Complete Invariant ⋆-Metrics on Semigroups and Groups

Abstract

In this paper, we study the complete ⋆-metric semigroups and groups and the Raǐkov completion of invariant ⋆-metric groups. We obtain the following. (1) Let $(X,d^{\star })$ be a complete ⋆-metric space containing a semigroup (group) G that is a dense subset of X. If the restriction of $d^{\star }$ on G is invariant, then X can become a semigroup (group) containing G as a subgroup, and $d^{\star }$ is invariant on X. (2) Let $(G,d^{\star })$ be a ⋆-metric group such that $d^{\star }$ is invariant on G. Then, $(G,d^{\star })$ is complete if and only if $(G,{\tau }_{d^{\star }})$ is Raǐkov complete.

1. Introduction

The combination of topological structure and algebraic structure is a very useful tool in modern mathematics research. This has stimulated the research enthusiasm of scholars (see [1,2]). In 1975, Kramosil and Michalek introduced a notion of metric fuzziness [3], which quickly became a hot topic of scholars (see [4,5,6]). In 2001, Romaguera and Sanchis introduced and studied the concept of the fuzzy metric group, extended the classical theorem of the metric group to the fuzzy metric group, and studied the properties of quotient subgroups of a fuzzy metric group [7]. Then, Sánchez and Sanchis found a sufficient condition for topological algebraic structures to become stronger topological structures [8].

Recently, Sánchez and Sanchis studied complete invariant fuzzy metrics on groups [9]. They proved that:

Theorem 1.

If $(G,M,\ast )$ is a fuzzy metric group such that $(M,\ast )$ is invariant, then a fuzzy metric completion $(\tilde{G},\tilde{M},\ast )$ of $(G,M,\ast )$ is a fuzzy metric group and $(\tilde{M},\ast )$ is invariant.

Theorem 2.

If $(M,\ast )$ is a complete fuzzy metric on $(G,\tau )$, then every compatible left invariant (or right invariant) fuzzy metric on G is complete.

At the same time, J.J. Tu and L.H. Xie further conducted further research on [9] and found that Theorem 1 is also valid for fuzzy metric semigroups [10]. In addition, they also found:

Theorem 3.

Let $(G,M,\ast )$ be a fuzzy metric group such that $(M,\ast )$ is invariant on G. Then, $(G,M,\ast )$ is complete if and only if $(G,{\tau }_M)$ is Raǐkov complete.

In 2020, Khatami and Mirzavaziri proposed the concept of the ⋆-metric and gave an example to illustrate that ⋆-metrics are not metrics ([11] (Example 2.4)). Some notations and definitions are as follows:

Recall that a t-definer is a function ⋆: $[0,\infty )\times [0,\infty )\to [0,\infty )$ satisfying the following conditions for each $a,b,c\in [0,\infty )$: (T1) $a\star b=b\star a$; (T2) $a\star (b\star c)=(a\star b)\star c$; (T3) if $a⩽b$, then $a\star c⩽b\star c$; (T4) $a\star 0=a$; (T5) ⋆ is continuous in its first component with respect to the Euclidean topology. In addition, ⋆ is continuous by [11].

Definition 1

([11] (Definition 2.2)). Let X be a nonempty set and ⋆ is a t-definer. If, for every $x,y,z\in X$, a function $d^{\star }:X\times X\to [0,\infty )$ satisfies the following conditions:

(M1) 

$d^{\star }(x,y)=0$ if and only if $x=y$;

(M2) 

$d^{\star }(x,y)=d^{\star }(y,x)$;

(M3) 

$d^{\star }(x,y)⩽d^{\star }(x,z)\star d^{\star }(z,y)$,

then $d^{\star }$ is called a ⋆-metric on X. The set X with a ⋆-metric is called ⋆-metric space, denoted by $(X,d^{\star })$.

Assume that $(X,d^{\star })$ is a ⋆-metric space. For any $a\in X$ and $r>0$, denote

$$B_{d^{\star }}(a,r)=\{x\in X:d^{\star }(a,x)<r\}$$

(1)

and

$${\tau }_{d^{\star }}=\{U\subseteq X:\mathrm{for}\mathrm{each}a\in U\mathrm{there}\mathrm{is}r>0\mathrm{such}\mathrm{that}{B}_{d^{\star }}(a,r)\subseteq U\}.$$

(2)

Let $\mathcal{B}=\{B_{d^{\star }}(x,\frac{1}{n})\mid x\in X,ϵ>0\}$ be a family of open balls on a ⋆-metric space $(X,d^{\star })$.

Lemma 1

([11] (Definition 3.1 and Lemma 3.3)). Let $(X,d^{\star })$ be a ⋆-metric space; then, ${\tau }_{d^{\star }}$ is a topology on X and $\mathcal{B}$ constitutes a topological basis of ${\tau }_{d^{\star }}$

According to the ⋆-metric proposed by Khatami and Mirzavaziri [11] and the ⋆-(quasi)-pseudometric semigroup proposed by S.Y. He, Y.Y. Jin, and L.H. Xie [12], we propose ⋆-metric semigroups and ⋆-metric groups. Referring to the research method in Refs. [9,10], we consider the following questions. (1) Is Theorem 1 still valid under ⋆-metric semigroups and ⋆-metric groups? (2) Let $(G,d^{\star })$ be a ⋆-metric group. Then, what is the relationship between $(G,d^{\star })$ completeness and $(G,{\tau }_{d^{\star }})$ Raǐkov completeness?

With this in mind, in Section 2, we prove that the topology induced by a invariant ⋆-metric can make an abstract semigroup become a topological semigroup, and give a characterization of invariant ⋆-metrics. Then, we obtain the following. Let $(X,d^{\star })$ be a complete ⋆-metric space containing a semigroup G that is a dense subset of X. If the restriction of $d^{\star }$ on G is invariant, then X can become a semigroup containing G as a subgroup and $d^{\star }$ is invariant on X. In Section 3, we discuss the Raǐkov completeness of ⋆-metrics and find the following. Let $(G,d^{\star })$ be a ⋆-metric group such that $d^{\star }$ is invariant on G. Then, $(G,d^{\star })$ is complete if and only if $(G,{\tau }_{d^{\star }})$ is Raǐkov complete.

2. Complete Invariant ⋆-Metrics on Semigroups and Groups

In this chapter, we first give the characterization of invariant ⋆-metrics on ⋆-metric groups. Then, using Sánchez’s method, we prove that Theorem 1 is still valid under ⋆-metric semigroups and ⋆-metric groups.

A topological semigroup $(G,\tau )$ is an algebraic semigroup G with a topology $\tau$ that makes the multiplication in G jointly continuous. A paratopological groupG is a topological semigroup such that G is an algebraic group. We say that a paratopological group $(G,\tau )$ is a topological group if the inverse is continuous.

A ⋆-metric $d^{\star }$ on a group G is left-invariant (respectively, right-invariant) if $d^{\star }(x,y)=d^{\star }(ax,ay)$ (respectively, $d^{\star }(x,y)=d^{\star }(xa,ya)$) whenever $a,x,y\in G$. We say that $d^{\star }$ is invariant if it is both left-invariant and right-invariant.

The notions and concepts of topological groups such as “group”, “semigroup”, “continuous function”, and so forth are defined as usual (e.g., see [13]).

Proposition 1.

Let S be an abstract semigroup and $d^{\star }$ be a ⋆-metric on S. If $d^{\star }$ is invariant, then $(S,{\tau }_{d^{\star }})$ is a topological semigroup.

Proof. 

According to Lemma 1, ${\tau }_{d^{\star }}$ is a topology on S. We will prove that the multiplication of semigroup S is joint continuous with respect to ${\tau }_{d^{\star }}$.

For every $y,z\in S$ and an open set U containing $yz$, according to the definition of ${\tau }_{d^{\star }}$, there exists an $ϵ>0$ such that $yz\in B_{d^{\star }}(yz,ϵ)\subseteq U$. Since the ⋆ is continuous, for the $ϵ$ above, we can find a $r>0$ such that $r\star r<ϵ$.

Now, we will prove that $B_{d^{\star }}(y,r)B_{d^{\star }}(z,r)\subseteq B_{d^{\star }}(yz,ϵ)$.

In fact, for every $a\in B_{d^{\star }}(y,r)$ and $b\in B_{d^{\star }}(z,r)$, $ab\in B_{d^{\star }}(y,r)B_{d^{\star }}(z,r)$. Thus, we have

$$d^{\star }(yz,ab)⩽d^{\star }(yz,yb)\star d^{\star }(yb,ab)=d^{\star }(z,b)\star d^{\star }(y,a)<r\star r<ϵ.$$

(3)

We have proved that $ab\in B_{d^{\star }}(yz,ϵ)$. Therefore, $(S,{\tau }_{d^{\star }})$ is a topological semigroup.

This completes the proof. □

Proposition 2.

Let G be an abstract group and $d^{\star }$ be a ⋆-metric on G. If $d^{\star }$ is invariant, then $(G,{\tau }_{d^{\star }})$ is a topological group.

Proof. 

Similar to the proof of Proposition 1, we see that the multiplication of group G is joint continuous with respect to ${\tau }_{d^{\star }}$. Therefore, $(G,{\tau }_{d^{\star }})$ is a paratopological group.

Now, we shall prove that the inverse mapping is continuous. To complete the proof, it is enough to show that $B_{d^{\star }}{(x,{ϵ}_0)}^{-1}\subseteq B_{d^{\star }}(x^{-1},ϵ)$ for $x\in G$. In fact, let ${ϵ}_0=ϵ$, and take $a\in B_{d^{\star }}(x,{ϵ}_0)$. Since $d^{\star }$ is invariant, we have

$$d^{\star }(x^{-1},a^{-1})=d^{\star }(x^{-1}a{a}^{-1},x^{-1}x{a}^{-1})=d^{\star }(a,x)<ϵ.$$

(4)

Therefore, we have proven that the inverse mapping is continuous.

Thus, $(G,{\tau }_{d^{\star }})$ is a topological group. □

Recall that a sequence ${\left(x_n\right)}_{n\in \mathbb{N}}$ in a ⋆-metric space $(X,d^{\star })$ is said to be a Cauchy sequence provided that, for each $ϵ>0$, there exists $n_0\in \mathbb{N}$ such that $d^{\star }(x_n,x_m)<ϵ$ for every $n,m⩾n_0$. A ⋆-metric space is complete if every Cauchy sequence is convergent. If $(X,d^{\star })$ is a complete ⋆-metric space, then the ⋆-metric $d^{\star }$ is said to be a complete ⋆-metric.

A topological space $(X,\tau )$ is called completely ⋆-metrizable if there exists a completely ⋆-metric $d^{\star }$ on X such that the topology ${\tau }_{d^{\star }}$ induced by $d^{\star }$ coincides with $\tau$; such a completely ⋆-metric is said to be compatible with $\tau$.

By a ⋆-metric group (resp., ⋆-metric semigroup), we mean a pair $(G,d^{\star })$ such that $(G,d^{\star })$ is a ⋆-metric space and $(G,{\tau }_{d^{\star }})$ is a topological group (resp., topological semigroup).

Now, we give the characterization of invariant ⋆-metrics.

Lemma 2.

Let $(G,d^{\star })$ be a ⋆-metric group. Then, $d^{\star }$ is invariant if and only if $d^{\star }(ab,cd)⩽d^{\star }(a,c)\star d^{\star }(b,d)$ for each $a,b,c,d\in G$.

Proof. 

If $d^{\star }$ is invariant, then

$$d^{\star }(ab,cd)⩽d^{\star }(ab,cb)\star d^{\star }(cb,cd)=d^{\star }(a,c)\star d^{\star }(b,d).$$

(5)

For the converse, if $a,b,g\in G$, then we have

$$d^{\star }(ga,gb)⩽d^{\star }(g,g)\star d^{\star }(a,b)=0\star d^{\star }(a,b)=d^{\star }(a,b).$$

(6)

Similarly,

$$d^{\star }(a,b)=d^{\star }(g^{-1}ga,g^{-1}gb)⩽d^{\star }(g^{-1},g^{-1})\star d^{\star }(ga,gb)=d^{\star }(ga,gb),$$

(7)

which means $d^{\star }$ is left invariant. With a similar argument, we can show that $d^{\star }$ is right invariant. Therefore, $d^{\star }$ is invariant. □

Remark 1.

Let $(G,d^{\star })$ be a ⋆-metric semigroup such that $d^{\star }$ is invariant. Then, $d^{\star }(ab,cd)⩽d^{\star }(a,c)\star d^{\star }(b,d)$ for each $a,b,c,d\in G$.

Theorem 4.

Let $(X,d^{\star })$ be a complete ⋆-metric space containing a semigroup S that is a dense subset of X. If the restriction of $d^{\star }$ on S is invariant, then X can become a semigroup containing S as a subsemigroup and $d^{\star }$ is invariant on X.

Proof. 

Take $a,b\in X$. Let ${\left(a_n\right)}_n$, ${\left(b_n\right)}_n\subseteq S$ be Cauchy sequences such that ${\left(a_n\right)}_n$ converges to a and ${\left(b_n\right)}_n$ converges to b.

We claim that ${\left(a_n{b}_n\right)}_n$ is a Cauchy sequence in $(X,d^{\star })$. For every $ϵ>0$, since ⋆ is a continuous t-definer, there is $r>0$ such that $r\star r<ϵ$. In addition, ${\left(a_n\right)}_n$, ${\left(b_n\right)}_n$ are Cauchy sequences in S, so there exists $n_0\in \mathbb{N}$ such that $d^{\star }(a_n,a_m)<r$ and $d^{\star }(b_n,b_m)<r$ whenever $n,m⩾n_0$. Since $d^{\star }$ is invariant in S, by Remark 1, we have

$$\begin{array}{cc}\hfill d^{\star }(a_n{b}_n,a_m{b}_m)& ⩽d^{\star }(a_n{b}_n,a_m{b}_n)\star d^{\star }(a_m{b}_n,a_m{b}_m)\hfill \\ \hfill & ⩽d^{\star }(a_n,a_m)\star d^{\star }(b_n,b_m)⩽r\star r<ϵ.\hfill \end{array}$$

(8)

This proves the claim.

Now we define a binary operation · on X as follows. Given two elements $a,b$ of X and two sequences ${\left(a_n\right)}_n$, ${\left(b_n\right)}_n\subseteq S$, as above, $a·b=x$ where x is the limit of ${\left(a_n{b}_n\right)}_n$ with respect to ${\tau }_{d^{\star }}$. Let us show that · is well defined.

In fact, choose two sequences ${\left(c_n\right)}_n$ and ${\left(d_n\right)}_n$ in S converging to a and b, respectively. We shall show that ${\left(c_n{d}_n\right)}_n$ converges to x. Take $ϵ>0$; there exists $r>0$ such that $r\star r\star r<ϵ$. According to assumptions, there exists $n_0\in \mathbb{N}$ such that $d^{\star }(x,a_n{b}_n)<r$, $d^{\star }(a_n,c_n)<r$, $d^{\star }(b_n,d_n)<r$, for $n⩾n_0$. Hence, for every $n⩾n_0$, Remark 1 implies that

$$\begin{array}{cc}\hfill d^{\star }(x,c_n{d}_n)& ⩽d^{\star }(x,a_n{b}_n)\star d^{\star }(a_n{b}_n,c_n{d}_n)\hfill \\ \hfill & <r\star d^{\star }(a_n,c_n)\star d^{\star }(b_n,d_n)\hfill \\ \hfill & <r\star r\star r<ϵ.\hfill \end{array}$$

(9)

Thus, the binary operation · is well defined.

Let us show that $(X,·)$ is a semigroup. Since

$$(a·b)·c=\underset{n\to \infty }{lim}\left(a_n{b}_n\right)c_n=\underset{n\to \infty }{lim}{a}_n\left(b_n{c}_n\right)=a·(b·c),$$

it follows that the operation · is associative.

Now, we will prove that $d^{\star }$ is invariant on X. Fix $a,x,y\in X$ and $ϵ>0$. We can choose $r>0$ such that $r\star r\star r\star r<ϵ$. Take sequences ${\left(a_n\right)}_n$, ${\left(x_n\right)}_n$, ${\left(y_n\right)}_n$ in S converging to a, x, and y, respectively. We can find $N\in \mathbb{N}$ such that $d^{\star }(ax,a_N{x}_N)<r$, $d^{\star }(a_N{y}_N,ay)<r$, $d^{\star }(x_N,x)<r$ and $d^{\star }(y,y_N)<r$. Since, $d^{\star }$ is left invariant on S, we have

$$\begin{array}{cc}\hfill d^{\star }(ax,ay)& ⩽d^{\star }(ax,a_N{x}_N)\star d^{\star }(a_N{x}_N,a_N{y}_N)\star d^{\star }(a_N{y}_N,ay)\hfill \\ \hfill & <r\star d^{\star }(x_N,y_N)\star r\hfill \\ \hfill & <r\star r\star d^{\star }(x_N,x)\star d^{\star }(x,y)\star d^{\star }(y,y_N)\hfill \\ \hfill & <r\star r\star r\star r\star d^{\star }(x,y)\hfill \\ \hfill & <ϵ\star d^{\star }(x,y).\hfill \end{array}$$

(10)

We have thus proven that $d^{\star }(ax,ay)<ϵ\star d^{\star }(x,y)$. Since the ⋆ is continuous, we have $d^{\star }(ax,ay)⩽d^{\star }(x,y)$. In a similar way, $d^{\star }(ax,ay)⩾d^{\star }(x,y)$. Hence, $d^{\star }$ is left invariant on X. Using a similar argument and the fact that $d^{\star }$ is right invariant on S, we can prove that $d^{\star }$ is right invariant on X. Therefore, $d^{\star }$ is invariant on X.

Finally, Proposition 1 permits us conclude that $(X,{\tau }_{d^{\star }})$ is a topological semigroup. □

Theorem 5.

Let $(X,d^{\star })$ be a complete ⋆-metric space containing a group G that is a dense subset of X. If the restriction of $d^{\star }$ on G is invariant, then X can become a group containing G as a subgroup and $d^{\star }$ is invariant on X.

Proof. 

Take $a,b\in X$. Let ${\left(a_n\right)}_n$, ${\left(b_n\right)}_n\subseteq G$ be Cauchy sequences such that ${\left(a_n\right)}_n$ converges to a and ${\left(b_n\right)}_n$ converges to b.

We claim that ${\left(a_n{b}_n\right)}_n$ is a Cauchy sequence in $(X,d^{\star })$. For every $ϵ>0$, since ⋆ is a continuous t-definer, there is $r>0$ such that $r\star r<ϵ$. In addition, ${\left(a_n\right)}_n$, ${\left(b_n\right)}_n$ are Cauchy sequences in G, so there exists $n_0\in \mathbb{N}$ such that $d^{\star }(a_n,a_m)<r$ and $d^{\star }(b_n,b_m)<r$ whenever $n,m⩾n_0$. Since $d^{\star }$ is invariant in G, by Lemma 2, we have

$$d^{\star }(a_n{b}_n,a_m{b}_m)⩽d^{\star }(a_n,a_m)\star d^{\star }(b_n,b_m)⩽r\star r<ϵ.$$

(11)

This proves the claim.

Now we define a binary operation · on X as follows: given two elements $a,b$ of X and two sequences ${\left(a_n\right)}_n$, ${\left(b_n\right)}_n\subseteq G$ as above, $a·b=x$ where x is the limit of ${\left(a_n{b}_n\right)}_n$ with respect to ${\tau }_{d^{\star }}$. Similar to the proof of Theorem 4, the operation · is well defined.

Let us show that $(X,·)$ is a group. First, notice that if e is the neutral element of G, then

$$a·e=\underset{n\to \infty }{lim}{a}_{n}e=a=\underset{n\to \infty }{lim}e{a}_n=e·a,$$

(12)

that is, e is the neutral element of $(X,·)$. Second, since

$$(a·b)·c=\underset{n\to \infty }{lim}\left(a_n{b}_n\right)c_n=\underset{n\to \infty }{lim}{a}_n\left(b_n{c}_n\right)=a·(b·c),$$

(13)

it follows that the operation · is associative. In addition, since $d^{\star }$ is invariant on G, we have

$$d^{\star }(a_m,a_n)=d^{\star }(a_m{a}_n^{-1}{a}_n,a_m{a}_m^{-1}{a}_n)=d^{\star }(a_n^{-1},a_m^{-1}),$$

(14)

and $e=\underset{n\to \infty }{lim}{a}_n{a}_n^{-1}$; the sequence ${\left(a_n^{-1}\right)}_{n\in \mathbb{N}}$ is a Cauchy sequence. Therefore, $a^{-1}=\underset{n\to \infty }{lim}{a}_n^{-1}$ is the inverse element of a.

According to the proof of Theorem 4, we see that $d^{\star }$ is invariant on X.

Finally, Proposition 2 permits us conclude that $(X,{\tau }_{d^{\star }})$ is a topological group. □

Theorem 6.

Suppose that $(G,d^{\star })$ is a ⋆-metric group such that $d^{\star }$ is invariant. If $d^{\star }$ is a complete ⋆-metric, then every compatible left invariant ⋆-metric on G is complete.

Proof. 

Let $\overline{d^{\star }}$ be a compatible left invariant ⋆-metric on G. Take a Cauchy sequence ${\left(x_n\right)}_n$ in $(G,\overline{d^{\star }})$. We claim that ${\left(x_n\right)}_n$ is a Cauchy sequence in $(G,d^{\star })$ as well.

Choose $ϵ>0$. By the compatibility of $\overline{d^{\star }}$, we can find $r>0$ such that $B_{\overline{d^{\star }}}(e,r)\subseteq B_{d^{\star }}(e,ϵ)$, where e is the neutral element of G. Since ${\left(x_n\right)}_n$ is a Cauchy sequence in $(G,\overline{d^{\star }})$, there exists $k\in \mathbb{N}$ such that $\overline{d^{\star }}(x_n,x_m)<r$ provided $n,m⩾k$, which means $x_m\in B_{\overline{d^{\star }}}(x_n,ϵ)$. Since $\overline{d^{\star }}$ is left invariant, according to the proof of ([12] Theorem 3.5), we have $B_{\overline{d^{\star }}}(x_n,ϵ)=x_n{B}_{\overline{d^{\star }}}(e,r)$, i.e., $x_n^{-1}{x}_m\in B_{\overline{d^{\star }}}(e,r)$ for each $n,m⩾k$. Therefore, $x_n^{-1}{x}_m\in B_{d^{\star }}(e,ϵ)$; equivalently, $d^{\star }(x_n,x_m)<ϵ$ for each $n,m⩾k$. We have thus proven our claim. Since $(G,d^{\star })$ is complete, the sequence ${\left(x_n\right)}_n$ converges with respect to ${\tau }_{d^{\star }}={\tau }_{\overline{d^{\star }}}$. Therefore, $\overline{d^{\star }}$ is complete as well. □

By Theorems 5 and 6, we easily see the following.

Corollary 1.

Let $(X,d^{\star })$ be a complete ⋆-metric space containing a group G that is a dense subset of X. If the restriction of $d^{\star }$ on G is invariant, then every left invariant ⋆-metric on the $(X,d^{\star })$ is complete.

3. The Raǐkov Completion of ⋆-Metrics Groups

A filter on a set X is a family $\eta$ of non-empty subsets of X satisfying the next two conditions. (i) If U and V are in $\eta$, then $U\cap V$ is also in $\eta$. (ii) If $U\in \eta$ and $U\subseteq W\subseteq X$, then $W\in \eta$.

Let G be a topological group with the identity e. A filter $\eta$ of a topological group G is said to be a Cauchy filter if for every open neighborhood V of e in G, there exist $a,b\in G$ and $A,B\in \eta$ such that $A\subseteq aV$ and $B\subseteq Vb$. In a topological space G, a filter $\eta$ on G converges to a point x of G if every neighborhood of x belongs to $\eta$. A topological group G such that every Cauchy filter on G converges is called Raǐkov complete. Next, we shall investigate the Raǐkov completeness of the group topologies induced by invariant ⋆-metrics on groups.

Theorem 7.

Let $(G,d^{\star })$ be a ⋆-metric group. If $(G,d^{\star })$ is complete, then $(G,{\tau }_{d^{\star }})$ is Raǐkov complete.

Proof. 

Suppose that $(G,d^{\star })$ is complete. Take an arbitrary Cauchy filter $\eta \subseteq G$. Then, for each $n\in \mathbb{N}$, there are $F_n^{\prime }\in \eta$ and $x_n\in G$ such that $F_n^{\prime }\subseteq B_{d^{\star }}(x_n,\frac{1}{n})$. Put $F_n={\bigcap }_{i=1}^n{F}_i^{\prime }$. Clearly, $F_n\subseteq B_{d^{\star }}(x_n,\frac{1}{n})$ holds for each $n\in \mathbb{N}$. Take $y_n\in F_n$ for each $n\in \mathbb{N}$. Then, the sequence ${\left(y_n\right)}_{n\in \mathbb{N}}$ is a Cauchy sequence in $(G,d^{\star })$. In fact, for each $n\in \mathbb{N}$, since ⋆ is continuous, there is $n_0\in \mathbb{N}$ such that $\frac{1}{n_0}\star \frac{1}{n_0}<\frac{1}{n}$. Obviously, $y_i,y_j\in F_{n_0}\subseteq B_{d^{\star }}(x_{n_0},\frac{1}{n_0})$, whenever $i,j\in \mathbb{N}$ and $i,j>n_0$. Thus,

$$d^{\star }(y_i,y_j)⩽d^{\star }(y_i,x_{n_0})\star d^{\star }(x_{n_0},y_j)<\frac{1}{n_0}\star \frac{1}{n_0}<\frac{1}{n}$$

(15)

whenever $i,j>n_0$. This shows that ${\left(y_n\right)}_{n\in \mathbb{N}}$ is a Cauchy sequence. Since $(G,d^{\star })$ is complete, ${\left(y_n\right)}_{n\in \mathbb{N}}$ converges to some $y\in G$. We shall show that the Cauchy filter $\eta$ converges to y, which implies that $(G,{\tau }_{d^{\star }})$ is Raǐkov complete.

Take any open neighborhood V of y. Without loss of generality, we assume that $V=B_{d^{\star }}(y,\frac{1}{n})$. Since ⋆ is continuous, there is $r\in \mathbb{N}$ such that $\frac{1}{n_0}\star \frac{1}{n_0}<\frac{1}{n}$. Note that ${\left\{y_n\right\}}_{n\in \mathbb{N}}$ converges to y; then, there is $n^{\prime }\in \mathbb{N}$ such that $\frac{1}{n^{\prime }}\star \frac{1}{n^{\prime }}<\frac{1}{n_0}$ and $d^{\star }(y,y_{n^{\prime }})<\frac{1}{n_0}$. Then, for each $x\in F_{n^{\prime }}$, noting that $y_{n^{\prime }}\in F_{n^{\prime }}\subseteq B_{d^{\star }}(x_{n^{\prime }},\frac{1}{n^{\prime }})$, we have

$$\begin{array}{cc}\hfill d^{\star }(y,x)& ⩽d^{\star }(y,y_{n^{\prime }})\star d^{\star }(y_{n^{\prime }},x_{n^{\prime }})\star d^{\star }(x_{n^{\prime }},x)\hfill \\ \hfill & <\frac{1}{n_0}\star \frac{1}{n^{\prime }}\star \frac{1}{n^{\prime }}\hfill \\ \hfill & <\frac{1}{n_0}\star \frac{1}{n_0}<\frac{1}{n}.\hfill \end{array}$$

(16)

This implies that $x\in B_{d^{\star }}(y,\frac{1}{n})$, i.e., $F_{n^{\prime }}\subseteq B_{d^{\star }}(y,\frac{1}{n})$. Clearly, $F_{n^{\prime }}\in \eta$; thus, we have proved = n that $\eta$ converges to y. □

Theorem 8.

Let $(G,d^{\star })$ be a ⋆-metric group with $d^{\star }$ being invariant. If $(G,{\tau }_{d^{\star }})$ is Raǐkov complete, then $(G,d^{\star })$ is complete.

Proof. 

Suppose that $(G,{\tau }_{d^{\star }})$ is Raǐkov complete. Take the arbitrary Cauchy sequence ${\left(x_n\right)}_{n\in \mathbb{N}}$ of $(G,d^{\star })$. Put $\eta =\{A\subseteq G:F_n\subseteq A\mathrm{for}\mathrm{some}{F}_n\}$, where $F_n=\{x_i:i\ge n\}$ for each $n\in \mathbb{N}$. Now, we shall prove that $\eta$ is a Cauchy filter of G. Take any $B_{d^{\star }}(e,\frac{1}{n})$, where e is the identity of G. Since ${\left(x_i\right)}_{i\in \mathbb{N}}$ is a Cauchy sequence, there is $n_0\in \mathbb{N}$ such that $d^{\star }(x_k,x_m)<\frac{1}{n}$ whenever $k,m⩾n_0$. This implies $x_k\in B_{d^{\star }}(x_{n_0},\frac{1}{n})$ whenever $k⩾n_0$. Hence, $F_{n_0}\subseteq B_{d^{\star }}(x_{n_0},\frac{1}{n})$. Noting that $d^{\star }$ is invariant, $B_{d^{\star }}(x_{n_0},\frac{1}{n})=x_{n_0}{B}_{d^{\star }}(e,\frac{1}{n})=B_{d^{\star }}(e,\frac{1}{n})x_{n_0}$, so $F_{n_0}\subseteq x_{n_0}{B}_{d^{\star }}(e,\frac{1}{n})$ and $F_{n_0}\subseteq B_{d^{\star }}(e,\frac{1}{n})x_{n_0}$. This implies that $\eta$ is a Cauchy filter. Since $(G,{\tau }_{d^{\star }})$ is Raǐkov complete, the Cauchy filter $\eta$ converges to a point g in G. Then, one can easily show that the Cauchy sequence ${\left(x_n\right)}_{n\in \mathbb{N}}$ converges to g. This implies that $(G,d^{\star })$ is complete. □

By Theorems 7 and 8, we have the following:

Corollary 2.

Let $(G,d^{\star })$ be a ⋆-metric group such that $d^{\star }$ is invariant on G. Then, $(G,d^{\star })$ is complete if and only if $(G,{\tau }_{d^{\star }})$ is Raǐkov complete.

4. Conclusions and Further Work

4.1. Conclusions

Our initial motivation was to study the invariant ⋆-metrics on semigroups and groups. In this paper, we first propose the concept of ⋆-metric semigroups (respectively, groups) and study some questions related to complete ⋆-metric topological semigroups. Invariant ⋆-metrics are characterized, which allow us to characterize this kind of ⋆-metric. We also study the ⋆-metric version of classical theorems in the framework of complete ⋆-metric groups. Our results fit in a long tradition of research in topological algebra. When applied to classical structures, they allow us to obtain new outcomes that generalize helpful results of the theory.

4.2. Limitations

In the Section 2, we discuss the conditions under which a complete metric space becomes a topological group. Since we have not yet proven that every ⋆-metric space has a completion, we first give a complete ⋆-metric space in Theorems 4 and 5. Otherwise, the expression of these theorems would be more concise and clearer.

4.3. Future Research Direction

It would be interesting to study the ⋆-metrics on the quotient groups. Future research could focus on the following topic. Giving a ⋆-metric group $(G,d^{\star })$ and a closed normal subgroup N, if one has information about two complete ⋆-metric groups $(G,d^{\star })$, $(N,d^{\star })$ and $(G/N,d^{\star })$, what can be said about the third group?