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How can I prove that all the a-cuts of any fuzzy set A defined on $R^n$ are convex if and only if

$$\mu_A(\lambda r + (1-\lambda)s) \geq min \{\mu_A(r), \mu_A(s)\}$$

such that $r, s \in R^n$, $\lambda \in [0, 1]$ ?

That's a fuzzy question on my assignment. Any idea on how to start with?

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We can assume without loss of generality that \begin{equation} \min\{\mu_A(r), \mu_A(s)\} = \mu_A(r) = \alpha. \end{equation} $\implies$

a-cut of fuzzy set $A$ is on $R^n$ is convex. A-cut can be defined as \begin{equation} A = \{x \in R^n| \mu_A(x) \geq \alpha\} \end{equation} If we take two elements $r$ and $s$, by the definition of convex set, number $\lambda r + (1 + \lambda)s$ is also an element of that set. Since it's an element of that set that means \begin{equation} \mu_A(\lambda r + (1 + \lambda)s) \geq \alpha \end{equation}

$\impliedby$

$\mu_A(\lambda r + (1 + \lambda)s) \geq \alpha$.

We know from $\min\{\mu_A(r), \mu_A(s)\} = \mu_A(r) = \alpha$ that $\mu_A(s) > \alpha$. We have an affine combination $\lambda r + (1 + \lambda)s$ for which also $\mu_A(\lambda r + (1 + \lambda)s) \geq \alpha$ so we know that all numbers $\lambda r + (1 + \lambda)s$ satisfy inequality $\mu_A(\cdot) \geq \alpha$ (belong to the same set as $r$ and $s$) which means this is a convex set again by the definition of a convex set.

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  • $\begingroup$ I posted an answer , it would be nice if you could tell me if it's true . In addition, since you know about Fuzzy Logic do you have any idea on this ? $\endgroup$ – estamos Dec 20 '19 at 12:13
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    $\begingroup$ @estamos sorry, I actually don't know about fuzzy logic, I googled a-cuts of fuzzy sets and I know some facts about convex sets which was enough to derive a proof. I can't really help you with other question, and I don't know if your answer is correct or not since it introduces some new things that I'm not familiar with. Sorry for misleading you that i know about fuzzy logic. You should probably post your answer as a separate question and if noone responds here consider posting your questions on Math StackExchange which is more math oriented website. $\endgroup$ – Brale Dec 20 '19 at 13:26
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A fuzzy set A in $R^n$ is said to be a convex fuzzy set if its $\alpha$-cuts $A_\alpha$ are (crisp) convex sets for all $A \in (0,1]$ .

Let A be a convex fuzzy set if and only if for all $r, s \in$ $R^n$, $\lambda \in [0, 1]$ .

Let $\alpha=\mu_A\leq\mu_B$

Then

\begin{equation} r\in A_{\alpha}, s\in A_{\alpha} \end{equation}

and also

\begin{equation} \lambda r + (1-\lambda)s \geq \alpha = min \{\mu_A(r), \mu_A(s)\} \end{equation}

Conversely, if the membership funciton $\mu_A$ of the fuzzy set A satisfies the inequality of Theorem 13.1 Convex fuzzy set, then taking $\alpha=\mu_A(r), A_\alpha$ may be regarded as set of all points $s$ for which $\mu_A(s)\geq\alpha=\mu_A(r)$. Therefore for all $r,s \in A_\alpha$,

\begin{equation} \mu_A(\lambda r + (1-\lambda)s) \geq min \{\mu_A(r), \mu_A(s)\} = \mu_A(r)=\alpha \end{equation}

which inplies that $\lambda r + (1-\lambda)s \in A_\alpha$. Hence $A_\alpha$ is a convex set for every $\alpha \in [0,1]$

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