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Prove that two fuzzy sets are disjoint if and only if their supports are disjoint.

Given two fuzzy sets $A,B$ of a reference set $X$,then :

$$ \begin{align} \\ &\text{Supp}(A) \cap \text{Supp}(B)= \emptyset\\ &\iff \nexists x \in X:x \in \text{Supp}(A) \wedge x \in \text{Supp}(B)\tag1\\ &\iff \nexists x \in X:\mu_A(x)>0 \wedge \mu_B(x)>0\tag2\\ &\iff A \cap B=\emptyset \tag3 \end{align} $$

$\hphantom{0}(1)$ Definition of $\cap$

$\hphantom{0}(2)$ Definition of $\text{Supp}$

$\hphantom{0}(3)$ Definition of disjoint fuzzy sets

The other way I came up with was:

$$\text{Supp}(A)=\{x \in X : \mu_{A}(x) >0\}=\{x \in X : x \in A\}=A \tag{I}$$

Hence if $A\cap B=\emptyset $ then it's enough to set $\text{Supp}(A)=A$ and $\text{Supp}(B)=B$, the other direction is the same, and then we conclude the result.

I want to know that how much of my work is correct, I think the first one is correct, but about the other one I doubt if for all fuzzy sets $(\text{I})$ does hold.

masaheb
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1 Answers1

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$\DeclareMathOperator{Supp}{Supp}$(I) does not make sense. $A$ is a fuzzy set, while $\Supp(A)$ is a set. So saying $\Supp(A) = A$ doesn't "type check".

Your proof is actually almost correct. Here is the complete proof.

$\begin{equation} \begin{split} \Supp(A) \cap \Supp(B) = \emptyset &\iff \forall x \in X . \neg (x \in \Supp (A) \land x \in \Supp (B)) \\ &\iff \forall x \in X . \neg (\mu_A(x) > 0 \land \mu_B(x) > 0) \\ &\iff \forall x \in X . \min(\mu_A(x), \mu_B(x)) \leq 0 \\ &\iff \forall x \in X . \min(\mu_A(x), \mu_B(x)) = 0 \\ &\iff \forall x \in X . \mu_{A \cap B}(x) = 0 \\ &\iff \forall x \in X . \mu_{A \cap B}(x) = \mu_\emptyset(x) \\ &\iff A \cap B = \emptyset \end{split} \end{equation} $

It's easier to do the proof using $\forall x \in X$ rather than $\nexists x \in X$.

Your proof is correct that $\Supp(A) \cap \Supp(B) = \emptyset \iff (1) \iff (2)$. But you did not actually prove that $(2) \iff A \cap B = \emptyset$. This is the gap which my proof fills in.

Mark Saving
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  • I don't understand how $$\forall x \in X . \neg (\mu_A(x) > 0 \land \mu_B(x) > 0) \iff \forall x \in X . \min(\mu_A(x), \mu_B(x)) \leq 0$$ – masaheb Oct 04 '21 at 17:16
  • @masaheb Suppose that $\min(\mu_A(x), \mu_B(x)) > 0$. Then we have $\mu_A(x) \geq \min(\mu_A(x), \mu_B(x)) > 0$, and similarly for $\mu_B(x)$. Conversely, if both $\mu_A(x) > 0$ and $\mu_B(x) > 0$, then $\min(\mu_A(x), \mu_B(x)) > 0$. So we see that $(\mu_A(x) > 0 \land \mu_B(x) > 0) \iff \min(\mu_A(x), \mu_b(x)) > 0$. Negating both sides gives us $(\neg (\mu_A(x) > 0 \land \mu_B(x) > 0)) \iff \min(\mu_A(x), \mu_b(x)) \leq 0$. – Mark Saving Oct 04 '21 at 17:18
  • About the part $(2) \iff A \cap B = \emptyset$, indeed I used the definition of disjoint fuzzy sets given by https://en.wikipedia.org/wiki/Fuzzy_set , So I thought that since this is a definition so we don't need to prove it. – masaheb Oct 05 '21 at 07:26
  • I understand that $\text{Supp}(A)$ is a non-fuzzy set and $A$ is a fuzzy set, so they cannot be equal, but assume for an arbitrary $x \in X:x \in \text{Supp}(A)$ then $\mu_A(x)>0$ which implies that $x \in A$, on the other hand assume that $x \in A$ then $\mu_A(x)>0$ which shows that $x \in \text{Supp}(A)$ so $\text{Supp}(A) \subseteq A$ and $\text{Supp}(A) \supseteq A$ which shows that $\text{Supp}(A) = A$. I know that this is a wrong proof which leads to a wrong conclusion, but where was I wrong in my proof? – masaheb Oct 05 '21 at 07:37
  • I'm giving you the bounty, but still I'm looking for the answers. – masaheb Oct 05 '21 at 16:33
  • @masaheb "but assume for an arbitrary $x \in X : x \in Supp(A)$ then $\mu_A(x) > 0$ which implies that $x \in A$" The statement "$x \in A$" is not well-defined. $A$ is a fuzzy set, not a set. The point of fuzzy sets is that instead of membership being binary, membership can take an intermediate value between $0$ and $1$. – Mark Saving Oct 05 '21 at 17:16
  • @masaheb There are multiple possible equivalent definitions of "disjoint". Because you wrote $A \cap B = \emptyset$, I assumed that this was the definition of disjointness. If you used a different definition of disjointness, it doesn't make much sense to write $A \cap B = \emptyset$ at all since this would be a statement which isn't necessary to prove disjointness. – Mark Saving Oct 05 '21 at 17:17