6
$\begingroup$

Give:

$fn(S)=\prod_{x\in S}x$

what is:

$fn(\emptyset)$

I can see reason that it would be defined as 1 or 0.


Note: I thought about restricting the domain of $S$ but that would make the problem less general. If there is no general answer then answers for restricted domains would be valid.

  • 0
    What is $S$? Generally you'll make some convention that's convenient.2012-02-18
  • 1
    The usual convention is that an empty product is equal to $1$, and that an empty sum is equal to $0$.2012-02-18
  • 2
    Product in *what*? As an element of a commutative ring?2012-02-18
  • 0
    @Jonas, maybe infinite products are also allowed.2012-02-18
  • 1
    @Dan: I would agree, but I would like to know, infinite products of what? I.e., what is BCS asking? Only BCS can tell us. By the way, you made an excellent point in the beginning of your answer.2012-02-18
  • 0
    @Jonas, well I was thinking that more generally we could define a function $M:(P(S)\setminus \{\emptyset\} )\rightarrow S$, and if there is some element $1 \in S$ that acts like an identity, then it is only natural to define $M(\emptyset) = 1$. My answer had too many mistakes so I deleted it. The wiki article looks like a good reference for this.2012-02-18

2 Answers 2

11

The empty product equals 1.

  • 1
    Would it be more correct to say the "multiplicative identity"?2012-02-18
3

If you want obvious relation $fn(A\cup B) = fn(A) \cdot fn(B)$ for disjoint $A,B$ to hold, then you don't have any choice, empy product must be multiplicative identity.