The fuzzy numbers are defined as fuzzy sets ($A$) defined over $\mathbb{R}$ which satisfy the following three properties:-
- $A$ is normal, i.e., the height of $A$ is $1$.
- $^{\alpha}A$ is a (non - empty) closed interval $\forall \alpha \in \left( 0, 1 \right]$, where $^{\alpha}A = \left\lbrace x \in \mathbb{R} | A(x) \ge \alpha \right\rbrace$ is called the alpha - cut of $A$.
- $^{0+}A$ is bounded, where $^{0+}A = \left\lbrace x \in \mathbb{R} | A(x) > 0 \right\rbrace$ is called the support of $A$.
Althought, intuitively this is very much correct, I think that for proving that a fuzzy set is a fuzzy number it is sufficient to prove the second condition is true.
For that, it will be sufficient to prove that the condition $2$ implies conditions $1$ and $3$. For the proof, consider the following:-
Let $^{\alpha}A$ be closed interval $\forall \alpha \in \left( 0, 1 \right]$.
Consider $^{1}A$,
$^{1}A = \left\lbrace x \in \mathbb{R} | A(x) \ge 1 \right\rbrace$.
But since $A(x) \le 1$, for $^{1}A$, we have $A(x) = 1$.
This is equal to a closed interval which is non - empty. This directly implies that the height of $A$ is $1$ and hence $A$ is normal.
Thus, $2$ implies $1$.
Now, consider $^{\alpha +}A = \left\lbrace x \in \mathbb{R} | A(x) > \alpha \right\rbrace$.
Consider any sequence of such $\alpha > 0$ and $^{\alpha +}A$ such that $\alpha$ approaches $0$ from the right. Thus, here I am wanting to have $\lim_{\alpha \rightarrow 0^{+}} .^{\alpha +}A$.
Since in the limiting case, we can have $\lim_{\alpha \rightarrow 0^{+}} \alpha \ge 0$. So, we have,
$$\lim_{\alpha \rightarrow 0^{+}} .^{\alpha +}A = \lim_{\alpha \rightarrow 0^{+}} \left\lbrace x \in \mathbb{R} | A(x) > \alpha \right\rbrace = \left\lbrace x \in \mathbb{R} | A(x) > \lim_{\alpha \rightarrow 0^{+}} \alpha \ge 0 \right\rbrace$$
$$= \left\lbrace x \in \mathbb{R} | A(x) > 0 \right\rbrace = ^{0+}A$$
Now, since each of the alpha cuts were closed intervals, they were bounded and so were strong cuts $^{\alpha +}A$. Therefore, for the limiting case as well, $^{0+}A$ is also bounded (from the fact that all strong alpha cuts are subsets of alpha cuts).
Thus, $3$ implies $1$.
Therefore, from the line of thought that I have, we only need to prove that all the alpha cuts of a fuzzy set are closed interval to prove that the fuzzy set is a fuzzy number.
But, I am not really getting the feel that the line of thought is right, especially in proving $3$ implies $2$. Any comments on this and a solution to my question, "whether we can just prove for the condition $2$ and prove that the fuzzy set is a fuzzy number is true or not" will be appreciated.
For condition 2 to be sufficient, it would need to imply condition 3. But surely this isn't true: suppose $A(x)=\exp(-x^2)$; then condition 2 holds but condition 3 doesn't.