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A family of fuzzy distance measures of fuzzy numbers

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Abstract

Recently, Guha and Chakraborty studied a new approach to fuzzy distance measure and similarity measure between two generalized fuzzy numbers. In this paper, we propose a family of fuzzy distance measures between arbitrary fuzzy numbers depending on a defuzzification, an interval similarity measure, and some real pseudosemimetrics, involving some of the most important possibilistic and geometric characteristics of any fuzzy number. Then, we study its metric properties and its relationship with a partial order on the set of all trapezoidal fuzzy numbers. Illustrative examples about the action of this class of metrics are shown, comparing them with the results obtained by other researchers. Finally, as application, we introduce a fuzzy ranking and we compare it with other techniques.

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Notes

  1. We use the term radius in the sense that if \(c\) is the center and \(r\) is the radius, then the interval is \(\left[ c-r,c+r\right] \).

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Acknowledgments

We are very thankful to the anonymous reviewers for their careful reading of our manuscript and for their constructive reports, which have been very useful to improve the paper.

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Correspondence to Antonio-Francisco Roldán-López de Hierro.

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Communicated by V. Loia.

Appendix A: Proofs

Appendix A: Proofs

For the sake of completeness, we include here the proofs of lemmas we have announced throughout the paper.

Proof

(Lemma 2.5, page 4) It is clear that:

$$\begin{aligned} \mathcal {D}(\mathcal {A}+\mathcal {B})&= \left( \beta _{i}~\underline{\mathcal {A}+\mathcal {B}}_{\alpha _{i}}+\gamma _{i}~\overline{\mathcal {A}+\mathcal {B}}_{\alpha _{i}}\right) \\&= \sum \limits _{i=1}^{n} \left( \beta _{i}(\underline{a}_{\alpha _{i}}+\underline{b}_{\alpha _{i} })+\gamma _{i}(\overline{a}_{\alpha _{i}}+\overline{b}_{\alpha _{i}})\right) \\&= \sum \limits _{i=1}^{n} \left( \beta _{i}\underline{a}_{\alpha _{i}}+\gamma _{i}\overline{a}_{\alpha _{i}}\right) + \textstyle \sum \limits _{i=1}^{n} \left( \beta _{i}\underline{b}_{\alpha _{i}}+\gamma _{i}\overline{b}_{\alpha _{i}}\right) \\&=\mathcal {DA}+\mathcal {DB}. \end{aligned}$$

Similarly,

$$\begin{aligned} \mathcal {D}(r\mathcal {A})&= \sum \limits _{i=1}^{n} \left( \beta _{i}\underline{r\mathcal {A}}_{\alpha _{i}}+\gamma _{i} \overline{r\mathcal {A}}_{\alpha _{i}}\right) \\&= \sum \limits _{i=1}^{n} \left( \beta _{i}r\underline{\mathcal {A}}_{\alpha _{i}}+\gamma _{i} r\overline{\mathcal {A}}_{\alpha _{i}}\right) \\&=r~ \sum \limits _{i=1}^{n} \left( \beta _{i}\underline{a}_{\alpha _{i}}+\gamma _{i}\overline{a}_{\alpha _{i}}\right) =r\,\mathcal {DA}. \end{aligned}$$

As \(\mathcal {D}_{c}\mathcal {A}=(1/2)\underline{a}_{1}+(1/2)\overline{a}_{1}\) for all \(\mathcal {A}\in \mathcal {F}\) (that is, \(n=1\) and \(\beta _{1}=\gamma _{1}=1/2\)), the same properties hold for \(\mathcal {D}_{c}\). \(\square \)

Proof

(Lemma 2.6, page 4) (1) By definition,

$$\begin{aligned} \ker (\mathcal {A}+\mathcal {B})&=[\underline{\mathcal {A}+\mathcal {B}} _{1},\overline{\mathcal {A}+\mathcal {B}}_{1}]=[\underline{a}_{1}+\underline{b}_{1},\overline{a}_{1}+\bar{b}_{1}]\\&=[\underline{a}_{1},\overline{a}_{1}]+[\underline{b}_{1},\bar{b}_{1} ]=\ker \mathcal {A}+\ker \mathcal {B}. \end{aligned}$$

(2) Similarly,

$$\begin{aligned} \hbox {supp}(\mathcal {A}+\mathcal {B})&=[\underline{\mathcal {A} +\mathcal {B}}_{0},\overline{\mathcal {A}+\mathcal {B}}_{0}]=[\underline{a} _{0}+\underline{b}_{0},\overline{a}_{0}+\bar{b}_{0}]\\&=[\underline{a}_{0},\overline{a}_{0}]+[\underline{b}_{0},\bar{b} _{0}]=\hbox {supp}\mathcal {A}+\hbox {supp}\mathcal {B}. \end{aligned}$$

(3) Similarly,

$$\begin{aligned} \hbox {spr}(\mathcal {A}\!+\!\mathcal {B})&\!=\!(\overline{\mathcal {A} \!+\!\mathcal {B}}_{1}\!+\!\underline{\mathcal {A}\!+\!\mathcal {B}}_{1})/2\!=\!(\overline{a} _{1}\!+\!\bar{b}_{1}\!-\!\underline{a}_{1}-\underline{b}_{1})/2\\&=(\overline{a}_{1}-\underline{a}_{1})/2+(\bar{b}_{1}-\underline{b} _{1})/2=\hbox {spr}\mathcal {A}+\hbox {spr}\mathcal {B}. \end{aligned}$$

\(\square \)

Proof

(Example 3.3, page 5) It is clear that \(\phi (x,y)=0\) iff \(x=y\) and \(\phi (x,y)=\phi (y,x)\). If \(p=1\), then \(q\left| x-z\right| \le q\left| x-y\right| +q\left| y-z\right| \). Suppose that \(0<p<1\) and we claim that

$$\begin{aligned} \left| x-z\right| ^{p}\le \left| x-y\right| ^{p}+\left| y-z\right| ^{p}. \end{aligned}$$
(8)

Fix \(x,z\in \mathbb {R}\). If \(x=z\), then (8) is trivial. Suppose that \(x<z\) without lost of generality. Consider the mapping \(h(y)=\left| x-y\right| ^{p}+\left| y-z\right| ^{p}-\left| x-z\right| ^{p}\) for all \(y\in \mathbb {R}\). Clearly, \(h\) is continuous on \(\mathbb {R}\) and \(h(x)=h(z)=0\). Let \(a=\left( z-x\right) ^{p}\). Then, we can decompose:

$$\begin{aligned} h(y)=\left\{ \begin{array}{ll} \left( x-y\right) ^{p}+\left( z-y\right) ^{p}-a, &{} \quad \hbox {if}\, y<x,\\ \left( y-x\right) ^{p}+\left( z-y\right) ^{p}-a, &{}\quad \hbox {if}\, x\le y\le z,\\ \left( y-x\right) ^{p}+\left( y-z\right) ^{p}-a, &{} \quad \hbox {if}\, y>z. \end{array}\right. \end{aligned}$$

It is not difficult to prove that \(h^{\prime }(y)<0\) for all \(y<x\) and \(h^{\prime }(y)>0\) for all \(y>z\). Therefore, \(h\) is strictly decreasing on \(\left] -\infty ,x\right[ \) and strictly increasing on \(\left] z,\infty \right[ \). Therefore, \(h\) has al least one absolute minimum, which is on \(\left[ x,z\right] \). As \(h^{\prime \prime }(y)<0\) for all \(y\in \left] x,z\right[ \), its minimum is not in \(\left] x,z\right[ \), but in \(\left\{ x,z\right\} \). As \(h(x)=h(z)=0\), then \(h(y)\ge 0\) for all \(y\in \mathbb {R}\), so (8) is true. \(\square \)

Proof

(Lemma 3.6, page 6) Suppose that \(I=[a,b]\), \(I^{\prime }=[a^{\prime },b^{\prime }]\), \(J=[c,e]\), \(J^{\prime }=[c^{\prime },e^{\prime }]\) and \(K=[f,g]\).

(1) As \(\underline{\delta }(I,J)=-\left| a-c\right| \le 0\) and \(\overline{\delta }(I,J)=\left| b-e\right| \ge 0\), then \(0_{\mathcal {I}}=[0,0]\subseteq \delta (I,J)\). Therefore, \(\delta (I,J)\in \mathcal {I}_{0_{\mathcal {I}},\subseteq }^{+}\). It is clear that:

$$\begin{aligned} \delta (I,J)&= 0_{\mathcal {I}}~\Leftrightarrow ~\left| a-c\right| =\left| b-e\right| =0\\&\Leftrightarrow \left[ ~a=c,~b=e~\right] ~\Leftrightarrow ~I=J. \end{aligned}$$

Moreover, \(\delta \) is symmetric. Finally, as \(\left| a-f\right| \le \left| a-c\right| +\left| c-f\right| \) (so \(-\left| a-f\right| \ge -\left| a-c\right| -\left| c-f\right| \)) and \(\left| b-g\right| \le \left| b-e\right| +\left| e-g\right| \)

$$\begin{aligned} \delta (I,K)&=[-\left| a-f\right| ,\left| b-g\right| ]\\&\subseteq [-\left| a-c\right| -\left| c-f\right| ,\left| b-e\right| +\left| e-g\right| ]\\&=[-\left| a-c\right| ,\left| b-e\right| ]+[-\left| c-f\right| ,\left| e-g\right| ]\\&=\delta (I,J)+\delta (J,K). \end{aligned}$$

(2) If \(I\subseteq I^{\prime }\) and \(J\subseteq J^{\prime }\), then \(a^{\prime }\le a\le b\le b^{\prime }\) and \(c^{\prime }\le c\le e\le e^{\prime }\). Then, \(a^{\prime }+c^{\prime }\le a+c\le b+e\le b^{\prime }+e^{\prime }\), so

$$\begin{aligned} \delta (I,I^{\prime })+\delta (J,J^{\prime })&=\delta ([a,b],[a^{\prime },b^{\prime }])+\delta ([c,e],[c^{\prime },e^{\prime }])\\&=[a^{\prime }-a,b^{\prime }-b]+[c^{\prime }-c,e^{\prime }-e]\\&=[a^{\prime }-a+c^{\prime }-c,b^{\prime }-b+e^{\prime }-e]\\&=[(a^{\prime }\!+\!c^{\prime })\!-\!(a\!+\!c),(b^{\prime }\!+\!e^{\prime })\!-\!(b\!+\!e)]\\&=\delta ([a+c,b+e],[a^{\prime }+c^{\prime },b^{\prime }+e^{\prime }])\\&=\delta ([a,b]+[c,e],[a^{\prime },b^{\prime }]+[c^{\prime },e^{\prime }])\\&=\delta (I+J,I^{\prime }+J^{\prime }). \end{aligned}$$

(3)

$$\begin{aligned} \delta (I+K,J+K)&=\delta ([a,b]+[f,g],[c,e]+[f,g])\\&=\delta ([a+f,b+g],[c+f,e+g])\\&=[-\left| (a+f)-(c+f)\right| ,\\&\quad \left| (b+g)-(e+g)\right| ]\\&=[-\left| a\!-\!c\right| ,\left| b-e\right| ]\\&=\delta ([a,b],[c,e])\\&=\delta (I,J). \end{aligned}$$

(4) If \(I\subseteq J\cap K\), then \(c\le a\le b\le e\) and \(f\le a\le b\le g\). Then, \([c-a,e-b]=\delta (I,J)=\delta (I,K)=[f-a,g-b]\), so \(J=[c,e]=[f,g]=K\).

(5) If \(I\subseteq J\), then \(c\le a\le b\le e\). Therefore, then \(c+f\le a+f\le b+f\le b+g\le e+g\), so then \(I+K=[a+f,b+g]\subseteq [c+f,e+g]=J+K\). \(\square \)

Proof

(Lemma 4.8, page 10) Since \(\phi _{1}\), \(\phi _{3}\), \(\phi _{4}\) and \(\psi _{2}\) are pseudosemimetrics, then

$$\begin{aligned} \phi (\mathcal {DA},\mathcal {DA})=0\quad \text {and}\quad \phi (\mathcal {DA} ,\mathcal {DB})=\phi (\mathcal {DB},\mathcal {DA}) \end{aligned}$$

for all \(\phi \in \{\phi _{1},\phi _{3},\phi _{4},\psi _{2}\}\) and all \(\mathcal {A},\mathcal {B}\in \mathcal {F}\). As a consequence, for all \(\alpha \in \mathbb {I}\),

$$\begin{aligned} \underline{D(\mathcal {A},\mathcal {A})}_{\,\alpha }&=q_{1}\phi _{1}(\mathcal {DA},\mathcal {DA})-q_{2}\psi _{2}\!~(\hbox {spr} \mathcal {A},\hbox {spr}\mathcal {A})-h_{1}(\alpha )\\&\phi _{3} \!~(\underline{\delta }(\ker \mathcal {A},\hbox {supp}\mathcal {A} ),\underline{\delta }(\ker \mathcal {A},\hbox {supp}\mathcal {A}))\\&=q_{1}0-q_{2}0-h_{1}(\alpha )~0=0, \end{aligned}$$

and similarly \(\overline{D(\mathcal {A},\mathcal {A})}_{\alpha }=0\). Therefore, \(D(\mathcal {A},\mathcal {A})=\tilde{0}\). Furthermore, as \(\delta \) is also symmetric,

$$\begin{aligned} \underline{D(\mathcal {B},\mathcal {A})}_{\,\alpha }&=q_{1}\phi _{1}(\mathcal {DB},\mathcal {DA})-q_{2}\psi _{2}\!~(\hbox {spr} \mathcal {B},\hbox {spr}\mathcal {A})-h_{1}(\alpha )\\&\quad \phi _{3} \!~(\underline{\delta }(\ker \mathcal {B},\hbox {supp}\mathcal {B} ),\underline{\delta }(\ker \mathcal {A},\hbox {supp}\mathcal {A}))\\&=q_{1}\phi _{1}(\mathcal {DA},\mathcal {DB})-q_{2}\psi _{2} \!~(\hbox {spr}\mathcal {A},\hbox {spr}\mathcal {B} )-h_{1}(\alpha )\\&\quad \phi _{3}\!~(\underline{\delta }(\ker \mathcal {A} ,\hbox {supp}\mathcal {A}),\underline{\delta }(\ker \mathcal {B} ,\hbox {supp}\mathcal {B}))\\&=\underline{D(\mathcal {A},\mathcal {B})}_{\,\alpha }, \end{aligned}$$

and similarly \(\overline{D(\mathcal {B},\mathcal {A})}_{\alpha }=\overline{D(\mathcal {A},\mathcal {B})}_{\alpha }\) for all \(\alpha \in \mathbb {I}\). Hence, \(D(\mathcal {B},\mathcal {A})=D(\mathcal {A},\mathcal {B})\).

Proof

(Lemma 4.10, page 10) (1) As \(h_{1}(1)=h_{2}(1)=0\), it is clear that:

$$\begin{aligned}&\ker D(\mathcal {A},\mathcal {B})=[\underline{D(\mathcal {A},\mathcal {B})} _{1},\overline{D(\mathcal {A},\mathcal {B})}_{1}]=[q_{1}\phi _{1}(\mathcal {DA} ,\mathcal {DB}) \\&-q_{2}\psi _{2}\!\,(\hbox {spr}\mathcal {A} ,\hbox {spr}\mathcal {B}),q_{1}\phi _{1}(\mathcal {DA},\mathcal {DB} )+q_{2}\psi _{2}\!~(\hbox {spr}\mathcal {A},\hbox {spr} \mathcal {B})]. \end{aligned}$$

(2) From (1):

$$\begin{aligned} \mathcal {D}_{c}(D(\mathcal {A},\mathcal {B}))&= \frac{\underline{D(\mathcal {A},\mathcal {B})}_{\,1}+\overline{D(\mathcal {A},\mathcal {B})}_{1}}{2}\\&= \frac{1}{2}\left[ {q_{1}\phi _{1}(\mathcal {DA},\mathcal {DB})-q_{2}\psi _{2} (\hbox {spr}\mathcal {A},\hbox {spr}\mathcal {B})}\right. \\&\quad \left. {+q_{1} \phi _{1}(\mathcal {DA},\mathcal {DB})+q_{2}\psi _{2} (\hbox {spr} \mathcal {A},\hbox {spr}\mathcal {B})}\right] \\&= q_{1}\phi _{1}(\mathcal {DA},\mathcal {DB}). \end{aligned}$$

(3) Similarly:

$$\begin{aligned} \hbox {spr}D(\mathcal {A},\mathcal {B})&= \frac{\overline{D(\mathcal {A},\mathcal {B})}_{1}-\underline{D(\mathcal {A},\mathcal {B})}_{1} }{2}\\&= \frac{1}{2}\left[ {q_{1}\phi _{1}(\mathcal {DA},\mathcal {DB})+q_{2}\psi _{2} (\hbox {spr}\mathcal {A},\hbox {spr}\mathcal {B})}\right. \\&\left. {-\left( q_{1}\phi _{1}(\mathcal {DA},\mathcal {DB})-q_{2}\psi _{2}\!\,(\hbox {spr} \mathcal {A},\hbox {spr}\mathcal {B})\right) }\right] \\&= q_{2}\psi _{2}\!~(\hbox {spr}\mathcal {A},\hbox {spr} \mathcal {B}). \end{aligned}$$

(4) If \(\mathcal {D}=\mathcal {D}_{c}\), then item 2 shows us that \(\mathcal {D}_{c}(D(\mathcal {A} ,\mathcal {B}))=q_{1}\phi _{1}(\mathcal {DA},\mathcal {DB})\ge 0\). Hence, by item 3 of Lemma 4.2, \(D(\mathcal {A} ,\mathcal {B})\succcurlyeq \tilde{0}\) w.r.t. \((\mathcal {D}_{c},\delta ,\subseteq )\). \(\square \)

Proof

(Statement before Theorem 4.13) Suppose that \(\mathcal {A}=(a/b/c/e)\) and \(\mathcal {B}=(a^{\prime }/b^{\prime }/c^{\prime }/e^{\prime })\). If \(D(\mathcal {A} ,\mathcal {B})\) is a crisp FN, then

$$\begin{aligned}&\psi _{2}\!~(\hbox {spr}\mathcal {A},\hbox {spr}\mathcal {B}) =\phi _{3}\!~(\underline{d}(\ker \mathcal {A},\hbox {supp}\mathcal {A} ),\underline{d}(\ker \mathcal {B},\hbox {supp}\mathcal {B}))\\&\quad =\phi _{4}\!~(\!~\overline{\delta }(\ker \mathcal {A},\hbox {supp} \mathcal {A}),\overline{\delta }(\ker \mathcal {B},\hbox {supp} \mathcal {B}))=0. \end{aligned}$$

This means that \(\hbox {spr}\mathcal {A}=\hbox {spr} \mathcal {B}\) and \(\delta (\ker \mathcal {A},\hbox {supp}\mathcal {A} )=\delta (\ker \mathcal {B},\hbox {supp}\mathcal {B})\), i.e.,

$$\begin{aligned} \frac{c-b}{2}=\frac{c^{\prime }-b^{\prime }}{2},\quad b-a=b^{\prime }-a^{\prime }\quad \text {and}\quad e-c=e^{\prime }-c^{\prime }. \end{aligned}$$

If \(r=a-a^{\prime }\), then \(b-b^{\prime }=c-c^{\prime }=e-e^{\prime }=r\), so \(a=r+a^{\prime }\), \(b=r+b^{\prime }\), \(c=r+c^{\prime }\) and \(e=r+e^{\prime }\). \(\square \)

Proof

(Theorem 4.13, page 12) Firstly, we are going to characterize:

$$\begin{aligned} N_{r}(\mathcal {A})&=\{\mathcal {B}\in \mathcal {T}:D(\mathcal {A} ,\mathcal {B})\prec \tilde{r}\}=\{\mathcal {B}\in \mathcal {T}:\mathcal {B}\\&=(\mathcal {D}_{c}\mathcal {B} -\mathcal {D}_{c}\mathcal {A})+\mathcal {A}\text { and }q_{1}\phi _{1} (\mathcal {D}_{c}\mathcal {A},\mathcal {D}_{c}\mathcal {B})<r\}. \end{aligned}$$

Indeed, let \(\mathcal {B}\in N_{r}(\mathcal {A})\). As \(D(\mathcal {A} ,\mathcal {B})\prec \tilde{r}\), then

$$\begin{aligned}&\bullet \quad q_{1}\phi _{1}(\mathcal {D}_{c}\mathcal {A},\mathcal {D} _{c}\mathcal {B})=\mathcal {D}_{c}(D(\mathcal {A},\mathcal {B}))\le \mathcal {D}_{c}\tilde{r}=r;\\&\bullet \quad 0\le q_{2}\psi _{2}\!~(\hbox {spr}\mathcal {A} ,\hbox {spr}\mathcal {B})=\hbox {spr}D(\mathcal {A} ,\mathcal {B})\le \hbox {spr}\tilde{r}=0\quad \\&\qquad \Rightarrow \quad \hbox {spr}\mathcal {A}=\hbox {spr}\mathcal {B};\\&\bullet \quad 0_{\mathcal {I}}\subseteq \delta (\ker D(\mathcal {A} ,\mathcal {B}),\hbox {supp}D(\mathcal {A},\mathcal {B}))\subseteq \delta (\ker \tilde{r},\hbox {supp}\tilde{r})\\&\qquad =\delta ([r,r],[r,r])=0_{\mathcal {I}}\\&\qquad \qquad \Rightarrow \quad \ker D(\mathcal {A},\mathcal {B} )=\hbox {supp}D(\mathcal {A},\mathcal {B})\\&\qquad \qquad \Rightarrow \quad D(\mathcal {A},\mathcal {B})\text { is rectangular.} \end{aligned}$$

As \(D(\mathcal {A},\mathcal {B})\) is rectangular but \(\hbox {spr} D(\mathcal {A},\mathcal {B})=0\), then \(D(\mathcal {A},\mathcal {B})\) is a crisp FN. In this case, we have just proved that \(\mathcal {B}\) is a translation of \(\mathcal {A}\), i.e., there exists \(s\in \mathbb {R}\) such that \(\mathcal {B} =s+\mathcal {A}\). Taking defuzzification, \(\mathcal {D}_{c}\mathcal {B} =\mathcal {D}_{c}(s+\mathcal {A})=\mathcal {D}_{c}s+\mathcal {D}_{c} \mathcal {A}=s+\mathcal {D}_{c}\mathcal {A}\), so \(s=\mathcal {D}_{c} \mathcal {B}-\mathcal {D}_{c}\mathcal {A}\). If \(q_{1}\phi _{1}(\mathcal {D} _{c}\mathcal {A},\mathcal {D}_{c}\mathcal {B})=r\), then \(D(\mathcal {A} ,\mathcal {B})\preccurlyeq \tilde{r}\) and \(\tilde{r}\preccurlyeq D(\mathcal {A} ,\mathcal {B})\), so \(D(\mathcal {A},\mathcal {B})=\tilde{r}\), but we suppose that \(D(\mathcal {A},\mathcal {B})\prec \tilde{r}\). Then, \(q_{1}\phi _{1} (\mathcal {D}_{c}\mathcal {A},\mathcal {D}_{c}\mathcal {B})<r\).

Next, we are going to see that \(\beta _{\mathcal {A}}=\{N_{r}(\mathcal {A} )\}_{r>0}\) is a neighborhood system at \(\mathcal {A}\). We have to prove three properties.

$$\begin{aligned}&\bullet \quad \mathcal {A}\in U\text { for all }U\in \beta _{\mathcal {A}}.\\&\bullet \quad \text {For all }U,V\in \beta _{\mathcal {A}}\text {, there exists }W\in \beta _{\mathcal {A}}\text { such that }\\&\qquad W\subseteq U\cap V.\\&\bullet \quad \text {For all }U\in \beta _{\mathcal {A}}\text {, there exists }U_{0}\in \beta _{\mathcal {A}}\text { such that for all }\\&\qquad \mathcal {B}\in U_{0}\text { there exists }V\in \beta _{\mathcal {B}}\text { such that }V\subseteq U. \end{aligned}$$

First two properties are trivial, since \(\mathcal {A}\in N_{r}(\mathcal {A})\) for all \(r>0\) and \(N_{r_{3}}(\mathcal {A})\subseteq N_{r_{1}}(\mathcal {A})\cap N_{r_{2}}(\mathcal {A})\) if \(0<r_{3}<\min (r_{1},r_{2})\). Let \(U=U_{0} =N_{r}(\mathcal {A})\) and let \(\mathcal {B}\in N_{r}(\mathcal {A})\). If \(\mathcal {B}=\mathcal {A}\), we can take \(V=U\). Suppose that \(\mathcal {B} \ne \mathcal {A}\). As \(q_{1}\phi _{1}(\mathcal {D}_{c}\mathcal {A},\mathcal {D} _{c}\mathcal {B})<r\), let \(s,\varepsilon >0\) such that \(q_{1}\phi _{1} (\mathcal {D}_{c}\mathcal {A},\mathcal {D}_{c}\mathcal {B})<s<s+\varepsilon <r\). Consider \(V=N_{\varepsilon }(\mathcal {B})\in \beta _{\mathcal {B}}\) and we have to prove that \(N_{\varepsilon }(\mathcal {B})\subseteq N_{r}(\mathcal {A})\). Indeed, let \(\mathcal {C}\in N_{\varepsilon }(\mathcal {B})\). We know that \(\mathcal {C} =(\mathcal {D}_{c}\mathcal {C}-\mathcal {D}_{c}\mathcal {B})+\mathcal {B}\) and \(q_{1}\phi _{1}(\mathcal {D}_{c}\mathcal {B},\mathcal {D}_{c}\mathcal {C} )<\varepsilon \). Then:

$$\begin{aligned} q_{1}\phi _{1}(\mathcal {D}_{c}\mathcal {A},\mathcal {D}_{c}\mathcal {C})&\le q_{1}\phi _{1}(\mathcal {D}_{c}\mathcal {A},\mathcal {D}_{c}\mathcal {B})+q_{1} \phi _{1}(\mathcal {D}_{c}\mathcal {B},\mathcal {D}_{c}\mathcal {C})\\&<s+\varepsilon <r. \end{aligned}$$

On the other hand,

$$\begin{aligned} \mathcal {C}&=(\mathcal {D}_{c}\mathcal {C}-\mathcal {D}_{c}\mathcal {B} )+\mathcal {B}\\&=(\mathcal {D}_{c}\mathcal {C}-\mathcal {D}_{c}\mathcal {B})+\left[ (\mathcal {D}_{c}\mathcal {B}-\mathcal {D}_{c}\mathcal {A})+\mathcal {A}\right] \\&=\left[ (\mathcal {D}_{c}\mathcal {C}-\mathcal {D}_{c}\mathcal {B} )+(\mathcal {D}_{c}\mathcal {B}-\mathcal {D}_{c}\mathcal {A})\right] +\mathcal {A}\\&=(\mathcal {D}_{c}\mathcal {C}-\mathcal {D}_{c}\mathcal {A})+\mathcal {A}, \end{aligned}$$

where we have used properties like the following

$$\begin{aligned} a+(b+\mathcal {A})=(a+b)+\mathcal {A},\quad a+\mathcal {A}=a+\mathcal {B} \Rightarrow \mathcal {A}=\mathcal {B}\end{aligned}$$

(that are easy to prove in general). Therefore, \(\mathcal {C}\in N_{r} (\mathcal {A})\) and we have proved that \(N_{\varepsilon }(\mathcal {B})\subseteq N_{r}(\mathcal {A})\). General Topology guarantees that there exists an unique topology on \(\mathcal {T}\) such that \(\beta _{\mathcal {A}}=\{N_{1/n} (\mathcal {A})\}_{n\in \mathbb {N}}\) is a (countable) neighborhood system at \(\mathcal {A}\). We only have to prove that it is Hausdorff.

Let \(\mathcal {A},\mathcal {B}\in \mathcal {T}\) such that \(\mathcal {A} \ne \mathcal {B}\). We distinguish two cases. Firstly, suppose that \(\mathcal {D}_{c}\mathcal {A}\ne \mathcal {D}_{c}\mathcal {B}\) and let \(r=q_{1}\phi _{1}(\mathcal {D}_{c}\mathcal {A},\mathcal {D}_{c}\mathcal {B})/4>0\). To see that \(N_{r}(\mathcal {A})\cap N_{r}(\mathcal {B})=\varnothing \), suppose that \(\mathcal {C}\in N_{r}(\mathcal {A})\cap N_{r}(\mathcal {B})\). In this case,

$$\begin{aligned} 4r&=q_{1}\phi _{1}(\mathcal {D}_{c}\mathcal {A},\mathcal {D}_{c}\mathcal {B})\\&\le q_{1}\phi _{1}(\mathcal {D}_{c}\mathcal {A},\mathcal {D}_{c} \mathcal {C})+q_{1}\phi _{1}(\mathcal {D}_{c}\mathcal {C},\mathcal {D} _{c}\mathcal {B})\\&<r+r=2r, \end{aligned}$$

but this is impossible. Therefore, \(N_{r}(\mathcal {A})\cap N_{r} (\mathcal {B})=\varnothing \).

Now suppose that \(\mathcal {D}_{c}\mathcal {A}=\mathcal {D}_{c}\mathcal {B}\). As \(\mathcal {C}\in N_{r}(\mathcal {A})\cap N_{r}(\mathcal {B})\), then \(\mathcal {C}=(\mathcal {D}_{c}\mathcal {C}-\mathcal {D}_{c}\mathcal {A} )+\mathcal {A}\) and \(\mathcal {C}=(\mathcal {D}_{c}\mathcal {C}-\mathcal {D} _{c}\mathcal {B})+\mathcal {B}\). Therefore, we can prove that \(\mathcal {A} =(\mathcal {D}_{c}\mathcal {A}-\mathcal {D}_{c}\mathcal {C})+\mathcal {C}\) and \(\mathcal {B}=(\mathcal {D}_{c}\mathcal {B}-\mathcal {D}_{c}\mathcal {C} )+\mathcal {C}\) (it is not difficult to prove that if \(\mathcal {A} =s+\mathcal {B}\), then \(\mathcal {B}=(-s)+\mathcal {A}\)). Since \(\mathcal {D} _{c}\mathcal {A}=\mathcal {D}_{c}\mathcal {B}\), it follows that \(\mathcal {A} =\mathcal {B}\), but this is impossible. Therefore, \(N_{r}(\mathcal {A})\cap N_{r}(\mathcal {B})=\varnothing \) in any case and the topology is Hausdorff. \(\square \)

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Aguilar-Peña, C., Roldán-López de Hierro, AF., Roldán-López de Hierro, C. et al. A family of fuzzy distance measures of fuzzy numbers. Soft Comput 20, 237–250 (2016). https://doi.org/10.1007/s00500-014-1497-0

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