$. Then the measure of the set $\{r\in J:f(r)\notin U\}$ is less than $1$. Take $r\in J$ such that $f(r)\in U$. Clearly, there exist elements $a,b\in P(f)$ such that and $f$ is constant on $[a,b)$. Then $f(a)=f(r)$, and the definition of $V(f)$ in $(\text{<a href="\#S2.E3" title="In 2. 𝐻𝑀-equivalence in semitopological groups ‣ The Hartman–Mycielski construction and topological monomorphisms" class="ltx\_ref"><span class="ltx\_text ltx\_ref\_tag">3</span></a>})$ implies that $f(r)\in V(f)$. Since $f\in D$, we conclude that $f(r)\in S\cap U\ne \mathrm{\varnothing }$. This proves that $S$ is dense in $X$, where $|S|\le \kappa$. Hence, $d(X)\le d(X^{\bullet })$.

(g) Assume that $X$ is a semitopological group and that the group $X^{\bullet }$ is $\kappa$-narrow, where $\kappa \ge \omega$. If $X$ were a topological group, it would be possible to deduce that the groups $i_X(X)\subset X^{\bullet }$ and $X$ are also $\kappa$-narrow by applying [1, Proposition 5.1.1 (a)]. However, in paratopological and semitopological groups, $\kappa$-narrowness is not inherited by subgroups. This is why we provide the following argument.

Let $U$ be an open neighborhood of the identity $e$ in $X$. Then $O=O(U,1)$ is an open neighborhood of $e^{\bullet }$ in $X^{\bullet }$ (see (2)), so there exists a set $A\subset X^{\bullet }$ with $|A|\le \kappa$ such that $AO=X^{\bullet }=OA$. Then the set $B=\bigcup \{V(f):f\in A\}$ satisfies $|B|\le \kappa$, and we claim that the equalities $BU=X=UB$ hold.

Indeed, take an arbitrary element $x\in X$. Then there exist elements $f,g\in A$ such that $x^{\bullet }\in fO\cap Og$, whence it follows that $f^{-1}{x}^{\bullet }\in O$ and $x^{\bullet }{g}^{-1}\in O$. The former inclusion implies that the measure of the set $\{r\in J:f^{-1}{x}^{\bullet }\notin U\}$ is less than $1$, so there exists $s\in J$ such that $f{(s)}^{-1}x\in U$, while the latter inclusion implies that $xg{(t)}^{-1}\in U$ for some $t\in J$. Hence, $x\in f(s)U\cap Ug(t)$. Arguing as in (f), we see that $f(s)\in V(f)$ and $g(t)\in V(g)$. Since $f,g\in A$, it follows that $f(s)\in B$ and $g(t)\in B$. Therefore, $x\in BU\cap UB$, which implies the equalities $BU=X=UB$. This concludes the proof of (f) and, as a result, the theorem.

Items (b)–(g) of Theorem 2.10 can be reformulated as follows:

Corollary 2.12.

The equalities $w(X^{\bullet })=w(X)$, $\chi (X^{\bullet })=\chi (X)$, $\psi (X^{\bullet })=\psi (X)$ (provided $X$ is a $T_1$-space), $nw(X^{\bullet })=nw(X)$ and $d(X^{\bullet })=d(X)$ are valid for every space $X$. If $G$ is a semitopological group, then $in(G^{\bullet })=in(G);$ furthermore, in this case, the equalities $\pi \chi (G^{\bullet })=\pi \chi (G)$ and $\pi w(G^{\bullet })=\pi w(G)$ hold.

Proof 2.13.

A proof is necessary only for the last two equalities, where $G$ is a semitopological group. We show first that $\pi \chi (G^{\bullet })\le \pi \chi (G)$. Let $\mu$ be a $\pi$-base at the identity $e$ of the group $G$. We claim that the family

$$\nu =\{O(U,1/n):U\in \mu ,n\in \mathbb{N}\}$$

is a $\pi$-base at the identity $e^{\bullet }$ of the group $G^{\bullet }$. Clearly, the elements of $\nu$ are nonempty open sets in $G^{\bullet }$. Let $O(V,ϵ)$ be an open neighborhood of the identity in $G^{\bullet }$, where $V$ is an open neighborhood of $e$ in $G$ and $ϵ>0$. Take $n\in \mathbb{N}$ such that $1/n\le ϵ$ and choose $U\in \mu$ such that $U\subset V$. An easy verification shows that $O(U,1/n)\subset O(V,ϵ)$, thus implying our claim. Since $|\nu |\le |\mu |\cdot \omega =|\mu |$, we infer that $\pi \chi (G^{\bullet })\le \pi \chi (G)$.

Conversely, let $\nu$ be a $\pi$-base at the identity $e^{\bullet }$ of the group $G^{\bullet }$. We can assume without loss of generality that $\nu =\{f_{\alpha }O(U_{\alpha },{ϵ}_{\alpha }):\alpha \in A\}$, where $f_{\alpha }\in G^{\bullet }$, $U_{\alpha }$ is an open neighborhood of $e$ in $G$ and ${ϵ}_{\alpha }>0$ for each $\alpha \in A$. Let us verify that the family

$$\mu =\{x{U}_{\alpha }:x\in V(f_{\alpha }),\alpha \in A\}$$

is a $\pi$-base at the identity of $G$. Take an arbitrary open neighborhood $W$ of $e$ in $G$. There exists $\alpha \in A$ such that $f_{\alpha }O(U_{\alpha },{ϵ}_{\alpha })\subset O(W,1)$. Let $P(f_{\alpha })=\{r_0,r_1,\mathrm{\dots },r_n\}$, where . If $x{U}_{\alpha }\setminus W\ne \mathrm{\varnothing }$ for each $x\in V(f_{\alpha })$, we choose $y_x\in x{U}_{\alpha }\setminus W$ for each $x\in V(f_{\alpha })$ and define an element $g\in f_{\alpha }O(U_{\alpha },{ϵ}_{\alpha })$ by letting $P(g)=P(f_{\alpha })$ and $g(r_i)=y_{x_i}$ for $i=0,1,\mathrm{\dots },n$, where $x_i=f_{\alpha }(r_i)$. It follows from the inclusions $y_{x_i}\in x_i{U}_{\alpha }$ and the definition of $g$ that $g\in f_{\alpha }O(U_{\alpha },{ϵ}_{\alpha })$. Also, our choice $y_{x_i}\notin W$ for each $i\le n$ implies that $g(r)\notin W$ for each $r\in J$, that is, $g\notin O(W,1)$. This contradicts the inclusion $f_{\alpha }O(U_{\alpha },{ϵ}_{\alpha })\subset O(W,1)$. Therefore, $x{U}_{\alpha }\subset W$ for some $x\in V(f_{\alpha })$, thus implying that $\mu$ is a $\pi$-base at the identity of the group $G$. Since $|\mu |\le |\nu |\cdot \omega$, it follows that $\pi \chi (G)\le \pi \chi (G^{\bullet })$. Combining the two inequalities, we conclude that $\pi \chi (G^{\bullet })=\pi \chi (G)$.

Finally, since $\pi w(X)=\pi \chi (X)\cdot d(X)$, for every space $X$ (see [18] or [10, 2.1(f)]), the last equality of the theorem follows from the equalities $\pi \chi (G^{\bullet })=\pi \chi (G)$ and $d(G^{\bullet })=d(G)$.

Theorem 2.10 and Corollary 2.12 imply the subsequent result concerning $\mathrm{𝐻𝑀}$-equivalence.

Corollary 2.14.

The relation of $\mathrm{𝐻𝑀}$-equivalence preserves metrizability, submetrizability, weight, $\pi$-weight, character, pseudocharacter, $\pi$-character, network weight, density and the index of narrowness in the class of semitopological groups.

The following theorem, which is similar to [1, Theorem 3.8.9], characterizes $\sigma$-compactness in the spaces of the form $X^{\bullet }$. The argument from [1] requires some changes.

Theorem 2.15.

Let $X$ be a Hausdorff space. Then $X^{\bullet }$ is $\sigma$-compact if and only if $X$ is $\sigma$-compact.

Proof 2.16.

Let $I$ be the closed unit segment with usual interval topology. For every $n,m\in \mathbb{N}$, let

and

$$A_{n,m}=\{(a_0,a_1,\mathrm{\dots },a_n)\in A_n:a_{k+1}-a_k\ge 1/m\text{ for each }k=0,\mathrm{\dots },n\},$$

where, as usual, $a_{n+1}=1$. It is clear that $A_n={\bigcup }_{m=1}^{\mathrm{\infty }}{A}_{n,m}$ and that each set $A_{n,m}$ is closed in $I^{n+1}$. Hence, the sets $A_{n,m}$ are compact.

For every integer $n\ge 1$, define a mapping ${\phi }_n:A_n\times X^{n+1}\to X^{\bullet }$ by the rule ${\phi }_n(a_0,\mathrm{\dots },a_n,x_0,\mathrm{\dots },x_n)=f$, where $f\in X^{\bullet }$ is constant on $[a_k,a_{k+1})$ and $f(a_k)=x_k$ for each $k\le n$.

Claim. The mapping ${\phi }_n:A_n\times X^{n+1}\to X^{\bullet }$ is continuous.

Indeed, take $p=(a_0,\mathrm{\dots },a_n,x_0,\mathrm{\dots },x_n)\in A_n\times X^{n+1}$ and put $f={\phi }_n(p)$. Choose an integer $m\ge 2$ such that $a_{k+1}-a_k\ge 1/m$ for each $i=0,1,\mathrm{\dots },n$. Then $(a_0,a_1,\mathrm{\dots },a_n)\in A_{n,m}$. Consider a subbasic open neighborhood $O(a,b,V,ϵ)$ of $f$ in $X^{\bullet }$, where , the set $V$ is open in $X$, and $ϵ>0$. Then the measure of the set $D=\{r\in [a,b):f(r)\notin V\}$ is less than $ϵ$.

Choose a positive real number $\delta$ satisfying and . Let $i$ be an integer, $0\le i\le n$. If $x_i\in V$, we assign $V_i=V$; otherwise, $V_i=X$. Further, we define a neighborhood $W$ of $p$ in ${ℝ}^{n+1}\times X^{n+1}$ by

$$W=(a_0-\delta ,a_0+\delta )\times \mathrm{\cdots }\times (a_n-\delta ,a_n+\delta )\times V_0\times \mathrm{\cdots }\times V_n.$$

Let us show that ${\phi }_n(W^{\ast })\subset O(a,b,V,ϵ)$, where $W^{\ast }=W\cap (A_n\times X^{n+1})$ is an open neighborhood of $p$ in $A_n\times X^{n+1}$.

Take an arbitrary element $q\in W^{\ast }$. Since , the family $\{(a_i-\delta ,a_i+\delta ):0\le i\le n\}$ is pairwise disjoint. Hence, it follows from the definition of $W$ that $q=(b_0,\mathrm{\dots },b_n,y_0,\mathrm{\dots },y_n)$, where $(b_0,\mathrm{\dots },b_n)\in A_n$, $b_i\in (a_i-\delta ,a_i+\delta )$ for each $i=0,\mathrm{\dots },n$ and $(y_0,\mathrm{\dots },y_n)\in V_0\times \mathrm{\cdots }\times V_n$. Set $h={\phi }_n(q)$. Our definition of the mapping ${\phi }_n$ implies that $h(r)=y_i$ provided that for some integer $i=0,1,\mathrm{\dots },n$ (again, we assume that $b_{n+1}=1$). Therefore, it follows from the choice of the set $W$ that $h(b_i)=y_i\in V_i=V$ provided that $x_i\in V$. In this case, $h(r)=h(b_i)\in V$ for each $r\in [b_i,b_{i+1})$.

Assume that $r\in [a,b)$ and $h(r)\notin V$. Take an integer $i\ge 0$ such that . As we have just demonstrated, our assumption implies that $x_i\notin V$. It follows from and that $r\in (a_i-\delta ,a_{i+1}+\delta )$. We see therefore that

$$E=\{r\in [a,b):h(r)\notin V\}\subset \bigcup _{i\in M}(a_i-\delta ,a_{i+1}+\delta ),$$ (4)

where $M$ is the set of integers $i\in \{0,1,\mathrm{\dots },n\}$ such that $x_i\notin V$. Notice that

$$D=\{r\in [a,b):f(r)\notin V\}=\bigcup \{[a_i,a_{i+1}):i\in M\}.$$ (5)

It follows from (4), (5) and our choice of $\delta$ that . According to the definition of $E$ in (4), the latter inequality implies that $h\in O(a,b,V,ϵ)$. We have thus proved that ${\phi }_n(W^{\ast })\subset O(a,b,V,ϵ)$. Since $O(a,b,V,ϵ)$ is an arbitrary subbasic neighborhood of ${\phi }_n(f)$ in $X^{\bullet }$, we infer that the mapping ${\phi }_n$ is continuous.

Assume that $X$ is $\sigma$-compact. Since $A_{n,m}$ is compact, the product $A_{n,m}\times X^{n+1}$ is $\sigma$-compact. The continuity of ${\phi }_n$ on $A_n\times X^{n+1}$ implies that the image ${\phi }_n(A_{n,m}\times X^{n+1})$ is also $\sigma$-compact for all integers $n,m\ge 1$. The equality $A_n={\bigcup }_{m=1}^{\mathrm{\infty }}{A}_{n,m}$, implies that the image ${\phi }_n(A_n\times X^{n+1})$ is $\sigma$-compact as well. Hence, the $\sigma$-compactness of $X^{\bullet }$ follows from the equality $X^{\bullet }={\bigcup }_{n=1}^{\mathrm{\infty }}{\phi }_n(A_n\times X^{n+1})$.

The converse implication follows from the fact that the homeomorphic copy $i_X(X)$ of $X$ is closed in $X^{\bullet }$.

The characterization of the Lindelöf property in $G^{\bullet }$ given in [12, Theorem 2.8] for a topological group $G$, as well as its proof, remain unaffected in the more general case of topological spaces.

Theorem 2.17.

Let $X$ be a Hausdorff space. Then $X^{\bullet }$ is Lindelöf if and only if $X^n$ is Lindelöf for each integer $n\ge 1$.

It is worth noting a difference between Theorems 2.15 and 2.17. In the latter one, the requirement on the space $X$ is stronger than the conclusion about $X^{\bullet }$. The explanation for that is fairly straightforward. According to [2, Theorem 2], the spaces $X^{\bullet }$ and ${(X^{\bullet })}^2$ are homeomorphic for each space $X$. Hence, by induction, one concludes that $X^{\bullet }$ and ${(X^{\bullet })}^k$ are also homeomorphic for each positive integer $k$. Since a Hausdorff space $X$ embeds as a closed subspace of $X^{\bullet }$ (so every finite power of $X$ is a closed subspace of $X^{\bullet }$), the Lindelöf property of $X^{\bullet }$ in Theorem 2.17 implies that all finite powers of $X$ are Lindelöf as well.

Corollary 2.7 allows us to show that the $\mathrm{𝐻𝑀}$-equivalence relation does not preserve various topological properties.

Corollary 2.18.

The following properties are not preserved by $\mathrm{𝐻𝑀}$-equivalence in the class of Hausdorff topological groups:

1. countable cellularity (adding a Cohen real);

2. countable compactness (under $\mathrm{𝐶𝐻}$);

3. Lindelöfness;

4. countable tightness (under $\mathrm{𝐶𝐻}$);

5. hereditary separability (under $\mathrm{𝐶𝐻}$);

6. the Fréchet-Urysohn property (adding a Cohen real).

Proof 2.19.

We know, as a result of Corollary 2.7, that every semitopological group $G$ is $\mathrm{𝐻𝑀}$-equivalent to $G^2$. Consequently, in order to deduce the corollary’s statement, it is sufficient to find a Hausdorff topological group $G_k$ for each $k=1,\mathrm{\dots },6$ that possesses the property in item (k) and such that $G_k^2$ does not possess the same property. A topological group $G_1$ of countable cellularity whose square has uncountable cellularity is constructed by V. I. Malykhin in [14] by adding a Cohen real number to an arbitrary model of $\mathrm{𝑍𝐹𝐶}$. A countably compact topological group $G_2$ whose square fails to be countably compact can be found in [7] or [13]. In both cases, the Continuum Hypothesis is used to construct the groups. The group $H$ from [13] is additionally hereditarily separable, but its square has uncountable tightness; hence, one can take $G_4=H$ and $G_5=H$. A Lindelöf topological group with non-Lindelöf square is constructed by Y. Peng and L. Wu in [15] without extra set-theoretic assumptions. Finally, according to [14, Corollary 2], adding a Cohen real to a model of $\mathrm{𝑍𝐹𝐶}$ forces the existence of a Fréchet-Urysohn topological group $G_6$ such that its square has uncountable tightness.

Remark 2.20.

Under Martin’s Axiom and the negation of $\mathrm{CH}$, the product of any family of countably cellular spaces is also countably cellular [11, Theorem 2.24]. This fact is used in [17] to show that, under the same set-theoretic assumptions, $\mathrm{𝐻𝑀}$-equivalence preserves countable cellularity in the class of semitopological groups. Combining this with item (1) of Corollary 2.18, we conclude that the statement “$\mathrm{𝐻𝑀}$-equivalence preserves countable cellularity in semitopological (or topological) groups” does not depend on $\mathrm{𝑍𝐹𝐶}$.

While the non-preservation of countable compactness under $\mathrm{𝐻𝑀}$-equivalence in item (2) of Corollary 2.18 could be a $\mathrm{𝑍𝐹𝐶}$ result, there is no $\mathrm{𝑍𝐹𝐶}$ example of a countably compact topological group whose square fails to be countably compact. Regarding items (4) through (6) of the corollary, a similar observation can be made. No topological group $G$ with a property in (4), (5), or (6) that does not have the same feature when squared is known to exist. Consequently, we are left with the possibility that some of the non-preservation results in (4)–(6) may be obtained in $\mathrm{𝑍𝐹𝐶}$.

3. Open Problems

According to Proposition 2.7, a semitopological group $G$ is $\mathrm{𝐻𝑀}$-equivalent to $G^k$, for each integer $k\ge 1$. This motivates the following two problems.

Problem 3.1.

Let $G$ and $H$ be topological (paratopological, semitopological) groups. Find general conditions on $G$ and $H$ guaranteeing that the groups $G^{\bullet }$ and $H^{\bullet }$ are topologically isomorphic.

Let us recall that a paratopological group $K$ is said to be $2$-pseudocompact if the intersection ${\bigcap }_{n\in \omega }\overline{U_n^{-1}}$ is nonempty, for every decreasing sequence $\{U_n:n\in \omega \}$ of nonempty open sets in $K$. According to [3, Proposition 3.13], every $2$-pseudocompact paratopological group is feebly compact. The next problem regarding $\mathrm{𝐻𝑀}$-equivalence contains $3\cdot 8+1=25$ (possibly) different subproblems.

Problem 3.2.

Let $G$ and $H$ be topological (paratopological, semitopological) groups, and suppose that $G^{\bullet }$ and $H^{\bullet }$ are topologically isomorphic. If the group $G$ has one of the following properties:

1. compactness (for Hausdorff $G$ and $H$);

2. local compactness (for Hausdorff $G$ and $H$);

3. pseudocompactness (for completely regular $G$ and $H$);

4. feeble compactness;

5. $2$-pseudocompactness;

6. precompactness;

7. Čech-completeness (for completely regular $G$ and $H$),

8. Raĭkov-completeness (assuming $G$ and $H$ are Hausdorff topological groups);

9. Dieudonné completeness (for completely regular $G$ and $H$),

must the group $H$ have the same property?

The properties in items (a)–(c) of the following problem are not finitely productive in topological groups. This does not imply, however, that the $\mathrm{𝐻𝑀}$-equivalence fails to preserve (some of) them:

Problem 3.3.

Does the relation of $\mathrm{𝐻𝑀}$-equivalence preserve

1. $o$-tightness (see [1, Section 5.5]);

2. sequentiality;

3. being a $k$-space

in topological (paratopological, semitopological) groups?

By [9, Theorem 9.11], every closed subgroup of ${ℝ}^n$, with $n\in \mathbb{N}$, is topologically isomorphic to ${ℝ}^k\times {\mathbb{Z}}^l$, where $k$ and $l$ are non-negative integers and $k+l\le n$. This, along with Corollary 2.8, gives rise to the following

Problem 3.4.

Let $K$ be a locally compact topological group and assume that the group $K^{\bullet }$ is topologically isomorphic to a subgroup of  ${ℝ}^{\bullet }$. Is $K$ topologically isomorphic to the product ${ℝ}^k\times {\mathbb{Z}}^l$, where $k$ and $l$ are non-negative integers?