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My question relates to a discussion I read showing the intersection of elements of the empty set is a universal set in ZFC. (I know this has been dealt with in several posts Why is the intersection of the empty set the universe? and Empty intersection and empty union for two.)

I would appreciate help as to why for a commutative and associative product operation, applying the operation to no elements gives the identity element $1$?

Then, how is the "product operation" related to intersection?

I would then think that $1$ corresponds to the universal set. Can I justify this surmise?

Thanks

EDIT Since having posted this, I have come to learn that one aspect of my difficulty is referred to as the "empty product." There are several questions here relating to it which can be found by searching on it.

Also, quite importantly, as Andres Caicedo points out in his comment, in ZFC there is no universal set, and I added a comment as to the relevance of my question.

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    If $V$ is the universal set, then for all $X$ we have $X\cap V=X.$ So $V$ is the neutral element for intersection, just as $1$ is for multiplication and $0$ for addition. – bof Aug 03 '17 at 21:21
  • If you are bothering to bring up $\mathsf{ZFC}$, then note that what you wrote makes no sense, as there is no universal set in this theory. – Andrés E. Caicedo Aug 04 '17 at 00:07
  • @AndrésE.Caicedo Yes I know that. Thank you for emphasizing that. The non-existence of the universe was established earlier, and my remarks are actually in the context of showing there is no need for an intersection axiom. In that regard ( I'm sure you are more than familiar with it) it said the set $S$ cannot be empty to avoid what you point out. I.e., the intersection of elements of the empty set (the universe) does not exist. Regards, –  Aug 04 '17 at 00:17

4 Answers4

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To see why $\displaystyle \bigcap_{i\,\in\,\varnothing} S_i = \text{the “universe''},$ it is not necessary to know how to view intersection as a product.

To see why $\displaystyle\sum_{i\,\in\,\varnothing} x_i = 0$ it is only necessary to see that if you add no numbers to the subtotal that you already have, the result is the same as if you had added the number $0.$

To see why $\displaystyle\prod_{i\,\in\,\varnothing} x_i = 1$ it is only necessary to see that if you multiply the product that you already have by no numbers, the result is the same as if you had multiplied it by the number $1.$

Likewise $\displaystyle \cdots\cap X\cap Y\cap Z\cap \bigcap_{i\,\in\,\varnothing} S_i = \cdots\cap X \cap Y\cap Z.$

One can also say that every time you add a new term to an intersection $\cdots\cap X\cap Y\cap Z$ you are excluding more things from membership in the intersection. If there are no such terms, then nothing has been excluded.

It is somewhat like the fact that $\sup\varnothing = -\infty.$ Every time you add more members to the set $A$, you add more things that $\sup$ cannot fall below. It falls as far downward as those constraints allow. That is why it is called the smallest upper bound. If there are no members, then there is nothing that it cannot fall below.

To view intersection as multiplication, think of indicator functions $\chi_A$ of sets $A$: $$ \chi_A(x) = \begin{cases} 1 & \text{if } x\in A, \\ 0 & \text{if } x\notin A. \end{cases} $$ Then the indicator function of the intersection of sets is the product of the indicator functions: $$ \chi_{A\,\cap\,B}(x) = \chi_A(x) \chi_B(x). $$

Michael Hardy
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  • Thanks you for this answer. I have been studying it. I recognize your opening remark - "not necessary to...view intersection as a product." Yet I would appreciate help as to how to approach it that way. In looking at the indicator functions, applying them to the elements of the empty set, I would think I get the product of $0$'s. I do see what you say about multiplying the product you already have by no numbers is equivalent to multiplying by $1$. But with the empty set itself, you only have no elements; so how do you get to the "universe." With regards, –  Aug 03 '17 at 22:01
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    @Andrew : Multiplying no numbers yields $1$ as the product, but multiplying by $0$ yields $0.$ Similarly, the intersection of no sets at all is the "universe", but in $A\cap B\cap C\cap\cdots,$ if one of $A,B,C,\ldots$ is the empty set, then the intersection is empty. $\qquad$ – Michael Hardy Aug 03 '17 at 23:32
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If $E$ is the universal set, then the power set $\mathcal P(E)$ is a complete lattice under set inclusion. The union of a family of subsets of $E$ is its least upper bound (or supremum), and the intersection is the greatest lower bound (or infimum). For the empty family, the least upper bound is $\emptyset$ (because every element of $\mathcal P(E)$ is an upper bound of the empty family), and the greatest lower bound is $E$ (because every element is a lower bound of the empty family).

bof
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  • Dear bof - This is a most illuminating answer. I accepted Michael's on the basis that he closely answered what I asked. I know you are extremely generous as on a prior occasion, you directed me to accept an answer that ostensibly said what you had earlier written in a comment. Thanks very much. Best regards, –  Aug 04 '17 at 00:07
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Intersection is a product. It's associative, commutative, and has an identity (the universal set, indeed). It has no inverse though.

The powerset of $X$ can be seen as a ring (a so-called Boolean ring) where the symmetric difference $A \mathbin \Delta B = (A \setminus B) \cup (B \setminus A)$ is the $+$, and $A\cap B$ is the $\times$. Neutral elements are $\emptyset = 0$ and $X = 1$ ,and the $+$ is indeed associative and $A + A = 0$ so every element is its own inverse. (Alternatively we get the same ring when we identify a set $A$ with its characteristic function $\chi_A :X \to \{0,1\}$ and use pointwise addition and multiplication modulo $2$ as operations.)

In older set theory and topology texts it was quite common to denote intersection of $A$ and $B$ by $A \cdot B$ (and union, confusingly, as $+$).

Michael Hardy
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Henno Brandsma
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  • Very elegant and clear. Would you please be kind enough to show how the product of no elements in fact equals $1$. Thanks. With regards, –  Aug 03 '17 at 21:31
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    The logarithm of the product of no elements is the sum of no elements, which equals $0.$ – bof Aug 03 '17 at 22:58
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This is just a convenient convention. If we have a binary operation $\circ : X \times X \to X$ on some set $X$ and $\circ$ happens to be associative, we want to define iterated versions of $\circ$: $\circ_n : X^n \to X$ that satisfy:

$$ (x_1 \circ_i \ldots \circ_i x_i) \circ (x_{i+1} \circ_{n-i} \ldots \circ_{n-i} x_n) = x_1 \circ_n \ldots \circ_n x_n$$

(where, in practice, we would always omit the subscripts on $\circ_i$ etc.). If $\circ$ happens to have a unit $e: X$ say, then it is very convenient to have the above equation hold when $i = 0$ or $i = n$, which means we want the zero-fold iteration of $\circ$, $x_1 \circ_0 \ldots \circ_0 x_0$ to be equal to $e$ so that:

$$ (x_1 \circ_0 \ldots \circ_0 x_0) \circ (x_1 \circ_{n} \ldots \circ_{n} x_n) = e \circ (x_1 \circ_n \ldots \circ_n x_n) = x_1 \circ_n \ldots \circ_n x_n$$

and similarly when $i = n$. (Here I'm pushing the "$\ldots$" notation a bit further than it really should go: I should write something like $\bigcirc_{i=1}^n x_i$ for $x_1 \circ_n \ldots \circ_n x_n$, so my $x_1 \circ_0 \ldots \circ_0 x_0$ above is the empty product $\bigcirc_{i=1}^0x_i$.)

Rob Arthan
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