Why is it that empty operations give the identity of that operation?

Wikipedia says that the empty product is taken to be 1 "by convention", but this convention is consistent with ideas that seem like an empty product (like $x^{0}$).

If I talk about the intersection of no sets (taking $U$ to be the "universal set") then logic demands the empty intersection be $U$ (because any element is vacuously in every set in an empty family, and is therefore in the empty intersection). $U$ indeed functions as the identity of intersections; for any $A \subseteq U$, $A \cap U = A$.

Same story for the union of sets. Logic demands that the union of no sets be the empty set, and this is indeed the identity of the union operation.

There seems to be much more than convention going on here, and it is consistent among binary operations of all types. So why is it that the operation on no operands tends to result in the identity of that operation?

  • 3
    $\begingroup$ You might want to check out en.wikipedia.org/wiki/Product_%28category_theory%29 Especially the part about the empty product under the discussion section. $\endgroup$
    – J126
    Nov 12, 2015 at 4:19
  • $\begingroup$ These seem more like operations involving the empty set as argument. $\endgroup$
    – mvw
    Nov 12, 2015 at 4:20
  • 7
    $\begingroup$ (1) It's a useful definition. (2) It's the only useful definition. $\endgroup$ Nov 12, 2015 at 4:34

3 Answers 3


The reason for the phenomenon is that we want our operation to be associative. We want the sum of two sums to be one big sum, for instance, and we want this to hold regardless of how many terms are in each sum.

$$\prod_{i\in \emptyset} f(i)\times \prod_{i\in A}f(i) = \prod_{i\in (\emptyset\cup A)} f(i)=\prod_{i\in A}f(i)$$

$$\sum_{i\in \emptyset} f(i)+ \sum_{i\in A}f(i) = \sum_{i\in (\emptyset\cup A)} f(i)=\sum_{i\in A}f(i)$$

$$\bigcup_{i\in \emptyset} f(i)\cup \bigcup_{i\in A}f(i) = \bigcup_{i\in (\emptyset\cup A)} f(i)=\bigcup_{i\in A}f(i) $$

$$\bigcap_{i\in \emptyset} f(i)\cap \bigcap_{i\in A}f(i) = \bigcap_{i\in (\emptyset\cup A)} f(i)=\bigcap_{i\in A}f(i)$$

In all cases, the desire to have the above hold uniquely determines what the empty operation must be.

Note: as filipos points out, we want $\prod_{i\in A}f(i)\times \prod_{i\in B}f(i)=\prod_{i\in (A\cup B)} f(i)$ for any disjoint sets $A,B$.

  • 1
    $\begingroup$ Just a note that unions here should be disjoint unions in general. In any case, the reasoning holds. $\endgroup$
    – filipos
    Nov 12, 2015 at 11:49
  • 1
    $\begingroup$ Example: $\sum_{i \in \{1, 2, 3\}} 1 + \sum_{i \in \{1, 2, 3\}} 1 \neq \sum_{i \in \{1, 2, 3\}} 1$. $\endgroup$
    – filipos
    Nov 12, 2015 at 14:30
  • $\begingroup$ Thank you for this. Does this imply that any associative operation must take (the appropriate) 1 as its empty operation? $\endgroup$ Nov 12, 2015 at 16:56
  • $\begingroup$ @JonathanHebert, let's not go too far here. Not every associative operation even has a neutral element (a "1"). But these important examples all do. $\endgroup$
    – vadim123
    Nov 12, 2015 at 17:13

Pardon, in advance, all the kooky notation I've made up in order to answer this question.

Everything you mention is an instance of a family of operators $\bigodot_{\alpha} : X^{\alpha} \to X$ on some set (or class) $X$ and some set (or class) of ordinals $\alpha$, such that:

  • $\bigodot_{i<1} x_i = x_1$ for all $x_1 \in X$, that is, the $1$-ary $\odot$ is the identity function; and
  • For all $\alpha \ge 1$, $\bigodot_{i<\alpha+1} x_i = \left( \bigodot_{i < \alpha} x_i \right) \odot x_{\alpha+1}$ for all $( x_i : i < \alpha ) \in X^{\alpha}$.

Here I've written $\bigodot_{i<\alpha} x_i$ as shorthand for $\bigodot_{\alpha} ( x_i : i < \alpha )$, and $x \odot y$ as shorthand for $\bigodot_2 ( x,y )$.

In order for the second property above to hold for all $\alpha$, not just all $\alpha \ge 1$, you need to have, for all $x \in X$, $$\left(\bigodot_0 ( ) \right) \odot x = \bigodot_1 \{ x \} = x$$ and likewise $x \odot \left( \bigodot_0 () \right) = x$.

Thus, if it exists, $\bigodot_0()$ (i.e. the 'empty $\odot$') must be the identity for $\odot$.

This can be justified at a more general level. The operators you mention are all instances of products and coproducts in particular categories. The empty product is the terminal object of the category (if it exists), and the empty coproduct is the initial object (if it exists). Indeed:

  • Multiplication of numbers. The natural numbers (thought of as von Neumann ordinals) form a category $\mathsf{FinOrd}$, whose morphisms are functions. In this category, the product coincides with multiplication of natural numbers as we know it, and $1$ is the terminal object, hence the empty product. The coproduct coincides with addition of natural numbers, and $0$ is the initial object, hence the empty sum.
  • Intersection of sets. The class of all sets forms a category, where a unique morphism $X \to Y$ exists if and only if $X \subseteq Y$. In this category, coproducts are unions and the empty set is the initial object, hence the empty union. Products are intersections, but there is no terminal object (since that would imply the existence of a universal set), so there is no empty intersection. However, in the subcategory consisting of subsets of some fixed set $U$, the terminal object is $U$ itself, hence $U$ is the empty intersection in this subcategory.
  • $\begingroup$ Adhering to this, what if we have magma $(M, *)$ with no identity element? You say that we must have the empty operation as the identity, if it exists. If it does not, must we leave it undefined? I am unfamiliar with category theory (that seems to be where this question is most relevant), so pardon if this is an ignorant question. $\endgroup$ Nov 12, 2015 at 4:55
  • 1
    $\begingroup$ @JonathanHebert: Right, if there is no identity element then there is really no sense to be made of the empty operation. This is the reason why, at least formally, the (global) empty intersection is undefined, because if we were to give it a definition then that definition would have to be the universal set, which doesn't exist (in ZFC at least). (Caveat: if we want to consider only the subsets of a fixed set, say a topological space, then we can consider the empty intersection to be that fixed set.) $\endgroup$ Nov 12, 2015 at 5:01
  • 1
    $\begingroup$ While this post does a great job in describing the phenomenon the OP refers to in a generalized way, I don't think it really answers the question. Why is it defined like that, even in this generality? In other words: why is it desirable that property number 2 holds for all $\alpha$ and not just for $\alpha\ge1$? $\endgroup$ Nov 12, 2015 at 5:07

It's a very useful convention, at very least; in some cases, though, it really is the correct value given the definitions involved.

As you note, the intersection of an empty collection of sets it's not empty; rather, it's everything. This can be seen by expanding the definition. Suppose $\mathcal{A}$ is empty. Then $$\begin{align} x \in \bigcap \mathcal{A} &\iff \forall A\,(A\in\mathcal{A} \to x\in A) \\ &\iff \forall A\,(A\in\emptyset \to x\in A) \\ &\iff \forall A\,(\mathsf{False} \to x\in A) \\ &\iff \forall A\,(\mathsf{True}) \\ &\iff \mathsf{True} \end{align}$$ So every $x$ is, in theory, a member of the intersection. This isn't a set, unless implicitly all $A\in \mathcal{A}$ are subsets of some set $X$, as is typical. *When that's the case, the empty intersection is $X$, because then implicitly $x$ is constrained to be in $X$.

A similar situation arises more generally in complete lattices — partially ordered sets $(X,\preceq)$ such that every $Y\subseteq X$ has a $\sup$ (least upper bound) and an $\inf$ (greatest lower bound) with respect to the ordering $\preceq$. Work through the definitions and you'll see that in a complete lattice, $$\begin{align} \sup \emptyset &= \mathbf{0} \quad\text{the 'bottom', least element, and} \\ \inf \emptyset &= \mathbf{1} \quad\text{the 'top', greatest element.} \end{align}$$ In fact, because the definition of a complete lattice implies that $\sup$ and $\inf$ exist for $\emptyset$, the definitions of $\sup$ and $\inf$ imply that for every element $x$ of the lattice, $\sup\emptyset \preceq x \preceq \inf \emptyset$, and this establishes that a complete lattice actually has least and greatest elements ($\mathbf{0}, \mathbf{1}$ respectively).

This is consistent with what happens with $\bigcup$ and $\bigcap$: these are $\sup$ and $\inf$ operations in powersets $\mathcal{P}(X)$, which are complete lattices with respect to the partial ordering $\subseteq$.

  • $\begingroup$ Unless I am mistaken, your warning regarding the empty intersection appears to be precisely what I said in the question. $\endgroup$ Nov 12, 2015 at 4:50
  • $\begingroup$ I must have misread you on that point. Anyway, the way the question analyzes it is just right. But that's only when $\bigcap$ is implicitly or explicitly relative to some "universe' or containing set; without such a constraint, $\bigcap \emptyset$ is the entire universe of set theory, which isn't a set :) I thought the complete lattice generalization might be enlightening. $\endgroup$
    – BrianO
    Nov 12, 2015 at 4:56
  • $\begingroup$ I will have to chew on it a bit, probably. Thank you. :) $\endgroup$ Nov 12, 2015 at 5:17
  • $\begingroup$ You're welcome. You'll see, the very same reasoning applies, as every $x$ in a lattice is, trivially, both an upper and a lower bound of $\emptyset$. For sup, as for $\bigcup$: $s=\sup\emptyset$ means — $(\forall a\in\emptyset) a\le s$ [vacuously true], and $\forall x$ in the lattice, if $(\forall a\in \emptyset) a\le x$ [vacuously true], then $s\le x$... This all simplifies to: $\forall x$ in the lattice, $s\le x$. Similarly for $\inf$. $\endgroup$
    – BrianO
    Nov 12, 2015 at 5:23

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