Answers.Body: <blockquote>
    <p>A fuzzy set A in <span class="math-container">$R^n$</span> is said to be a convex fuzzy set if its
    <span class="math-container">$\alpha$</span>-cuts <span class="math-container">$A_\alpha$</span> are (crisp) convex sets for all <span class="math-container">$A \in (0,1]$</span>
    .</p>
  </blockquote>
  
  <p>Let A be a convex fuzzy set if and only if for all <span class="math-container">$r, s \in$</span> <span class="math-container">$R^n$</span>, <span class="math-container">$\lambda \in [0, 1]$</span> .</p>
  
  <p>Let <span class="math-container">$\alpha=\mu_A\leq\mu_B$</span></p>
  
  <p>Then</p>
  
  <p><span class="math-container">\begin{equation}
      r\in A_{\alpha}, s\in A_{\alpha}
  \end{equation}</span></p>
  
  <p>and also</p>
  
  <p><span class="math-container">\begin{equation}
      \lambda r + (1-\lambda)s \geq \alpha = min \{\mu_A(r), \mu_A(s)\}
  \end{equation}</span></p>
  
  <p>Conversely, if the membership funciton <span class="math-container">$\mu_A$</span> of the fuzzy set A satisfies the inequality of Theorem 13.1 <strong>Convex fuzzy set</strong>, then taking <span class="math-container">$\alpha=\mu_A(r), A_\alpha$</span> may be regarded as set of all points <span class="math-container">$s$</span> for which <span class="math-container">$\mu_A(s)\geq\alpha=\mu_A(r)$</span>. Therefore for all <span class="math-container">$r,s \in A_\alpha$</span>,</p>
  
  <p><span class="math-container">\begin{equation}
      \mu_A(\lambda r + (1-\lambda)s) \geq min \{\mu_A(r), \mu_A(s)\} = \mu_A(r)=\alpha
  \end{equation}</span></p>
  
  <p>which inplies that <span class="math-container">$\lambda r + (1-\lambda)s \in A_\alpha$</span>. Hence <span class="math-container">$A_\alpha$</span> is a convex set for every <span class="math-container">$\alpha \in [0,1]$</span></p>
  
Answers.ContentLicense: CC BY-SA 4.0
Answers.CreationDate: 2019-12-20T12:12:35.933Z
Answers.LastActivityDate: 2019-12-20T12:12:35.933Z
Answers.UserDisplayName: estamos
Answers.Body: <p>We can assume without loss of generality that
  <span class="math-container">\begin{equation}
  \min\{\mu_A(r), \mu_A(s)\} = \mu_A(r) = \alpha.
  \end{equation}</span>
  <span class="math-container">$\implies$</span></p>
  
  <p>a-cut of fuzzy set <span class="math-container">$A$</span> is on <span class="math-container">$R^n$</span> is convex. A-cut can be defined as
  <span class="math-container">\begin{equation}
  A = \{x \in R^n| \mu_A(x) \geq \alpha\}
  \end{equation}</span>
  If we take two elements <span class="math-container">$r$</span> and <span class="math-container">$s$</span>, by the definition of convex set, number <span class="math-container">$\lambda r + (1 + \lambda)s$</span> is also an element of that set. Since it's an element of that set that means
  <span class="math-container">\begin{equation}
  \mu_A(\lambda r + (1 + \lambda)s) \geq \alpha
  \end{equation}</span></p>
  
  <p><span class="math-container">$\impliedby$</span></p>
  
  <p><span class="math-container">$\mu_A(\lambda r + (1 + \lambda)s) \geq \alpha$</span>.</p>
  
  <p>We know from <span class="math-container">$\min\{\mu_A(r), \mu_A(s)\} = \mu_A(r) = \alpha$</span> that <span class="math-container">$\mu_A(s) &gt; \alpha$</span>. We have an affine combination <span class="math-container">$\lambda r + (1 + \lambda)s$</span> for which also <span class="math-container">$\mu_A(\lambda r + (1 + \lambda)s) \geq \alpha$</span> so we know that all numbers <span class="math-container">$\lambda r + (1 + \lambda)s$</span> satisfy inequality <span class="math-container">$\mu_A(\cdot) \geq \alpha$</span> (belong to the same set as <span class="math-container">$r$</span> and <span class="math-container">$s$</span>) which means this is a convex set again by the definition of a convex set.</p>
  
Answers.ContentLicense: CC BY-SA 4.0
Answers.CreationDate: 2019-12-20T12:13:56.513Z
Answers.Text: 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](https://ai.stackexchange.com/q/17170/32076) ?
Answers.UserDisplayName: estamos
Answers.ContentLicense: CC BY-SA 4.0
Answers.CreationDate: 2019-12-20T13:26:21.933Z
Answers.Text: @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.
Answers.UserDisplayName: Brale
Answers.ContentLicense: CC BY-SA 4.0
Answers.CreationDate: 2019-12-20T01:20:18.683Z
Answers.LastActivityDate: 2019-12-20T01:20:18.683Z
Answers.UserDisplayName: Brale
Body: <p>How can I prove that all the a-cuts of any fuzzy set A defined on <span class="math-container">$R^n$</span>
  are convex if and only if</p>
  
  <p><span class="math-container">$$\mu_A(\lambda r + (1-\lambda)s) \geq min \{\mu_A(r), \mu_A(s)\}$$</span></p>
  
  <p>such that <span class="math-container">$r, s \in R^n$</span>, <span class="math-container">$\lambda \in [0, 1]$</span> ?</p>
  
  <p>That's a fuzzy question on my assignment. Any idea on how to start with?</p>
  
Tags: <fuzzy-logic>
Title: How can I prove that all the a-cuts of any fuzzy set A defined on $R^n$ are convex?
UserDisplayName: estamos
ThreadText: # How can I prove that all the a-cuts of any fuzzy set A defined on $R^n$ are convex?
  
  (Asked by: username_0 on 2019-12-19)
  
  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?
  
  **Tags:** fuzzy-logic
  
  
  ---
  
  ## 2 answers
  
  ### Answer by username_0 on 2019-12-20. Score: 0 | Accepted answer
  
  > 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 $\alpha \in [0,1]$
  
  
  ---
  
  ### Answer by username_1 on 2019-12-20. Score: 2
  
  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.
  
  > **Comments:**
  > **username_0**: 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](https://ai.stackexchange.com/q/17170/32076) ?  
  > **username_1**: @username_0 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.  
  
  ---
  
  