Does iterating the Union operator yield the empty set?

Question

Background

In elementary set theory the union is taught as a binary operator on two sets, which contains all the elements of both those sets.

$$ A = \{1,2,3\}\\ B = \{2,3,4\}\\ A\cup B = \{1,2,3,4\} $$

In formal set theory this is generalized to a unary operator on one collection of sets, acting on each of the sets in that collection.

$$ \bigcup \{A,B\} = \{1,2,3,4\} $$

And now we can apply the union operator to a collection of not only two, but an arbitrary number of sets …

$$ C = \{1,4,5\}\\ \bigcup \{A,B,C\} = \{1,2,3,4,5\} $$

… even infinitely many.

$$ N_1 = \{1\} \land N_2 = \{2\} \land N_3 = \{3\} \land \cdots\\ \bigcup \{N_1,N_2,N_3,\dots\} = \{1,2,3,\dots\} $$

We can even take the union of a union of sets. $$ D = \{A,B\}\\ E = \{B,C\}\\ \bigcup \bigcup \{D,E\} = \bigcup \{A,B,C\} = \{1,2,3,4,5\} $$

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Intermission

Advanced axiomatic set theory teaches the idea that everything is a set, and that the first set guaranteed to exist is the empty set. Other sets must be built from that starting point.

For example the natural number $0$ is the empty set $\emptyset$, the natural number $1$ is constructed as the singleton $\{0\}$, the natural number $2$ can be constructed as the pair $\{0,1\}$ (depending on the author), and so on.

The Axiom of Existence is the only axiom that guarantees that a set—the empty set—exists without the input of other sets. Other sets like ordered pairs and unions can only exist given an input.

The Axiom of Union states, “given a collection of sets, there exists a set who owns any element of any set in that collection” (as opposed to “all elements of all sets in the collection,” which would yield the intersection). The point is, a union set can exist, but the only way to create it is by giving, as an input, an existing set. That existing set could be a pair or a subset or even another union, which also require inputs; or it could be the empty set, which requires no previous set to exist. The bottom line is that the empty set is the foundation upon which all other sets are built.

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Question

So finally here’s my question. If you take the union of any set $S$, and then take the union of that set, and keep going, will this process eventually yield the empty set?

$$ \text{hypothesis: } \left(\forall S\right)\left(\bigcup\bigcup\cdots\bigcup S = \emptyset\right) $$

Answer 1

First of all, “eventually” should be rigorously defined. Given a set $S$, and its union $\bigcup S$, and that union $\bigcup\bigcup S$, and so on… let’s say that at some point in that process you do get the empty set $\emptyset$. Then when you take its union, you’ll get the empty set again, and this process will go on forever. Suffice it to say that this process will never stop, because you can take the union of any set.

We’re looking for a set for which this process continues indefinitely, and fails to yield the empty set. It seems that an infinite set is a good candidate. Let’s try the set of Natural Numbers.

Use the method of construction given

\begin{align} 0 &= \emptyset\\ 1 &= \{0\} = \{\emptyset\}\\ 2 &= \{0,1\} = \{\emptyset, \{\emptyset\}\}\\ 3 &= \{0,1,2\} = \{\emptyset, \{\emptyset\}, \{\emptyset, \{\emptyset\}\}\}\\ &\cdots \end{align}

and the set that contains all of these

$$\mathbb{N} = \{0,1,2,3,\dots\}$$

Now the union of $\mathbb{N}$ must contain any element in any set inside $\mathbb{N}$. That is, $\bigcup\mathbb{N}$ contains all the elements in $0$, all the elements in $1$, and in $2$, and in $3$, and so on. Keeping in mind that every set in $\mathbb{N}$ contains all the elements of the set “before” it, let’s start a list:

$$ \bigcup \mathbb{N} = \{0,1,2,3,\dots\} $$

What do you know! This is exactly the set of Natural Numbers! The union of $\mathbb{N}$ really contains all the elements of $\mathbb{N}$.

$$ \bigcup \mathbb{N} = \mathbb{N} $$

This is a perfect counterexample to the hypothesis. You can take as many unions as you want of $\mathbb{N}$ and you will always get $\mathbb{N}$.

$$ \bigcup\bigcup\cdots\bigcup \mathbb{N} = \mathbb{N} \not= \emptyset $$

Answer 2

(This refers to an earlier version of the question, where it was said that $\bigcup\emptyset$ does not exist.)

Your question has been answered. This is just a note to point out that there is no such caveat; there's no problem with $\bigcup\emptyset$. By definition $\bigcup S$ is the set of $x$ such that there exists $y\in S$ with $x\in y$; this applies perfectly well to the empty set and shows that $\bigcup\emptyset=\emptyset$.

As Noah points out, there's no such thing as $\bigcap\emptyset$. Why is that? Well, the definition is that $\bigcap S$ is the set of all $x$ such that $x\in y$ for every $y\in S$. If $x$ is anything at all then $x\in y$ for every $y\in\emptyset$, since there are no such $y$. So if $\bigcap\emptyset$ existed it would be the set of all sets, which leads to contradiction.

(No, there's no mention of this in the axioms. There's no mention of $\bigcap$ at all in the axioms. It's a theorem that if $S$ is nonempty then $\bigcap S$ exists, and the proof doesn't work if $S=\emptyset$.)

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The following is really part of a comment that doesn't fit very well into the comment box:

def WrongSum(data):
  res = data[0]
  for j = 1 to len(data)-1:
     res = res + data[j]
  return res

def RightSum(data):
  res = 0
  for j = 0 to len(data)-1:
     res = res + data[j]
  return res

They're the same, except when data is empty, in which case WrongSum crashes and RightSum gives the right answer.