8

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.

BCS
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  • What is $S$? Generally you'll make some convention that's convenient. – Dylan Moreland Feb 18 '12 at 04:17
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    The usual convention is that an empty product is equal to $1$, and that an empty sum is equal to $0$. – André Nicolas Feb 18 '12 at 04:18
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    Product in what? As an element of a commutative ring? – Jonas Meyer Feb 18 '12 at 04:22
  • @Jonas, maybe infinite products are also allowed. – Dan Brumleve Feb 18 '12 at 04:30
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    @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. – Jonas Meyer Feb 18 '12 at 04:35
  • @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. – Dan Brumleve Feb 18 '12 at 04:46

2 Answers2

13

The empty product equals 1.

Daniel Pietrobon
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4

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.

Rafael Mrden
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