Empty set and empty sum

Question

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)

Answer 1

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 $$\sum_{x \in \emptyset} x = 0$$

Answer 2

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.

Answer 3

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 $\mathcal{S}$!}

Answer 4

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.

Answer 5

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).

Answer 6

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).

Answer 7

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).

Answer 8

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)