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?


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}


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

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.

  • $\begingroup$ 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
  • 1
    $\begingroup$ @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

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 $\alpha=\mu_A\leq\mu_B$


\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 $\alpha \in [0,1]$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.