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Chapter 3 • Theory & Derivations

Unit 3: α-Cuts, Decomposition Theorems & The Extension Principle

The fundamental bridge between crisp mathematics and fuzzy set theory: crisp α-cuts A_α and strong α-cuts A_{α^+}, cut monotonicity, preservation of set intersections and unions across cuts, the First Decomposition Theorem (representation via special fuzzy sets α · A_α), the Second Decomposition Theorem (continuous integral representation A = ∪_{α ∈ [0, 1]} α A_α), and Zadeh's Extension Principle for lifting crisp point mappings f: X → Y and multi-argument functions f: X_1 × ... × X_n → Y to fuzzy domains.

§3.1 Crisp $\alpha$-Cuts and Strong $\alpha$-Cuts ($A_\alpha$ and $A_{\alpha^+}$)

1. Connecting Fuzzy Sets to Classical Subsets

One of the most powerful and fruitful methodologies in fuzzy mathematics is the $\alpha$-cut representation. Rather than treating a fuzzy set as an exotic non-classical entity, $\alpha$-cuts decompose any fuzzy set into an ordered family of standard, crisp subsets.

Definition 3.1 ($\alpha$-Cut and Strong $\alpha$-Cut): Let $A$ be a fuzzy set on universe $X$ with membership function $\mu_A(x)$, and let $\alpha \in [0, 1]$.

  1. The $\alpha$-cut (or weak $\alpha$-cut / $\alpha$-level set) of $A$, denoted $A_\alpha$ or $[A]_\alpha$, is the crisp set of all elements whose membership grade is at least $\alpha$:
$$A_\alpha = \{x \in X \mid \mu_A(x) \ge \alpha\}$$
  1. The strong $\alpha$-cut of $A$, denoted $A_{\alpha^+}$ or $(A)_\alpha$, is the crisp set of all elements whose membership grade strictly exceeds $\alpha$:
$$A_{\alpha^+} = \{x \in X \mid \mu_A(x) > \alpha\}$$
Boundary Cases:
  • For $\alpha = 0$: $A_0 = \{x \in X \mid \mu_A(x) \ge 0\} = X$ (the entire universe).
  • The strong 0-cut is the Support: $A_{0^+} = \{x \in X \mid \mu_A(x) > 0\} = \text{Supp}(A)$.
  • The 1-cut is the Core: $A_1 = \{x \in X \mid \mu_A(x) = 1\} = \text{Core}(A)$.
  • The strong 1-cut is always empty: $A_{1^+} = \{x \in X \mid \mu_A(x) > 1\} = \emptyset$.

2. Level Set of a Fuzzy Set

The set of all distinct values assumed by the membership function $\mu_A(x)$ is called the level set (or spectrum) of $A$:

$$\Lambda(A) = \{\alpha \in [0, 1] \mid \exists x \in X, \; \mu_A(x) = \alpha\}$$

If $X$ is finite, $\Lambda(A) = \{\alpha_1, \alpha_2, \dots, \alpha_k\}$ is a finite ordered subset of $[0, 1]$.

§3.2 Algebraic Properties of $\alpha$-Cuts: Monotonicity, Intersection & Union Preservation

1. The Monotonicity Property of Cuts

Theorem 3.1 (Cut Monotonicity): For any fuzzy set $A$ and any $\alpha, \beta \in [0, 1]$:

$$\alpha \le \beta \implies A_\beta \subseteq A_\alpha \quad \text{and} \quad A_{\beta^+} \subseteq A_{\alpha^+}$$

Furthermore, for any $\alpha \in [0, 1)$:

$$A_\alpha \supseteq A_{\alpha^+}$$
Proof:

Let $x \in A_\beta$. Then $\mu_A(x) \ge \beta$. Since $\beta \ge \alpha$, transitivity yields $\mu_A(x) \ge \alpha$, which means $x \in A_\alpha$. Hence $A_\beta \subseteq A_\alpha$. Similarly, if $x \in A_{\alpha^+}$, then $\mu_A(x) > \alpha \ge \alpha$, so $x \in A_\alpha$. $\blacksquare$


2. Commutativity with Set Theoretic Operations

Theorem 3.2 (Preservation of Intersections and Unions): For any fuzzy sets $A$ and $B$ under standard Zadeh operations, and for any $\alpha \in [0, 1]$:

  1. $(A \cap B)_\alpha = A_\alpha \cap B_\alpha$
  2. $(A \cup B)_\alpha = A_\alpha \cup B_\alpha$
  3. $(A \cap B)_{\alpha^+} = A_{\alpha^+} \cap B_{\alpha^+}$
  4. $(A \cup B)_{\alpha^+} = A_{\alpha^+} \cup B_{\alpha^+}$
Proof of (1):
$$\begin{aligned} x \in (A \cap B)_\alpha &\iff \mu_{A \cap B}(x) \ge \alpha \\ &\iff \min(\mu_A(x), \mu_B(x)) \ge \alpha \\ &\iff \mu_A(x) \ge \alpha \text{ and } \mu_B(x) \ge \alpha \\ &\iff x \in A_\alpha \text{ and } x \in B_\alpha \\ &\iff x \in A_\alpha \cap B_\alpha \qquad \blacksquare \end{aligned}$$
Non-Preservation of Complements:

Does $(A^c)_\alpha = (A_\alpha)^c$? NO! In fact:

$$x \in (A^c)_\alpha \iff 1 - \mu_A(x) \ge \alpha \iff \mu_A(x) \le 1 - \alpha \iff x \notin A_{(1-\alpha)^+}$$

Therefore:

$$(A^c)_\alpha = (A_{(1-\alpha)^+})^c$$

Complementation inverts the cut level and swaps weak cuts with strong cuts!

§3.3 First and Second Decomposition Theorems (Representation of Fuzzy Sets via Cuts)

1. Scaling a Crisp Set by a Scalar $\alpha$

To reconstruct a fuzzy set from its crisp cuts, we define the product of a scalar $\alpha \in [0, 1]$ and a crisp set $C \subseteq X$:

Definition 3.2 (Special Fuzzy Set $\alpha C$): For $\alpha \in [0, 1]$ and $C \subseteq X$, the fuzzy set $\alpha C$ (or $\alpha \cdot C$) is defined by the membership function:

$$\mu_{\alpha C}(x) = \alpha \cdot \chi_C(x) = \begin{cases} \alpha, & \text{if } x \in C \\ 0, & \text{if } x \notin C \end{cases}$$

2. The First Decomposition Theorem

Theorem 3.3 (First Decomposition Theorem): For any fuzzy set $A$ on universe $X$:

$$A = \bigcup_{\alpha \in [0, 1]} \alpha A_\alpha$$

where $\bigcup$ denotes the standard fuzzy union ($\sup$ operator). That is, for all $x \in X$:

$$\mu_A(x) = \sup_{\alpha \in [0, 1]} \mu_{\alpha A_\alpha}(x) = \sup_{\alpha \in [0, 1]} \left( \alpha \cdot \chi_{A_\alpha}(x) \right)$$
Proof:

For any chosen point $x \in X$, let $\mu_A(x) = a \in [0, 1]$. Evaluate the right-hand supremum:

$$\sup_{\alpha \in [0, 1]} \left( \alpha \cdot \chi_{A_\alpha}(x) \right)$$

Recall that $x \in A_\alpha \iff \mu_A(x) \ge \alpha \iff a \ge \alpha$. Therefore:

$$\chi_{A_\alpha}(x) = \begin{cases} 1, & \text{if } \alpha \le a \\ 0, & \text{if } \alpha > a \end{cases}$$

Substituting into the product:

$$\alpha \cdot \chi_{A_\alpha}(x) = \begin{cases} \alpha, & \text{if } \alpha \le a \\ 0, & \text{if } \alpha > a \end{cases}$$

Taking the supremum over all $\alpha \in [0, 1]$:

$$\sup_{\alpha \in [0, 1]} \left( \alpha \cdot \chi_{A_\alpha}(x) \right) = \sup_{\alpha \in [0, a]} \alpha = a = \mu_A(x) \qquad \blacksquare$$

3. The Second Decomposition Theorem

Theorem 3.4 (Second Decomposition Theorem): For any fuzzy set $A$ on universe $X$:

$$A = \bigcup_{\alpha \in [0, 1]} \alpha A_{\alpha^+} = \bigcup_{\alpha \in \Lambda(A)} \alpha A_\alpha$$

where $\Lambda(A)$ is the level set of $A$.

This proves that ANY fuzzy set is completely and losslessly determined by its crisp $\alpha$-cuts! Every theorem in fuzzy set theory can be proved by proving it for its crisp cuts.

§3.4 Zadeh's Extension Principle: Mapping Fuzzy Sets Through Crisp Functions $f: X \to Y$

1. Motivation of the Extension Principle

Suppose we have a crisp mathematical function $f: X \to Y$ (for example, $f(x) = x^2$ or $f(x) = \sin x$), and we wish to evaluate $f$ when the input is not a crisp number, but a fuzzy set $A$ on $X$. How do we compute the induced fuzzy set $B = f(A)$ on $Y$? In 1975, Lotfi A. Zadeh formulated the Extension Principle, which is considered the single most important operational engine in fuzzy mathematics.


2. Formal Definition of the Extension Principle

Definition 3.3 (Zadeh's Extension Principle): Let $f: X \to Y$ be a crisp mapping from universe $X$ to universe $Y$. Let $A$ be a fuzzy set on $X$ with membership function $\mu_A(x)$. The mapping $f$ induces a fuzzy set $B = f(A)$ on $Y$ whose membership function $\mu_B(y)$ is defined for all $y \in Y$ by:

$$\mu_B(y) = \mu_{f(A)}(y) = \begin{cases} > \sup_{x \in f^{-1}(y)} \mu_A(x), & \text{if } f^{-1}(y) \ne \emptyset \\ > 0, & \text{if } f^{-1}(y) = \emptyset > \end{cases}$$

where $f^{-1}(y) = \{x \in X \mid f(x) = y\}$ is the preimage of $y$.

Special Cases:

1. One-to-One (Invertible) Mapping: If $f$ is injective, each $y \in f(X)$ has a unique preimage $x = f^{-1}(y)$, so:

$$\mu_B(y) = \mu_A(f^{-1}(y))$$

2. Many-to-One Mapping: If multiple distinct points $x_1, x_2, \dots$ map to the same $y$, the principle assigns $y$ the maximum (or supremum) of their membership grades:

$$\mu_B(y) = \max_{x \in f^{-1}(y)} \mu_A(x)$$

(Optimistic principle: an outcome $y$ is as possible as the most possible input that can produce it!)

§3.5 Extended Functions on Cartesian Products and Verification via $\alpha$-Cuts

1. Extension Principle for Multi-Argument Functions

Let $f: X_1 \times X_2 \times \dots \times X_n \to Y$ be a crisp multi-variable function (e.g. addition $f(x_1, x_2) = x_1 + x_2$, or multiplication $f(x_1, x_2) = x_1 \cdot x_2$). Let $A_1, A_2, \dots, A_n$ be fuzzy sets defined on universes $X_1, X_2, \dots, X_n$.

Definition 3.4 (Multi-Variable Extension Principle): The image fuzzy set $B = f(A_1, \dots, A_n)$ on $Y$ has membership function:

$$\mu_B(y) = \begin{cases} > \sup_{(x_1, \dots, x_n) \in f^{-1}(y)} \min\left( \mu_{A_1}(x_1), \mu_{A_2}(x_2), \dots, \mu_{A_n}(x_n) \right), & \text{if } f^{-1}(y) \ne \emptyset \\ > 0, & \text{if } f^{-1}(y) = \emptyset > \end{cases}$$

where $f^{-1}(y) = \{(x_1, \dots, x_n) \mid f(x_1, \dots, x_n) = y\}$.


2. Cut-Level Preservation of the Extension Principle

Theorem 3.5 (Nguyen's Theorem / Cut Preservation): If $X$ and $Y$ are Euclidean spaces and $f: X \to Y$ is continuous, and if $A$ has compact $\alpha$-cuts, then the $\alpha$-cut of the extended fuzzy set $f(A)$ is identically the crisp image of the $\alpha$-cut of $A$:

$$[f(A)]_\alpha = f(A_\alpha) \quad (\forall \alpha \in (0, 1])$$

Similarly, for multi-argument functions:

$$[f(A_1, \dots, A_n)]_\alpha = f([A_1]_\alpha, \dots, [A_n]_\alpha)$$
Profound Consequence:

This theorem guarantees that evaluating complicated non-linear functions of fuzzy sets via the extension principle is completely mathematically equivalent to evaluating standard interval analysis on each $\alpha$-cut level! This is the foundational cornerstone of all fuzzy arithmetic.

Rigorous Tiered Solved Examination Problems

Step-by-step unskipped derivations, complete proofs, and verification across Foundational, Advanced, and Honors tiers.

Fundamentals Example 3.1: Problem 3.1: Explicit Calculation of $\alpha$-Cuts for Triangular and Trapezoidal Fuzzy Sets

Consider the real universe $X = \mathbb{R}$.

  1. Let $A = \text{trimf}(x; 2, 5, 9)$ be a triangular fuzzy set.

Derive the closed-form expression for the $\alpha$-cut interval $A_\alpha = [a_1(\alpha), a_2(\alpha)]$ as a function of $\alpha \in [0, 1]$. Evaluate $A_0$, $A_{0.5}$, and $A_1$.

  1. Let $B = \text{trapmf}(x; 1, 4, 7, 10)$ be a trapezoidal fuzzy set.

Derive the closed-form expression for $B_\alpha$. Evaluate $B_{0.25}$ and $B_{0.8}$.

  1. For $A$, find the strong $\alpha$-cut $A_{0.5^+}$ and verify that $A_{0.5^+} \subset A_{0.5}$.

1. Triangular Fuzzy Set $A = \text{trimf}(x; 2, 5, 9)$

The membership function is:

$$\mu_A(x) = \begin{cases} \frac{x - 2}{5 - 2} = \frac{x - 2}{3}, & 2 \le x \le 5 \\ \frac{9 - x}{9 - 5} = \frac{9 - x}{4}, & 5 \le x \le 9 \\ 0, & \text{otherwise} \end{cases}$$
Derivation of $\alpha$-cut $A_\alpha$:
  • Left bound: $\frac{x - 2}{3} = \alpha \implies x = 2 + 3\alpha$.
  • Right bound: $\frac{9 - x}{4} = \alpha \implies x = 9 - 4\alpha$.

Thus, for $\alpha \in (0, 1]$:

$$A_\alpha = [2 + 3\alpha, \; 9 - 4\alpha] \qquad \blacksquare$$
Specific Evaluations:
  • $\alpha = 0$: $A_0 = [2, 9]$ (the support closure).
  • $\alpha = 0.5$: $A_{0.5} = [2 + 3(0.5), 9 - 4(0.5)] = [3.5, 7.0]$.
  • $\alpha = 1.0$: $A_1 = [2 + 3(1), 9 - 4(1)] = [5, 5] = \{5\}$ (the core). $\blacksquare$

2. Trapezoidal Fuzzy Set $B = \text{trapmf}(x; 1, 4, 7, 10)$

  • Left branch ($1 \le x \le 4$): $\frac{x - 1}{4 - 1} = \frac{x - 1}{3} = \alpha \implies x = 1 + 3\alpha$.
  • Core ($4 \le x \le 7$): $\mu_B(x) = 1 \ge \alpha$.
  • Right branch ($7 \le x \le 10$): $\frac{10 - x}{10 - 7} = \frac{10 - x}{3} = \alpha \implies x = 10 - 3\alpha$.

Thus:

$$B_\alpha = [1 + 3\alpha, \; 10 - 3\alpha] \qquad \blacksquare$$
Specific Evaluations:
  • $\alpha = 0.25$: $B_{0.25} = [1 + 3(0.25), 10 - 3(0.25)] = [1.75, 9.25]$.
  • $\alpha = 0.80$: $B_{0.80} = [1 + 3(0.8), 10 - 3(0.8)] = [3.4, 7.6]$. $\blacksquare$

3. Strong $\alpha$-Cut $A_{0.5^+}$

The strong cut requires $\mu_A(x) > 0.5$:

  • For $x \in [2, 5]$: $\frac{x - 2}{3} > 0.5 \implies x > 3.5$.
  • For $x \in [5, 9]$: $\frac{9 - x}{4} > 0.5 \implies 9 - x > 2 \implies x < 7.0$.

Therefore, the strong cut is the open interval:

$$A_{0.5^+} = (3.5, 7.0)$$

Comparing with the weak cut $A_{0.5} = [3.5, 7.0]$, the boundary points $\{3.5, 7.0\}$ where $\mu_A(x) = 0.5$ belong to $A_{0.5}$ but not to $A_{0.5^+}$. Thus $A_{0.5^+} \subset A_{0.5}$ strictly. $\blacksquare$

Computational Triad Example 3.2: Problem 3.2: Applying the Extension Principle to Non-Linear Functions

Let $X = [-4, 4]$ and let $A$ be a triangular fuzzy set on $X$ defined by $A = \text{trimf}(x; -2, 1, 3)$. Consider the non-linear crisp function $f(x) = x^2$.

  1. Using Zadeh's Extension Principle, determine the membership function $\mu_B(y)$ of the induced fuzzy set $B = f(A)$ on $Y = [0, 16]$.
  2. Identify the core, support, and height of $B$.
  3. Verify your result using the $\alpha$-cut method ($B_\alpha = f(A_\alpha)$).

1. Evaluation via Extension Principle

$$\mu_B(y) = \sup_{x \in f^{-1}(y)} \mu_A(x) = \sup_{x: x^2 = y} \mu_A(x)$$

For $y \ge 0$, the preimages are $x = +\sqrt{y}$ and $x = -\sqrt{y}$. The membership function of $A$ is:

$$\mu_A(x) = \begin{cases} \frac{x - (-2)}{1 - (-2)} = \frac{x + 2}{3}, & -2 \le x \le 1 \\ \frac{3 - x}{3 - 1} = \frac{3 - x}{2}, & 1 \le x \le 3 \\ 0, & \text{otherwise} \end{cases}$$
Case 1: $y \in [0, 1]$

Here $x = +\sqrt{y} \in [0, 1]$ and $x = -\sqrt{y} \in [-1, 0]$. Both preimages fall in the rising branch $[-2, 1]$:

  • $\mu_A(+\sqrt{y}) = \frac{\sqrt{y} + 2}{3}$
  • $\mu_A(-\sqrt{y}) = \frac{-\sqrt{y} + 2}{3}$

Since $\frac{\sqrt{y} + 2}{3} \ge \frac{-\sqrt{y} + 2}{3}$ for all $y \ge 0$:

$$\mu_B(y) = \frac{\sqrt{y} + 2}{3}, \quad y \in [0, 1]$$

At $y = 1$: $\mu_B(1) = \frac{1 + 2}{3} = 1.0$.

Case 2: $y \in [1, 4]$

Here $x = +\sqrt{y} \in [1, 2]$ falls in the falling branch:

$$\mu_A(+\sqrt{y}) = \frac{3 - \sqrt{y}}{2}$$

While $x = -\sqrt{y} \in [-2, -1]$ falls in the rising branch:

$$\mu_A(-\sqrt{y}) = \frac{-\sqrt{y} + 2}{3}$$

Compare the two values:

$$\frac{3 - \sqrt{y}}{2} - \frac{2 - \sqrt{y}}{3} = \frac{3(3 - \sqrt{y}) - 2(2 - \sqrt{y})}{6} = \frac{9 - 3\sqrt{y} - 4 + 2\sqrt{y}}{6} = \frac{5 - \sqrt{y}}{6}$$

Since $y \le 4 \implies \sqrt{y} \le 2$, we have $5 - \sqrt{y} \ge 3 > 0$. Thus $\mu_A(+\sqrt{y}) > \mu_A(-\sqrt{y})$ everywhere on $[1, 4]$! Therefore:

$$\mu_B(y) = \frac{3 - \sqrt{y}}{2}, \quad y \in [1, 4]$$
Case 3: $y \in [4, 9]$

Here $-\sqrt{y} \le -2$, so $\mu_A(-\sqrt{y}) = 0$. Only $+\sqrt{y} \in [2, 3]$ has positive membership:

$$\mu_B(y) = \frac{3 - \sqrt{y}}{2}, \quad y \in [4, 9]$$
Case 4: $y > 9$

$\mu_B(y) = 0$.

Final Piecewise Expression:
$$\mu_B(y) = \begin{cases} \frac{\sqrt{y} + 2}{3}, & 0 \le y \le 1 \\ \frac{3 - \sqrt{y}}{2}, & 1 \le y \le 9 \\ 0, & \text{otherwise} \end{cases} \qquad \blacksquare$$

2. Core, Support, and Height

  • Core: $\{y \in [0, 16] \mid \mu_B(y) = 1\} = \{1\}$.
  • Support: $\{y \in [0, 16] \mid \mu_B(y) > 0\} = [0, 9)$.
  • Height: $h(B) = \mu_B(1) = 1.0$ (normal fuzzy set). $\blacksquare$

3. Verification via $\alpha$-Cuts

For $A = \text{trimf}(x; -2, 1, 3)$:

$$A_\alpha = [-2 + 3\alpha, \; 3 - 2\alpha]$$

Applying $f(x) = x^2$ to the interval $[a_1, a_2] = [-2 + 3\alpha, 3 - 2\alpha]$: Since the interval contains $x = 0$ for $\alpha \le 2/3$, the minimum square is 0 for small $\alpha$, and $(-2 + 3\alpha)^2$ for $\alpha > 2/3$. The maximum square is achieved at $(3 - 2\alpha)^2$. Thus:

  • Right boundary: $y = (3 - 2\alpha)^2 \implies \sqrt{y} = 3 - 2\alpha \implies \alpha = \frac{3 - \sqrt{y}}{2}$.
  • Left boundary (when $\alpha > 2/3$): $y = (3\alpha - 2)^2 \implies \sqrt{y} = 3\alpha - 2 \implies \alpha = \frac{\sqrt{y} + 2}{3}$.

This matches the Extension Principle derivation identically! $\blacksquare$

Honors / Proof Challenge Example 3.3: Problem 3.3: Complete Proof of the Second Decomposition Theorem

The Second Decomposition Theorem states that any fuzzy set $A$ on universe $X$ can be reconstructed from its strong $\alpha$-cuts:

$$A = \bigcup_{\alpha \in [0, 1]} \alpha A_{\alpha^+}$$

and also from the finite set of distinct levels if $X$ is finite:

$$A = \bigcup_{i=1}^k \alpha_i A_{\alpha_i}$$

where $\Lambda(A) = \{\alpha_1, \alpha_2, \dots, \alpha_k\}$ with $0 = \alpha_0 < \alpha_1 < \dots < \alpha_k \le 1$.

  1. Provide a rigorous, line-by-line mathematical proof that for all $x \in X$:
$$\mu_A(x) = \sup_{\alpha \in [0, 1]} \left( \alpha \cdot \chi_{A_{\alpha^+}}(x) \right)$$
  1. For a discrete universe $X = \{x_1, \dots, x_n\}$, prove that:
$$\mu_A(x) = \max_{1 \le i \le k} \left( \alpha_i \cdot \chi_{A_{\alpha_i}}(x) \right)$$
  1. Illustrate this theorem explicitly for the fuzzy set:
$$A = \frac{0.2}{x_1} + \frac{0.5}{x_2} + \frac{0.8}{x_3} + \frac{1.0}{x_4}$$

by expressing $A$ as an explicit union of four crisp cut sets.

1. Proof of the Second Decomposition Theorem (Strong Cuts)

Let $x \in X$ be arbitrary, and let $a = \mu_A(x) \in [0, 1]$. We must evaluate:

$$S(x) \equiv \sup_{\alpha \in [0, 1]} \mu_{\alpha A_{\alpha^+}}(x) = \sup_{\alpha \in [0, 1]} \left( \alpha \cdot \chi_{A_{\alpha^+}}(x) \right)$$

Recall Definition 3.1:

$$x \in A_{\alpha^+} \iff \mu_A(x) > \alpha \iff a > \alpha \iff \alpha \in [0, a)$$

Therefore, the characteristic function is:

$$\chi_{A_{\alpha^+}}(x) = \begin{cases} 1, & \text{if } 0 \le \alpha < a \\ 0, & \text{if } \alpha \ge a \end{cases}$$

Substituting this into the product:

$$\alpha \cdot \chi_{A_{\alpha^+}}(x) = \begin{cases} \alpha, & \text{if } 0 \le \alpha < a \\ 0, & \text{if } \alpha \ge a \end{cases}$$

Now take the supremum over all $\alpha \in [0, 1]$:

  • If $a = 0$: the set $\{\alpha \in [0, 1] \mid \alpha < 0\} = \emptyset$, so the product is 0 for all $\alpha$, and the supremum is 0, which equals $a = \mu_A(x)$.
  • If $a > 0$:
$$S(x) = \sup_{\alpha \in [0, a)} \alpha = a = \mu_A(x)$$

In all cases, $S(x) = \mu_A(x)$. Since $x \in X$ was arbitrary, we have proven:

$$A = \bigcup_{\alpha \in [0, 1]} \alpha A_{\alpha^+} \qquad \blacksquare$$

2. Proof for Finite Level Sets

Let $\Lambda(A) = \{\alpha_1, \alpha_2, \dots, \alpha_k\}$ be the set of distinct non-zero membership grades, with:

$$0 < \alpha_1 < \alpha_2 < \dots < \alpha_k = h(A) \le 1$$

For any $x \in X$, there exists some $j \in \{1, \dots, k\}$ such that $\mu_A(x) = \alpha_j$ (or $\mu_A(x) = 0$). If $\mu_A(x) = \alpha_j > 0$: Then $x \in A_{\alpha_i} \iff \mu_A(x) \ge \alpha_i \iff \alpha_j \ge \alpha_i \iff i \le j$. Therefore:

$$\chi_{A_{\alpha_i}}(x) = \begin{cases} 1, & 1 \le i \le j \\ 0, & i > j \end{cases}$$

Then:

$$\max_{1 \le i \le k} \left( \alpha_i \cdot \chi_{A_{\alpha_i}}(x) \right) = \max_{1 \le i \le j} \alpha_i = \alpha_j = \mu_A(x)$$

If $\mu_A(x) = 0$, $x \notin A_{\alpha_i}$ for all $i \ge 1$, so the maximum is 0. Thus:

$$A = \bigcup_{i=1}^k \alpha_i A_{\alpha_i} \qquad \blacksquare$$

3. Explicit Demonstration

Let $A = \frac{0.2}{x_1} + \frac{0.5}{x_2} + \frac{0.8}{x_3} + \frac{1.0}{x_4}$. The level set is $\Lambda(A) = \{0.2, 0.5, 0.8, 1.0\}$.

Step A: Determine the Crisp Cuts:
  • $A_{0.2} = \{x_1, x_2, x_3, x_4\}$
  • $A_{0.5} = \{x_2, x_3, x_4\}$
  • $A_{0.8} = \{x_3, x_4\}$
  • $A_{1.0} = \{x_4\}$
Step B: Construct the Scaled Fuzzy Sets $\alpha_i A_{\alpha_i}$:
  • $0.2 A_{0.2} = \frac{0.2}{x_1} + \frac{0.2}{x_2} + \frac{0.2}{x_3} + \frac{0.2}{x_4}$
  • $0.5 A_{0.5} = \frac{0.0}{x_1} + \frac{0.5}{x_2} + \frac{0.5}{x_3} + \frac{0.5}{x_4}$
  • $0.8 A_{0.8} = \frac{0.0}{x_1} + \frac{0.0}{x_2} + \frac{0.8}{x_3} + \frac{0.8}{x_4}$
  • $1.0 A_{1.0} = \frac{0.0}{x_1} + \frac{0.0}{x_2} + \frac{0.0}{x_3} + \frac{1.0}{x_4}$
Step C: Take the Standard Union (Max):
$$\begin{aligned} \bigcup_{i=1}^4 \alpha_i A_{\alpha_i} &= \frac{\max(0.2, 0, 0, 0)}{x_1} + \frac{\max(0.2, 0.5, 0, 0)}{x_2} + \frac{\max(0.2, 0.5, 0.8, 0)}{x_3} + \frac{\max(0.2, 0.5, 0.8, 1.0)}{x_4} \\ &= \frac{0.2}{x_1} + \frac{0.5}{x_2} + \frac{0.8}{x_3} + \frac{1.0}{x_4} = A \end{aligned}$$

The reconstructed fuzzy set equals $A$ exactly! $\blacksquare$