I'm trying to prove the following theorem (and please, don't give me a proof, this is a conceptual question):

(Negoita and Ralescu's Representation Theorem) Let there be $A_{\alpha}$, $\alpha \in [0,1]$, a family of crisp subsets of $U$ such that the following are true:

  1. $\displaystyle \bigcup_{\alpha \in [0,1]} A_{\alpha} \subset A_{0}$;

  2. $A_{\alpha} \subset A_{\beta}$ if $\beta \leq \alpha$

  3. $\displaystyle A_{\alpha} = \bigcap_{k \geq 0} A_{\alpha_{k}}$ se $\alpha_{k}$ convergir para $\alpha$ com $\alpha_{k} \leq \alpha$, para todo $k \geq 0$.

On these conditions, there exists one, and only one, fuzzy set $A$ in $U$ on which $$A_{\alpha} = [A]^{\alpha}.$$

This theorem is just to contextualize my question. So, I've always struggled with the proof of this type of statements. The uniqueness part is pretty straightforward, just assume that such conditions create two things (sets, groups, etc.) and then show that those two things are actually the same. Done, as simple as that.

The existence part that always have made me "stutter", because in maths when you want something to exist you just say that it exists and then derive the other results (I'm just oversimplifying hahaha). But when one of the results is an existence itself, I can't seem to find a way of showing it. And this really compromise my confidence as an aspiring mathematician.

So, to the point, what are the logical devices that I can use for those kind of problems? (that's the most important question).

And, if I show that $[A]^{\alpha}$ satisfies the $A_{\alpha}$ properties, I'm showing that there exists a fuzzy set $A$ (because $[A]^{\alpha}$ derives from $A$), or I'm just showing the obvious that those two families just share those 3 properties?


1 Answer 1


One way to show existence is by construction. For instance, you might want to show that for any triple $(a, b, c)$ of integers with $$ 0 \le a < 2\\ 0 \le b < 3\\ 0 \le c < 5 $$ there's an integer $n$ with $$ n \bmod 2 = a\\ n \bmod 3 = b \\ n \bmod 5 = c. $$ You might, with a little work, be able to say something like "Let $$ n = (2 \cdot 3 + c) + (3 \cdot 5 + a) + (2 \cdot 5 + b)" $$ and then show that $n$ has the three required properties. (This particular formula for $n$ doesn't work at all, of course, but I can't remember enough of the chinese remainder theorem to recall how to construct such a thing.)

Alternatively, sometimes injectivity/surjectivity can help. For instance, you might observe that if $k$ and $p$ are relatively prime, then for any $s$, the pairs $$ (s \bmod k, s \bmod p)\\ (s+1 \bmod k, s+1 \bmod p)\\ \ldots\\ (s+pk-1 \bmod k,s+ bk-1 \bmod p) $$ are all distinct. That lets you say that there's a number $s$ between $0$ and $2\cdot 3 - 1$ whose remainders mod $2$ and mod $3$ are $a$ and $b$ (by the pigeonhole principle). And furthermore, adding $6$ to $s$ doesn't change its remainders, so among the numbers $$ s + 0 s + 1\cdot 6 \\ \ldots\\ s + 5 \cdot 6) $$ which all have the same residues mod 2 and mod 3, the residues mod 5 are all distinct numbers (by the same little theorem), so you have five distinct numbers between 0 and 4, hence each (by the pigeonhole principle) must appear at least one, so one of them is $c$.

In this case existence is shown indirectly (via pigeonhole) rather than constructively.

  • $\begingroup$ Thanks for the help. $\endgroup$ Commented Sep 10, 2020 at 20:06
  • $\begingroup$ Also, In this case you can just show one example, can't you? Just let $n = 34$ and the triple be $(a,b,c) = (0,1,4)$. $\endgroup$ Commented Sep 10, 2020 at 20:10
  • $\begingroup$ Sort of. The claim is that for EVERY triple $(a,b,c)$, there's a number $n$. It's not enough to show that for some particular triple there's a number $n$ -- otherwise you'd just pick $(0,0,0)$ and $n = 0$. But you COULD do the proof by writing down all 30 possible triples and exhibiting an $n$ that works for each one of them. $\endgroup$ Commented Sep 10, 2020 at 20:13
  • $\begingroup$ Oh yes, that makes sense, I just misread it hahaha. The book suggest the constructive way of proving it, constructing for every $x \in U$, $\varphi_{A}(x) = \sup \{ \alpha \in [0,1]; x \in A_{\alpha}\}$. I'll give it a shot. $\endgroup$ Commented Sep 10, 2020 at 20:18
  • $\begingroup$ Thank you very much man, it was such an enlightenment for me. $\endgroup$ Commented Sep 10, 2020 at 20:20

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .