A Note on Extension of Fuzzy Metric Spaces

Abstract

In this note, we prove that for two compatible fuzzy metrics $M_H$ and $M_K$ on H and K, respectively, there exists a fuzzy metric M on $H\bigcup K$ such that ${M|}_H=M_H$ and ${M|}_K=M_K$ under the conditions that t-norm ∗ is positive and fuzzy metrics $M_H,M_K$ are strong, or t-norm ∗ is positive and t-norm ∗, fuzzy metrics $M_H,M_K$ satisfy the Lipschitz conditions.

1. Introduction and Preliminaries

In this note, fuzzy metric refers to fuzzy metric in the sense of George and Veeramani [1]. In [2], Gregori et al. made a detailed summary of the theory of fuzzy metric space in recent years, including completeness, completion, continuity, extension, contractivity, and application (see [2,3,4] and the references therein). At the end of each direction, the authors put forward an open question, which points out the directions for the research of fuzzy metric spaces. Some of these problems have been solved, but the following problem is still open.

Problem 1. 

Let H and K be two distinct sets with $H\bigcap K\ne \varnothing$. Let $(M_H,\ast )$ and $(M_K,\ast )$ be two non-stationary fuzzy metrics on H and K, respectively, that agree in $H\bigcap K$. Does a fuzzy metric M exist on $H\bigcup K$ such that ${M|}_H=M_H$ and ${M|}_K=M_K$?

The fact is well known [5]: Let H and K be two distinct sets with $H\bigcap K\ne \varnothing$. Let $d_H$ and $d_K$ be metrics on H and K, respectively, that agree in $H\bigcap K$, then there exists a metric d on $H\bigcup K$ such that ${d|}_H=d_H$ and ${d|}_K=d_K$.

It can be seen that the above problem is whether the conclusion of metric space is valid in fuzzy metric space.

Because the parameters t and t-norm ∗ are used in the definition of fuzzy metric space, the structure of fuzzy metric space is much more complex than that of metric space. Many methods and techniques used in metric space often fail in fuzzy metric space, which forces researchers to introduce novel concepts, methods, and techniques into fuzzy metric space.

If a fuzzy metric does not involve parameters t, the fuzzy metric is called stationary [6]. Gregori et al. obtained the following result in stationary fuzzy metric [7]:

Let H and K be two distinct sets with $H\bigcap K\ne \varnothing$. Let $(M_H,\ast )$ and $(M_K,\ast )$ be two stationary fuzzy metrics on H and K, respectively, that agree in $H\bigcap K$. Then, there exists a fuzzy metric M on $H\bigcup K$ such that ${M|}_H=M_H$ and ${M|}_K=M_K$.

Although the influence of parameters t should not be considered, the authors ignored or underestimated the influence of t-norm ∗ on the construction of fuzzy metrics, resulting in a small gap in the proof.

Example 1. 

Let $H=\{1,2\}$ and $K=\{2,3\}$. Let $M_H(1,1)=M_H(2,2)=1,M_H(1,2)=M_H(2,1)=0.3$ and $M_K(2,2)=M_K(3,3)=1,M_K(2,3)=M_K(3,2)=0.5$. Let ∗ be the Łukasievicz t-norm ($a\ast b=max\{a+b-1,0\}$). Then $(M_H,\ast )$ and $(M_K,\ast )$ are two stationary fuzzy metrics on H and K, respectively, and they agree in $H\bigcap K$.

If the construction of fuzzy metric in the proof of the above result is adopted, then $M(1,3)=M_H(1,2)\ast M_K(2,3)=0$. This contradicts the definition of fuzzy metric.

In order to overcome the above gap, we adopt positive t-norm ∗. A t-norm ∗ is called positive, if $a\ast b>0$ for all $a,b\in (0,1]$. The revised theorem is described as follows.

Theorem 1. 

Let H and K be two distinct sets with $H\bigcap K\ne \varnothing$. Let $(M_H,\ast )$ and $(M_K,\ast )$ be two stationary fuzzy metrics on H and K, respectively, that agree in $H\bigcap K$. If ∗ is positive, then there exists a fuzzy metric M on $H\bigcup K$ such that ${M|}_H=M_H$ and ${M|}_K=M_K$.

Definition 1 

(See George and Veeramani [1]). A fuzzy metric space is an ordered triple $(X,M,\ast )$ such that X is a (non-empty) set, ∗ is a continuous t-norm, and M is a fuzzy set on $X\times X\times (0,+\infty )$, satisfying the following conditions, for all $x,y,z\in X,s,t>0$:

(GV1) $M(x,y,t)>0$;

(GV2) $M(x,y,t)=1$ if and only if $x=y$;

(GV3) $M(x,y,t)=M(y,x,t)$;

(GV4) $M(x,y,t)\ast M(y,z,s)\le M(x,z,t+s)$;

(GV5) $M(x,y,\_):(0,+\infty )\to (0,1]$ is continuous.

If $(X,M,\ast )$ is a fuzzy metric space, we will say that $(M,\ast )$, or simply M, is a fuzzy metric on X.

In Definition 1, M is a fuzzy set on $X\times X\times (0,+\infty )$, and $M(x,y,t)$ can be thought of as the degree of nearness between x and y with respect to t. A quite natural interpretation to the conditions can see [1].

Remark 1 

(See Grabiec [8]). $M(x,y,\_)$ is non-decreasing for all $x,y\in X.$

George and Veeramani proved in [1] that every fuzzy metric M on X generates a topology ${\tau }_M$ on X which has as a base the family of open sets of the form $\{B_M(x,ϵ,t):x\in X,0<ϵ<1,t>0\}$, where $B_M(x,ϵ,t)=\{y\in X:M(x,y,t)>1-ϵ\}$ for all $x\in X,ϵ\in (0,1)$ and $t>0$.

Definition 2 

(See Gregori and Romaguera [6]). A fuzzy metric M on X is said to be strong if it satisfies for each $x,y\in X$ and $t>0$,

$$\begin{array}{c}M(x,z,t)\ge M(x,y,t)\ast M(y,z,t).\end{array}$$

Let $(X,d)$ be a metric space and let $M_d$ be a function on $X\times X\times (0,+\infty )$ defined by

$$M_d(x,y,t)=\frac{t}{t+d(x,y)}$$

Then, $(X,M_d,.)$ is a fuzzy metric space [1] and $M_d$ is called the standard fuzzy metric induced by d, where . is the product t-norm. The topology ${\tau }_{M_d}$ coincides with the topology ${\tau }_d$ on X deduced from d. It is easy to see that $M_d$ is strong.

To prove the continuity of the constructed fuzzy metric, we need definitions of the Lipschitz condition regarding t-norm and fuzzy metric.

Definition 3. 

(i) 

We say t-norm ∗ satisfies the Lipschitz condition, if

$$\begin{array}{c}{a}_1\ast b-a_2\ast b\le a_1-a_2, whenever a_1\ge a_2.\end{array}$$

(ii) 

We say fuzzy metric M satisfies the Lipschitz condition, if

$$\begin{array}{c}M(x,y,t_1)-M(x,y,t_2)\le t_1-t_2, whenever t_1\ge t_2.\end{array}$$

Lemma 1. 

If t-norm ∗ and fuzzy metrics $M_H,M_K$ satisfy the Lipschitz conditions, then

$$\begin{array}{c}{M}_H(x,y,a_1+a_2)\ast M_K(y,z,b)\le M_H(x,y,a_1)\ast M_K(y,z,b)+a_2\end{array}$$

and

$$M_H(x,y,a)\ast M_K(y,z,b_1+b_2)\le M_H(x,y,a)\ast M_K(y,z,b_1)+b_2,$$

where $M_H$ and $M_K$ are two fuzzy metrics on H and K, respectively, $x,y\in H,y,z\in K,a,a_1,a_2,b,b_1,b_2>0.$

Proof. 

We only prove the first inequality. The second inequality can be proved similarly.

(i)

If $M_H(x,y,a_1)+a_2\le 1$, then by the Lipschitz condition of $M_H$, we have

$$\begin{array}{c}{M}_H(x,y,a_1+a_2)\le M_H(x,y,a_1)+a_2.\end{array}$$

Furthermore, by the Lipschitz condition of ∗,

$$\begin{array}{c}(M_H(x,y,a_1)+a_2)\ast M_K(y,z,b)\le M_H(x,y,a_1)\ast M_K(y,z,b)+a_2.\end{array}$$

Thus,

$$M_H(x,y,a_1+a_2)\ast M_K(y,z,b)\le M_H(x,y,a_1)\ast M_K(y,z,b)+a_2.$$

(ii)

If $M_H(x,y,a_1)+a_2\ge 1$, then $a_2\ge 1-M_H(x,y,a_1)$.

$$\begin{array}{cc}\hfill M_H(x,y,a_1+a_2)\ast M_K(y,z,b)& \le M_K(y,z,b)\hfill \\ \hfill & =(M_H(x,y,a_1)+1-M_H(x,y,a_1))\ast M_K(y,z,b)\hfill \\ \hfill & \le M_H(x,y,a_1)\ast M_K(y,z,b)+1-M_H(x,y,a_1)\hfill \\ \hfill & \le M_H(x,y,a_1)\ast M_K(y,z,b)+a_2.\hfill \end{array}$$

Thus, we also have

$$M_H(x,y,a_1+a_2)\ast M_K(y,z,b)\le M_H(x,y,a_1)\ast M_K(y,z,b)+a_2.$$

□

In this paper, we prove that Theorem 1 still holds on certain conditions even if $M_H$ and $M_K$ do not have the stationary property.

2. Main Results

In this section, we answer Problem 1 on certain conditions. Fisrtly, the follow theorem is an answer to Problem 1 under the conditions that $M_H$ and $M_K$ are two strong fuzzy metrics.

Theorem 2. 

Let H and K be two distinct sets with $H\bigcap K\ne \varnothing$. Let $(M_H,\ast )$ and $(M_K,\ast )$ be two strong fuzzy metrics on H and K, respectively, that agree in $H\bigcap K$. If ∗ is positive, then there exists a fuzzy metric M on $H\bigcup K$ such that ${M|}_H=M_H$ and ${M|}_K=M_K$.

Proof. 

Define the fuzzy set M as follows:

$$M(x,y,t)=\left\{\begin{array}{cc}{M}_H(x,y,t),\hfill & \mathrm{i}\mathrm{f} x,y\in H;\hfill \\ M_K(x,y,t),\hfill & \mathrm{i}\mathrm{f} x,y\in K;\hfill \\ sup\{M_H(x,a,t)\ast M_K(a,y,t):a\in H\bigcap K\},\hfill & \mathrm{i}\mathrm{f} x\in H\setminus K,y\in K\setminus H.\hfill \end{array}\right.$$

Now, we prove that M is a fuzzy metric on $H\bigcup K$. For the definition of fuzzy metric space, the first three axioms are obvious. So, we just prove the triangular inequality and continuity of $M(x,y,\_)$, where $M(x,y,\_):(0,+\infty )\to (0,1]$. Firstly, we show the triangular inequality.

Let $x,y,z\in H\bigcup K$. If $x,y,z\in H$ or $x,y,z\in K$, the proof is obvious. So, we distinguish three non-trivial cases:

(i)

Suppose that $x,y\in H\setminus K$ and $z\in K\setminus H$. Then, given $\epsilon >0$, we can find $a\in H\bigcap K,s_1\in (0,s)$ such that

$$\begin{array}{c}M(y,z,s)-\epsilon <M_H(y,a,s)\ast M_K(a,z,s).\end{array}$$

Thus,

$$\begin{array}{c}M(x,y,t)\ast (M(y,z,s)-\epsilon )\le M(x,y,t)\ast M_H(y,a,s)\ast M_K(a,z,s)\hfill \\ =M_H(x,y,t)\ast M_H(y,a,s)\ast M_K(a,z,s)\hfill \\ \le M_H(x,a,t+s)\ast M_K(a,z,s)\hfill \\ \le M_H(x,a,t+s)\ast M_K(a,z,t+s)\hfill \\ \le M(x,z,t+s)\hfill \end{array}$$

Since $\epsilon$ is arbitrary and ∗ is continuous, we have

$$\begin{array}{c}M(x,y,t)\ast \left(M\right(y,z,s)\le M(x,z,t+s).\end{array}$$

(The proof of the case $x,y\in K\setminus H$ and $z\in H\setminus K$ is similar).

(ii)

Suppose that $x,z\in H\setminus K$ and $y\in K\setminus H$. Then, given $\epsilon >0$, we can find $a,b\in H\bigcap K$ such that

$$\begin{array}{c}M(x,y,t)-\epsilon <M_H(x,a,t)\ast M_K(a,y,t)\end{array}$$

and

$$\begin{array}{c}M(z,y,s)-\epsilon <M_H(z,b,s)\ast M_K(b,y,s).\end{array}$$

Thus,

$$\begin{array}{c}\left(M\right(x,y,t)-\epsilon )\ast \left(M\right(z,y,s)-\epsilon )\hfill \\ \le M_H(x,a,t)\ast M_K(a,y,t)\ast M_H(z,b,s)\ast M_K(b,y,s)\hfill \\ =M_H(x,a,t)\ast M_H(z,b,s)\ast M_K(a,y,t)\ast M_K(b,y,s)\hfill \\ \le M_H(x,a,t)\ast M_H(z,b,s)\ast M_K(a,b,t+s)\hfill \\ =M_H(x,a,t)\ast M_H(z,b,s)\ast M_H(a,b,t+s)\hfill \\ =M_H(x,a,t)\ast M_H(a,b,t+s)\ast M_H(z,b,s)\hfill \\ \le M_H(x,b,t+s)\ast M_H(z,b,s)\hfill \\ \le M_H(x,z,t+s)\hfill \\ =M(x,z,t+s)\hfill \end{array}$$

Since $\epsilon$ is arbitrary and ∗ is continuous, we have

$$\begin{array}{c}M(x,y,t)\ast \left(M\right(y,z,s)\le M(x,z,t+s).\end{array}$$

(The proof of the case $x,z\in K\setminus H$ and $y\in H\setminus K$ is similar).

(iii)

Suppose that $x\in H\bigcap K,y\in H\setminus K$ and $z\in K\setminus H$ or $x\in H\bigcap K,z\in H\setminus K$ and $y\in K\setminus H$. As proof of (i), we have

$$\begin{array}{c}M(x,y,t)\ast \left(M\right(y,z,s)\le M(x,z,t+s).\end{array}$$

Suppose that $x\in H\setminus K,y\in H\cap K,z\in K\setminus H$ or $x\in K\setminus H,y\in H\cap K,z\in H\setminus K$. The inequality still holds by definition of M.

Now, we shall prove that $M(x,y,\_)$ is continuous.

If $x,y\in H$ or $x,y\in K$, then $M(x,y,\_)$ is obviously continuous. Suppose that $x\in H\setminus K,y\in K\setminus H$, $△t>0$, we have

$$\begin{array}{c}\hfill M(x,y,t+△t)\ge M_H(x,a,t+△t)\ast M_K(a,y,t+△t)\ge M_H(x,a,t)\ast M_K(a,y,t)\end{array}$$

Taking sup on both sides for a, $M(x,y,t+△t)\ge M(x,y,t)$, which implies that $M(x,y,\_)$ is increasing.

Then, given $\epsilon >0$, we can find $a\in H\cap K$, such that

$$\begin{array}{c}\hfill M(x,y,t+△t)<M_H(x,a,t+△t)\ast M_K(a,y,t+△t)+\epsilon .\end{array}$$

Since $M_H(x,a,\_),M_K(a,y,\_)$ and ∗ are continuous, taking the limit as $△t\to 0^{+}$ in the last inequality, we obtain

$$\begin{array}{c}\hfill \underset{△t\to 0^{+}}{lim}M(x,y,t+△t)\le M_H(x,a,t)\ast M_K(a,y,t)+\epsilon \le M(x,y,t)+\epsilon .\end{array}$$

Since $\epsilon$ is arbitrary, that is

$$\begin{array}{c}\hfill \underset{△t\to 0^{+}}{lim}M(x,y,t+△t)\le M(x,y,t).\end{array}$$

Furthermore, $M(x,y,\_)$ is increasing, and we obtain

$$\begin{array}{c}\hfill \underset{△t\to 0^{+}}{lim}M(x,y,t+△t)=M(x,y,t).\end{array}$$

That is, $M(x,y,\_)$ is right continuous.

Next, we prove that $M(x,y,\_)$ is left continuous.

Let $△t<0$. Similarly, from the definition of M, we have

$$M(x,y,t+△t)\le M(x,y,t).$$

For each $a\in H\bigcap K$,

$$\begin{array}{c}\hfill M(x,y,t+△t)\ge M_H(x,a,t+△t)\ast M_K(a,y,t+△t).\end{array}$$

Since $M_H(x,a,\_),M_K(a,y,\_)$ and ∗ are continuous, taking the limit as $△t\to 0^{-}$ in the last inequality, we obtain

$$\begin{array}{c}\hfill \underset{△t\to 0^{-}}{lim}M(x,y,t+△t)\ge M_H(x,a,t)\ast M_K(a,y,t).\end{array}$$

Taking sup on both sides for $a\in H\bigcap K$, we have

$$\begin{array}{c}\hfill \underset{△t\to 0^{-}}{lim}M(x,y,t+△t)\ge M(x,y,t).\end{array}$$

Therefore, we proved that

$$\begin{array}{c}\hfill \underset{△t\to 0^{-}}{lim}M(x,y,t+△t)=M(x,y,t).\end{array}$$

Which means $M(x,y,\_)$ is also left continuous. Thus, $M(x,y,\_)$ is continuous. Therefore, M is a fuzzy metric on $H\bigcup K$.

□

Theorem 3. 

Let H and K be two distinct sets with $H\bigcap K\ne \varnothing$. Let $(M_H,\ast )$ and $(M_K,\ast )$ be two fuzzy metrics on H and K, respectively, that agree in $H\bigcap K$. If ∗ is positive, t-norm ∗ and fuzzy metrics $M_H,M_K$ satisfy the Lipschitz conditions; then, there exists a fuzzy metric M on $H\bigcup K$ such that ${M|}_H=M_H$ and ${M|}_K=M_K$.

Proof. 

Define the fuzzy set M as follows:

$M(x,y,t)=M_H(x,y,t)$, if $x,y\in H$; $M(x,y,t)=M_K(x,y,t)$, if $x,y\in K$; $M(x,y,t)=sup\{M_H(x,a,t_1)\ast M_K(a,y,t-t_1):a\in H\bigcap K,t_1\in (0,t)\}$, if $x\in H\setminus K,y\in K\setminus H$ or $x\in K\setminus H,y\in H\setminus K$.

Now, we prove that M is a fuzzy metric on $H\bigcup K$. For the definition of fuzzy metric space, the first three axioms are obvious. So, we just prove the triangular inequality and continuity of $M(x,y,\_)$, where $M(x,y,\_):(0,+\infty )\to (0,1]$. Firstly, we show the triangular inequality.

Let $x,y,z\in H\bigcup K$. If $x,y,z\in H$ or $x,y,z\in K$, the proof is obvious. So, we distinguish three non-trivial cases:

(i)

Suppose that $x,y\in H\setminus K$ and $z\in K\setminus H$. Then, given $\epsilon >0$, we can find $a\in H\bigcap K,s_1\in (0,s)$ such that $M(y,z,s)-\epsilon <M_H(y,a,s_1)\ast M_K(a,z,s-s_1)$. Thus,

$$\begin{array}{c}M(x,y,t)\ast (M(y,z,s)-\epsilon )\le M(x,y,t)\ast M_H(y,a,s_1)\ast M_K(a,z,s-s_1)\hfill \\ =M_H(x,y,t)\ast M_H(y,a,s_1)\ast M_K(a,z,s-s_1)\hfill \\ \le M_H(x,a,t+s_1)\ast M_K(a,z,s-s_1)\hfill \\ \le M(x,z,t+s)\hfill \end{array}$$

Since $\epsilon$ is arbitrary and ∗ is continuous, we have

$$\begin{array}{c}M(x,y,t)\ast \left(M\right(y,z,s)\le M(x,z,t+s).\end{array}$$

(The proof of the case $x,y\in K\setminus H$ and $z\in H\setminus K$ similar).

(ii)

Suppose that $x,z\in H\setminus K$ and $y\in K\setminus H$. Then, given $\epsilon >0$, we can find $a_1,a_2\in H\bigcap K,t_1\in (0,t),s_1\in (0,s)$ such that

$$\begin{array}{c}M(x,y,t)-\epsilon <M_H(x,a_1,t_1)\ast M_K(a_1,y,t-t_1)\end{array}$$

and

$$\begin{array}{c}M(z,y,s)-\epsilon <M_H(z,a_2,s_1)\ast M_K(a_2,y,s-s_1).\end{array}$$

Thus,

$$\begin{array}{c}\left(M\right(x,y,t)-\epsilon )\ast \left(M\right(z,y,s)-\epsilon )\hfill \\ \le M_H(x,a_1,t_1)\ast M_K(a_1,y,t-t_1)\ast M_H(z,a_2,s_1)\ast M_K(a_2,y,s-s_1)\hfill \\ =M_H(x,a_1,t_1)\ast M_H(z,a_2,s_1)\ast M_K(a_1,y,t-t_1)\ast M_K(a_2,y,s-s_1)\hfill \\ \le M_H(x,a_1,t_1)\ast M_H(z,a_2,s_1)\ast M_K(a_1,a_2,t-t_1+s-s_1)\hfill \\ =M_H(x,a_1,t_1)\ast M_H(z,a_2,s_1)\ast M_H(a_1,a_2,t-t_1+s-s_1)\hfill \\ =M_H(x,a_1,t_1)\ast M_H(a_1,a_2,t-t_1+s-s_1)\ast M_H(z,a_2,s_1)\hfill \\ \le M_H(x,a_2,t+s-s_1)\ast M_H(z,a_2,s_1)\le M_H(x,z,t+s)\hfill \\ =M(x,z,t+s)\hfill \end{array}$$

Since $\epsilon$ is arbitrary and ∗ is continuous, we have

$$\begin{array}{c}M(x,y,t)\ast \left(M\right(y,z,s)\le M(x,z,t+s).\end{array}$$

(The proof of the case $x,z\in K\setminus H$ and $y\in H\setminus K$ similar).

(iii)

Suppose that $x\in H\bigcap K,y\in H\setminus K$ and $z\in K\setminus H$ or $x\in H\bigcap K,z\in H\setminus K$ and $y\in K\setminus H$. As proof of (i), we have

$$\begin{array}{c}M(x,y,t)\ast \left(M\right(y,z,s)\le M(x,z,t+s).\end{array}$$

Secondly, we show that $M(x,y,\_)$ is continuous for $x,y\in H\bigcup K$.

If $x,y\in H$ or $x,y\in K$, then $M(x,y,\_)$ is obviously continuous.

Suppose that $x\in H\setminus K,y\in K\setminus H$, $△t>0$,

$$\begin{array}{cc}\hfill & M(x,y,t+\Delta t)\hfill \\ \hfill & =sup\{M_H(x,a,t_1)\ast M_K(a,y,t+\Delta t-t_1):a\in H\cap K,t_1\in (0,t+\Delta t)\}\hfill \\ \hfill & \ge sup\{M_H(x,a,t_1)\ast M_K(a,y,t+\Delta t-t_1):a\in H\cap K,t_1\in (0,t)\}\hfill \\ \hfill & \ge sup\{M_H(x,a,t_1)\ast M_K(a,y,t-t_1):a\in H\cap K,t_1\in (0,t)\}\hfill \\ \hfill & =M(x,y,t)\hfill \end{array}$$

thus, $M(x,y,t+△t)\ge M(x,y,t)$, which implies that $M(x,y,\_)$ is increasing.

Then, given $\epsilon >0$, we can find $a\in H\cap K,t_1\in (0,t+△t)$, such that

$$\begin{array}{c}\hfill M(x,y,t+△t)<M_H(x,a,t_1)\ast M_K(a,y,t+△t-t_1)+\epsilon .\end{array}$$

If $t_1\in (0,t)$, then by Lemma 1,

$$\begin{array}{cc}\hfill M(x,y,t+△t)& <M_H(x,a,t_1)\ast M_K(a,y,t-t_1+△t)+\epsilon \hfill \\ \hfill & \le M_H(x,a,t_1)\ast M_K(a,y,t-t_1)+△t+\epsilon \hfill \\ \hfill & \le M(x,y,t)+△t+\epsilon \hfill \end{array}$$

If $t_1\in [t,t+△t)$, let $t_2=t_1-△t$, then $t_2\in (0,t)$. By Lemma 1,

$$\begin{array}{cc}\hfill M(x,y,t+△t)& <M_H(x,a,t_2+△t)\ast M_K(a,y,t-t_2)+\epsilon \hfill \\ \hfill & \le M_H(x,a,t_2)\ast M_K(a,y,t-t_2)+△t+\epsilon \hfill \\ \hfill & \le M(x,y,t)+△t+\epsilon \hfill \end{array}$$

Thus, for each $△t>0$, we have

$$M(x,y,t+△t)\le M(x,y,t)+△t+\epsilon$$

Taking the limit as $△t\to 0^{+}$, we obtain

$$\begin{array}{c}\hfill \underset{△t\to 0^{+}}{lim}M(x,y,t+△t)\le M(x,y,t)+\epsilon .\end{array}$$

Since $\epsilon$ is arbitrary, that is

$$\begin{array}{c}\hfill \underset{△t\to 0^{+}}{lim}M(x,y,t+△t)\le M(x,y,t).\end{array}$$

Furthermore, $M(x,y,\_)$ is increasing, we obtain

$$\begin{array}{c}\hfill \underset{△t\to 0^{+}}{lim}M(x,y,t+△t)=M(x,y,t).\end{array}$$

That is, $M(x,y,\_)$ is right continuous.

Next, we prove that $M(x,y,\_)$ is left continuous.

Let $△t<0$, then $-△t>0$. Similarly, from the definition of M, we have

$$M(x,y,t+△t)\le M(x,y,t).$$

Then, given $\epsilon >0$, we can find $a\in H\cap K,t_1\in (0,t)$, such that

$$\begin{array}{c}\hfill M(x,y,t)<M_H(x,a,t_1)\ast M_K(a,y,t-t_1)+\epsilon .\end{array}$$

If $t_1\in (0,t+△t)$, then by Lemma 1,

$$\begin{array}{cc}\hfill M(x,y,t)& <M_H(x,a,t_1)\ast M_K(a,y,t-t_1)+\epsilon \hfill \\ \hfill & =M_H(x,a,t_1)\ast M_K(a,y,t+△t-t_1-△t)+\epsilon \hfill \\ \hfill & \le M(x,y,t+△t)-△t+\epsilon \hfill \end{array}$$

If $t_1\in [t+△t,t)$, let $t_2=t_1+△t$, then $t_2\in (0,t+△t)$. By Lemma 1,

$$\begin{array}{cc}\hfill M(x,y,t)& <M_H(x,a,t_1)\ast M_K(a,y,t-t_1)+\epsilon \hfill \\ \hfill & =M_H(x,a,t_2-△t)\ast M_K(a,y,t-t_2+△t)+\epsilon \hfill \\ \hfill & =M_H(x,a,t_2+(-△t))\ast M_K(a,y,t+△t-t_2)+\epsilon \hfill \\ \hfill & \le M_H(x,a,t_2)\ast M_K(a,y,t+△t-t_2)-△t+\epsilon \hfill \\ \hfill & \le M(x,y,t+△t)-△t+\epsilon \hfill \end{array}$$

Thus, for each $△t<0$, we have

$$M(x,y,t)\le M(x,y,t+△t)-△t+\epsilon .$$

That is,

$$M(x,y,t+△t)\ge M(x,y,t)+△t-\epsilon .$$

Taking the limit as $△t\to 0^{-}$, we obtain

$$\begin{array}{c}\hfill \underset{△t\to 0^{-}}{lim}M(x,y,t+△t)\ge M(x,y,t)-\epsilon .\end{array}$$

Since $\epsilon$ is arbitrary, then

$$\begin{array}{c}\hfill \underset{△t\to 0^{-}}{lim}M(x,y,t+△t)\ge M(x,y,t).\end{array}$$

Furthermore, $M(x,y,\_)$ is increasing, we obtain

$$\begin{array}{c}\hfill \underset{△t\to 0^{-}}{lim}M(x,y,t+△t)=M(x,y,t).\end{array}$$

Which means $M(x,y,\_)$ is also left continuous. Thus, $M(x,y,\_)$ is continuous.

Therefore, M is a fuzzy metric on $H\bigcup K$.

□

Next, we will provide an example to support our results.

Example 2. 

Let $H=[1,3]$ and $K=[2,4]$ with the usual Euclidean metric. Let

$$M_H(x,y,t)=\frac{t}{t+d(x,y)},ifx,y\in H$$

$$M_K(x,y,t)=\frac{t}{t+d(x,y)},ifx,y\in K$$

where $d(x,y)=\mid x-y\mid$, then $(M_H,.)$, $(M_K,.)$ are strong fuzzy metric spaces, where · is the product t-norm.

If $x\in H\setminus K$, $y\in K\setminus H$, then $x<2$, $y>3$. $\forall a\in H\bigcap K=[2,3]$

$$d(x,a)+d(a,y)=(a-x)+(y-a)=y-x=d(x,y)$$

$$d(x,a)·d(a,y)=(a-x)·(y-a)=-{(a-\frac{y+x}{2})}^2+{\left(\frac{y-x}{2}\right)}^2.$$

Since $\frac{y+x}{2}\in [2,3]$ and $a\in [2,3]$, then

$$d(x,a)·d(a,y)\ge min\left\{\right(2-x\left)\right(y-2),(3-x\left)\right(y-3\left)\right\}\triangleq s(x,y).$$

It is obvious that the equal sign in the above inequality can be obtained.

$$\begin{array}{cc}\hfill M_H(x,a,t)\ast M_K(a,y,t)& =\frac{t}{t+d(x,a)}·\frac{t}{t+d(a,y)}\hfill \\ \hfill & =\frac{t^2}{t^2+t(d(x,a)+d(a,y))+d(x,a)d(a,y)}\hfill \\ \hfill & =\frac{t^2}{t^2+t\mid y-x\mid +d(x,a)d(a,y)}\hfill \\ \hfill & \le \frac{t^2}{t^2+td(x,y)+s(x,y)}\hfill \end{array}$$

Now, define the fuzzy set M as follows:

$$\begin{array}{cc}M(x,y,t)\hfill & =\left\{\begin{array}{cc}{M}_H(x,y,t),\hfill & ifx,y\in H;\hfill \\ M_K(x,y,t),\hfill & ifx,y\in K;\hfill \\ sup\{M_H(x,a,t)\ast M_K(a,y,t):a\in H\bigcap K\},\hfill & ifx\in H\setminus K,y\in K\setminus H.\hfill \end{array}\right.\hfill \\ \hfill & =\left\{\begin{array}{cc}\frac{t}{t+d(x,y)},\hfill & ifx,y\in H;\hfill \\ \frac{t}{t+d(x,y)},\hfill & ifx,y\in K;\hfill \\ \frac{t^2}{t^2+td(x,y)+s(x,y)},\hfill & ifx\in H\setminus K,y\in K\setminus H.\hfill \end{array}\right.\hfill \end{array}$$

Then by Theorem 2, M is a fuzzy metric on $H\bigcup K$ such that ${M|}_H=M_H$ and ${M|}_K=M_K$.

3. Conclusions

For two compatible fuzzy metrics $M_H$ and $M_K$ on H and K, respectively, two sufficient additions for the existence of a fuzzy metric M on $H\bigcup K$ such that ${M|}_H=M_H$ and ${M|}_K=M_K$ are given. One is that t-norm ∗ is positive and fuzzy metrics $M_H,M_K$ are strong, and the other is that t-norm ∗ is positive and t-norm ∗, fuzzy metrics $M_H,M_K$ satisfy the Lipschitz conditions.