# Empty set and empty sum

Empty set is closed under addition. If not, there would be at least one element in the empty set whose addition with itself is not there, contradiction. Thus the result follows. This not a convention, but a claim, right?

Empty sum is defined to be zero. This is a convention.

But these two contradict each other. Empty set should contain zero then. What am I missing? Can a convention contradict a claim?

(By the way, this is not my own question, I heard it over from some student in BU, Turkey)

• In the von Neumann construction of the natural numbers, zero is the empty set, so the "empty sum" is empty.... Jun 5, 2017 at 14:30
• Your argument in the first paragraph does not work, by the way. The set $\{1,2,4,8,16,32,\ldots\}$ is not closed under addition, but you still cannot find any element "whose addition with itself is not there". Jun 5, 2017 at 14:39
• @HenningMakholm So, should I say, ... if not, there would be at least two elements in the empty set whose sum is not there? Jun 22, 2017 at 19:49
• @Serpenche: Yes. Or, more precisely, "at least one pair of two elements". Jun 22, 2017 at 20:09
• @HenningMakholm What is the difference between "one pair of two elements" and "two elements"? Jun 22, 2017 at 20:20

I would say that $\sum_{x \in \emptyset} x = 0$ is more than just a convention.

If we split a set $A$ into two disjoint subsets $A_1, A_2$, i.e. $A_1 \cup A_2 = A$ and $A_1 \cap A_2 = \emptyset$, then we expect $$\sum_{x \in A} x = \sum_{x \in A_1} x + \sum_{x \in A_2} x$$

Now we can take $A_1 = A$ and $A_2 = \emptyset$ which gives $$\sum_{x \in A} x = \sum_{x \in A} x + \sum_{x \in \emptyset} x$$ For this to be valid, we must have $$\sum_{x \in \emptyset} x = 0$$

Why would that be a contradiction?

The claim

$$\forall x, y\in \emptyset: x+y\in\emptyset$$

is a true statement, and it is independent of the fact that $$\sum_{x\in\emptyset} x$$

is defined as $0$ or anything else.

• In what way independent? Why empty set does not contain zero then? Jun 22, 2017 at 19:38
• @Serpenche Independent in the sense that no matter what you define the sum $\sum_{x\in\emptyset} x$ to be equal to, the claim $\forall x,y\in\emptyset: x+y\in\emptyset$ will always be true.
– 5xum
Jun 23, 2017 at 6:43

This is a clever paradox! I like it. The trick lies in conflating the following two statements:

1. "$$\mathcal{S}$$ is closed under addition"
2. "For any subset† $$\mathcal{U} \subseteq \mathcal{S}$$, $$\sum_{x \in \mathcal{U}} x \in \mathcal{S}$$"

If these were both equivalent, then the paradox would go through, because the empty set $$\mathcal{S} = \emptyset$$ is trivially closed under addition, so taking $$\mathcal{U} = \mathcal{S}$$ above gives us $$\sum_{x \in \mathcal{S}}x \in \mathcal{S}$$. But $$\sum_{x \in \mathcal{S}}x$$ is the empty sum, and the empty sum equals‡ $$\sum_{x \in \mathcal{S}}x = 0 \not \in \mathcal{S}$$ (since $$\mathcal{S}$$ is empty and has no elements), which is a contradiction.

However, the first is strictly weaker than the second, because the definition of "closed under addition" is specifically:

$$\text{For all *two-element* subsets } \mathcal{U} = \{x_1, x_2 \} \subseteq \mathcal{S}, \sum_{x \in \mathcal{U}} x = x_1 + x_2 \in \mathcal{S}.$$

By induction, we can show that

$$\mathcal{S} \text{ is closed under addition} \longrightarrow \sum_{x \in \mathcal{U}} x \in \mathcal{S} \text{ for any finite subset } \mathcal{U} \subseteq \mathcal{S} \text{ of size } |\mathcal{U}| \geq 2.$$

But the empty set $$\mathcal{U} = \emptyset$$ has size $$|\mathcal{U}| = 0$$, and we can't inductively apply the definition of "closed under addition" to prove $$\sum_{x \in \mathcal{U}} x \in \mathcal{S}$$ for subsets of $$\mathcal{S}$$ with so few elements§. So the apparent contradiction disappears.

†Actually, $$\mathcal{U}$$ should be a multiset whose elements are from $$\mathcal{S}$$, in order to allow repetition of summands, such as adding an element of the set to itself.

‡The exact value of the empty sum depends on what our empty set $$\mathcal{S} = \emptyset$$ is considered to be a set of. If $$\mathcal{S}$$ is a set of real or complex numbers, the empty sum is zero; if $$\mathcal{S}$$ is a set of vectors or matrices, the empty sum is the zero vector or matrix; if $$\mathcal{S}$$ is a set of n-manifolds under connected sum, the empty sum is the n-sphere; and so on.

§It is true that if $$|\mathcal{U}| = 1$$ (that is, $$\mathcal{U} = \{ a \} \subseteq \mathcal{S}$$ is a singleton set),

$$\sum_{x \in \mathcal{U}} x = \sum_{x \in \{ a \}} x= a \in \mathcal{S},$$ but this is not a consequence of $$\mathcal{S}$$ being closed under addition--in fact, it's true for any set $$\mathcal{S}$$!

I don't think it's a contradiction.

The sum of the numbers in the empty set is $0$, by convention. Why should that imply that the set itself contains $0$, its sum? The sum of the set $\{ 1,2\}$ isn't in the set.

The convention is a good one. It makes particular sense in computer programs, where you compute a sum as

sum = 0
for i in S
sum = sum + i


If the loop body never executes the sum is $0$.

The analogous convention for products is

product = 1
for i in S
product = product * i


so the empty product should be $1$.

This model has far reaching generalizations (in lisp in particular).

• "The sum of the numbers in the empty set is 0, by convention. Why should that imply that the set itself contains 0, its sum?" Because empty set is closed under addition. Jun 22, 2017 at 19:51
• @Serpenche You could require that, since the empty sum is $0$. But I don't think you have to. " Closed under addition" means "whenever $x$ and $y$ are there, so is $x+y$". For the empty set you never have and such $x$ and $y$ so you never have to evaluate a sum, so you don't even need the empty sum. Jun 22, 2017 at 20:35

You appear to think that closed under addition means that any sum of elements is also in the set. That's why you think the empty sum must be in the set.

But closed under addition means the sum of any two elements (could be the same) is also in the set.

The empty sum is the sum of zero elements. Not of two. So it doesn't need to be in the set in order for the set to be closed under addition.

• (Yes, this was already hidden in previous answers, but that's exactly how I perceive it - they embedded this in lengthy complicated talk so that this is rather hidden there.) Mar 28 at 10:55
• Thank you very much! The sum of $\textbf {zero}$ elements can not be anywhere let alone in the set. Mar 29 at 11:13

It is a convention in rings that the empty sum is the zero element of the ring and the empty product is the unit element of the ring (if the ring has a unit element).

When we ask whether something is closed under the binary operation $+$, the only thing we're interested in is that operation applied to exactly two (not necessarily different) elements of the set.

This is useful because we can then talk about being closed under operations that don't have neutral elements, or are not commutative or associative. For such operations it doesn't even make sense to ask whether the operation applied to a different number of elements than two yields something in the set.

So the empty set is closed under addition.

But it is arguably (!) not closed under finite sums. (That depends on whether we consider the empty sum a finite sum. There are other associative and commutative operations where it makes sense to talk about the operation applied to any non-empty finite families of input -- for example the $\min$ operation -- but where there is still no neutral element).

• So, we admit that empty set does not contain infinitely many elements and when we add infinitely many non-elements we get zero? :) I do not quite get it...If empty set is not closed under infinite sums then empty set should contain infinitely many elements whose sum is not there? Jun 22, 2017 at 20:17
• @Serpenche: I have no idea what you're talking about here. Did you misread "finite" as "infinite"? Jun 22, 2017 at 21:01
• Exactly, I misuderstood. Jun 23, 2017 at 8:25

The argument in "Empty set is closed under addition"

"The sum of every element in the empty set is zero (i.e. Empty sum is zero)"

"Therefore $0=1$" is similar to

"All the elements of the empty set is black", "All the elements of the empty set is white" therefore black is white. (By the valuable help of instructor Burak Kaya, METU)