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This is such a simple question but I couldn't find the answer on the internet. What is the default result of the Big Pi notation when it happens to be applied to an empty set? Is it 1 or 0, or even undefined until I specifically define it?

$$\prod_{x \in X} x,\quad X=\emptyset.$$

LSpice
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Gwangmu Lee
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    It is $1$, just as empty sums are $0$. This has the advantage that you can break the product (or summation) into disjoint pieces without changing the product (or sum) even if some if the pieces are empty. At least in some cases. – lulu Dec 01 '22 at 16:32
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    https://en.wikipedia.org/wiki/Empty_product – Thomas Dec 01 '22 at 16:34
  • Thank you all! I just couldn't come up with the term "empty product" before I see it. :) – Gwangmu Lee Dec 01 '22 at 16:38
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    Cf. this question and this question – J. W. Tanner Dec 01 '22 at 16:39
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    See https://math.stackexchange.com/questions/1951490/when-indexing-set-is-empty-how-come-the-union-of-an-indexed-family-of-subsets-o/1952303#1952303 – Ethan Bolker Dec 01 '22 at 16:40
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    Semantically speaking, one should distinguish $\phi$ \phi from $\emptyset$ \emptyset. (There is also $\varnothing$ \varnothing.) I have edited accordingly. \ I would understand the notation $\prod_{x \in \emptyset} x$ (which is $1$, as @lulu says), but I do not understand $\prod_{x \in X} x, X$. What is the "$, X$" doing? – LSpice Dec 01 '22 at 16:42
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    I understand OP asking what is $\prod_{x \in X} x$ when $ X=\emptyset$ – J. W. Tanner Dec 01 '22 at 16:47
  • @J.W.Tanner, ah, right! I thought the formula was claiming that $\prod_{x \in X} x, X$ equalled $\emptyset$, whatever that meant, rather than asking about $\prod_{x \in X} x$ when $X = \emptyset$. Your interpretation makes much more sense, and I will edit accordingly. – LSpice Dec 01 '22 at 18:30
  • The way I say it is we need a default for doing nothing or doing something no times. And that default has to be the identity. In this case... the multiplicative identity which is $1$. This is pretty much how it has to be. After all of $\prod_{a\in A}\prod_{b\in B} =\prod_{c\in A\cup B} c$ (assuming $A,B$ are multisets [or disjoint]) then we kind of have to have $\prod_{a\in A}a = \prod_{a\in \emptyset \cup A} a=\prod_{b\in \emptyset} b\times \prod_{a\in A}a$. I think of this as "background noise"; background noise is the starting point; the identity. This is also why $b^0=1$. – fleablood Dec 01 '22 at 18:45
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    I used to teach exponentiation to inner-city first graders via consensus. Whether $b^0= 0$ or $1$ was always the first stumbling block with half the eight-year olds insisting $b^ab^0=\underbrace{b\times ....\times b}{a\text{ times}}\times \underbrace{}{0\text {times}}=b^{a+0}=b^a$ meant it had to be $1$ and the other half insisting it had be that if you multiply $b$ times itself zero you can't get anything and that had to be zero. We'd let the kids fight it out for a week with no guidance. The $b^0=1$ side would always win out eventually. – fleablood Dec 01 '22 at 18:56

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