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I've recently taken on interest for Fuzzy Set Theory and I've been reading George J. Klir and Bo Yuan. 1994. Fuzzy sets and fuzzy logic: theory and applications. Prentice-Hall, Inc., USA.

Where the authors define the standard union of two fuzzy sets $A, B \in \mathcal{F}(X)$ where $X$ is the universe set and $\mathcal{F}(X)$ is the fuzzy power set of $X$ as $$ (A \cup B)(x) = max[A(x), B(x)] \; \; \forall x \in X $$

Where $A(x)$ denotes the membership function of fuzzy set $A$ and similarly for $B$.

In another book (Gottwald, Siegfried & Bandemer, Hans. (1995). Fuzzy Sets, Fuzzy Logic, Fuzzy Methods with Applications), they define the standard union of two fuzzy sets the same as above, but they also define a generalised union for a family of fuzzy sets $(A_i| i \in \mathcal{I})$ where $\mathcal{I}$ is the index set as

$$ (\bigcup_{i \in \mathcal{I}} A_i)(x) = \sup_{i \in \mathcal{I}} A_i(x) \; \forall x \in X $$

I'm under the assumption that this definition is the generalisation of the standard union of 2 fuzzy sets and under this assumption, I have a few questions

Does replacing the max operator by the supremum operator alter the definition of the Union operation?

I understand that if a set has a maximum then it will have a supremum whose value is the maximum. And if a set has a supremum and if this supremum belongs to the set, then this supremum is also the maximum. But as I understand it, it is possible for sets to have a supremum and not a maximum, so does replacing the max operator by the supremum operator alter the original definition of the standard union (defined using the max operator)?

Or was the max operator only used in the definition of the standard union of 2 fuzzy sets because the max of a finite number of elements (or the max of a finite set) exists and hence would also anyway be the supremum?

Before posting I did read through the following answers, but they didn't quite help me:

Also in the second link, the author states and I quote

So, one can replace $max$ by $sup$ in any context and never use $max$

Is this actually true?

I apologise in advance if this question seems trivial, simple or non-intellectual, I just don't think I've completely grasped the concept of this generalisation.

Jayitha Reddy
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    Yes, it’s true, though when the supremum is guaranteed to be the maximum, using $\max$ instead of $\sup$ probably contributes to the reader’s ready understanding. That’s why you will generally see $\max$ when a finite set of numbers is at issue. In the case of the union of an arbitrary collection of fuzzy sets, you really do need $\sup$, since there’s no reason to think that the set of $A_i(x)$ has a largest element. – Brian M. Scott Sep 01 '20 at 23:27
  • Welcome to MSE, and let me congratulate you on a well-posed first question. – saulspatz Sep 01 '20 at 23:33

1 Answers1

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Consider the example where we have $A_i(x)=-\frac 1 i$ for all $x$ in $X$ and for $i$ in $\mathcal I=\mathbb N$.

Note that $\max[A_i(x),A_j(x)]=\max[-\frac 1i, -\frac 1j]$ is well defined.

However $\max[A_1(x),A_2(x),\ldots]$ is not defined since the maximum can get as close to $0$ as we want, but there is no $A_i(x)$ that actually is $0$.

So instead we introduce $\sup[A_1(x),A_2(x),\ldots]=0$, which is the generalization of $\max$ that includes the limit.

Klaas van Aarsen
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