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
    @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.