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

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 $$\begin{equation} \mu_A(\lambda r + (1 + \lambda)s) \geq \alpha \end{equation}$$

$$\impliedby$$

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

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.

• 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 ? – estamos Dec 20 '19 at 12:13
• @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. – 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 $$\alpha=\mu_A\leq\mu_B$$

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