Why can't we define arbitrarily large sets yet after defining these axioms? (Tao's Analysis I)

Question

In Tao's Analysis I I am very confused why he says we do not have the rigor to define arbitrarily large sets after defining the below 2 axioms:

Axiom 3.4 If $a$ is an object, then there exists a set $\{ a \}$ whose only element is $a$, i.e., for every object $y$, we have $y \in \{ a\}$ if and only if $y = a$

Axiom 3.5 (Pairwise union). Given any two sets $A, B$, there exists a set $A \cup B$, called the union of $A$ and $B$, which consists of all the elements which belong to $A$ or $B$ or both. In other words, for any object $x$, $$ x \in A \cup B \Longleftrightarrow(x \in A \text { or } x \in B) . $$

(Prior to this point Tao mentioned the axiom of extension and axiomatized the existence of the empty set).

With these axioms, it seems that we are in a position to define sets which have $n$ objects for any natural number- we just apply this union operation $n$ times. However, after these axioms have both been defined, Tao (after seemingly acknowledging this) says this is not the case:

This axiom allows us to define triplet sets, quadruplet sets, and so forth: if $a, b, c$ are three objects, we define $\{a, b, c\}:=\{a\} \cup\{b\} \cup\{c\}$; if $a, b, c, d$ are four objects, then we define $\{a, b, c, d\}:=\{a\} \cup\{b\} \cup\{c\} \cup\{d\}$, and so forth. On the other hand, we are not yet in a position to define sets consisting of $n$ objects for any given natural number $n$; this would require iterating the above construction " $n$ times", but the concept of $n$-fold iteration has not yet been rigorously defined. For similar reasons, we cannot yet define sets consisting of infinitely many objects, because that would require iterating the axiom of pairwise union infinitely often, and it is not clear at this stage that one can do this rigorously. Later on, we will introduce other axioms of set theory which allow one to construct arbitrarily large, and even infinite, sets.

Tao first seems like he is saying we can construct arbitrarily large sets, as by saying "so forth" in his first sentence, he seems like he is allowing that progression to get arbitrarily large through some sort of inductive progression. However, he then says that we are not able to do so yet, which confuses me. I do not understand why we need more axioms to define these. These are the axioms he introduces after this (I do not understand why we need any of them to define arbitrarily large sets):

Axiom 3.6 (Axiom of specification). Let $A$ be a set, and for each $x \in A$, let $P(x)$ be a property pertaining to $x$ (i.e., for each $x \in A, P(x)$ is either a true statement or a false statement). Then there exists a set, called $\{x \in A: P(x)$ is true $\}$ (or simply $\{x \in A: P(x)\}$ for short), whose elements are precisely the elements $x$ in $A$ for which $P(x)$ is true.

Axiom 3.7 (Replacement). Let $A$ be a set. For any object $x \in A$, and any object $y$, suppose we have a statement $P(x, y)$ pertaining to $x$ and $y$, such that for each $x \in A$ there is at most one $y$ for which $P(x, y)$ is true. Then there exists a set $\{y: P(x, y)$ is true for some $x \in A\}$, such that for any object $z$, $$ \begin{aligned} z \in\{y & : P(x, y) \text { is true for some } x \in A\} \\ & \Longleftrightarrow P(x, z) \text { is true for some } x \in A . \end{aligned} $$

Axiom 3.8 (Infinity). There exists a set $\mathbf{N}$, whose elements are called natural numbers, as well as an object 0 in $\mathbf{N}$, and an object $n++$ assigned to every natural number $n \in \mathbf{N}$, such that the Peano axioms (Axioms 2.1-2.5) hold.

(Prior to Axiom $3.8$, Tao in the previous chapter included the assumption, marked as informal, that "There exists a number system $\mathbf{N}$, whose elements we will call natural numbers, for which Axioms 2.1–2.5 [The Peano Axioms] are true", which makes me confused about the necessity of this axiom- are we not formally using this assumption?)

Axiom 3.10 (Regularity). If $A$ is a non-empty set, then there is at least one element $x$ of $A$ which is either not a set, or is disjoint from $A$.

Axiom 3.11 (Power set axiom). Let $X$ and $Y$ be sets. Then there exists a set, denoted $Y^X$, which consists of all the functions from $X$ to $Y$, thus $$ f \in Y^X \Longleftrightarrow(f \text { is a function with domain } X \text { and codomain } Y) \text {. } $$

Axiom 3.12 (Union). Let $A$ be a set, all of whose elements are themselves sets. Then there exists a set $\bigcup A$ whose elements are precisely those objects which are elements of the elements of $A$, thus for all objects $x$ $$ x \in \bigcup A \Longleftrightarrow(x \in S \text { for some } S \in A) \text {. } $$

I do not understand why we needed further axioms than the ones we had at the point at which Tao said we needed more axioms in order to define arbitrarily large sets. What is the reason why we weren't able to construct arbitrarily large sets with only axioms 3.4 and 3.5?

EDIT: I now have 2 further things I am confused by with respect to the new axioms allowing us to define arbitrarily large sets:

1. why it is that even with the axioms, we can define arbitrarily large sets- we haven't yet defined the cardinality of sets and haven't defined an association between the size of a set and the natural numbers; we also haven't shown a system satisfying the Peano axioms can describe the amount of elements in a set. Why is it the case that somehow axiom 3.8 allows us to do this and all the sudden induct on the number of elements in a set? We would need to show that elements of $\mathbf{N}$ can describe the amount of elements in a set first, right?

2. @Henry and @MichaelCarey said we can construct sets of large finite numbers,such as sets with a million elements, but not arbitrarily large sets- and I was thinking that we could just have a generic finite number $n$ which is a placeholder for any arbitrary number, and reason with $n$ the same we do any specific number, thereby constructing a set of arbitrarily large size; but @MichaelCarey said we need an axiom to generalize the use of our logic with a specific finite number to a generic number $n$- which axiom here does this, and why would we need an axiom to do this- it seems like just common sense?

Answer 1

Notice the difference between:

• the external statement "for every $n$, we can prove that there is a set with $n$ elements";
• the internal statement "we can prove that, for every $n$, there is a set with $n$ elements".

The point of the induction principle/axiom is to internalise the first statement.

Answer 2

Ask yourself: can you write down a proof there exists a set with $1$ element? I'm sure that the answer is yes. Okay, so go write that proof down......

Now that you're done with that, ask yourself: can you write down a proof that there exists a set with $2$ elements? I'm sure that the answer is yes. So, okay, go write down that proof......

Next up: can you write down a proof that there exists a set with $3$ elements? I'm confident that you can, I'll wait while you write that proof down....

Okay, so, let's keep doing this until noon tomorrow? Okay? Then I have to go to a meeting. I'm sure you'll get up to some large number.

Oh? This is boring? Well, okay, here's a very different job for you: write down a proof that for every natural number $n$, there exists a set with $n$ elements.

Answer 3

Recall that the axioms tell us exactly what are sets: finite proofs resting on the axioms give us sets. Nothing else is a set. You will see that I use collection (and "not-set") for things which are not proven sets (at the various points of the development of the axioms).

So, you can make a sequence of finite proofs:

• (I think you have the empty set...) $\varnothing$ is a set.
• $\{a\}$ is a set.
• $\{a,b\}$ is a set.
• ...

Note that the ellipses are not a proof in the set theory. They are an argument in a meta-theory, one which actually contains a copy ("model") of the naturals to act as a spine for the induction.

Now here's a mind-bender. What proof can we write in the set theory up to Axiom 3.5 that the sequence of sets elided (by the ellipses) is actually modelled by the natural numbers. It's some collection of sets only finitely many of which can appear in any proof. Do we know that the naturals are so universal that the presumably infinite collection of sets parallels the naturals? How do we know that? How do we prove that using the axioms so far? We don't. Nothing in the theory up to Axiom 3.5 allows us to conclude any of that.

We have introduced a meta-theoretic/informal copy of the naturals in the preceding of Axiom 3.8, but it is worthwhile to note that we only ever use a finite initial segment of these informal naturals in any proof. Nothing we do prior to Axiom 3.8 allows us to call the naturals a set. Nor can we compare the entire infinite not-set of meta-theoretic natural numbers to the collection of sets we started to explicitly write down above. (This collection of sets may or may not be a set. Can you prove that it is using only the axioms up to Axiom 3.5?)

If we want our theory of sets to be able to reason about the entire collection of natural numbers, we need to import that collection into the theory. And, it will be a big step since to that point, all sets have been finite. So we introduce Axiom 3.8. The naturals are now a set. They weren't a set until we made an Axiom that said so. Recall: the axioms tell us exactly what are sets: finite proofs resting on the axioms give us sets. Since nothing in the preceding axioms gave us an infinite set, nothing in the preceding could prove that the naturals are an infinite set nor could reason about the informal collection of naturals. We could reason about finite segments, but no finite proof about finite segments of the naturals will address the set-ness of the collection of all the informal naturals.

Once you have that the naturals are a set, then you can do most of the things you write that you want to do with them. Until the naturals are a set, all the set theory theorems about sets do not apply to the naturals.

Then we can show that the usual models of the naturals as sets actually are models of the set of naturals from Axiom 3.8. There has to be a set-thing that, from the axioms, is the naturals before this set theory can prove anything about that set-thing.

Answer 4

You can prove any of these statements:

1. There is a set with 0 elements
2. There is a set with 1 element
3. There is a set with 2 elements
4. There is a set with 3 elements
5. There is a set with 4 elements
6. $\ldots$

You cannot prove this statement:

• Given $n$, there is a set with $n$ elements

In that statement, the input $n$ is an abstract element that has to be dealt with in the proof. You don't have the ability yet to handle this abstract quantity $n$.

Answer 5

Most answers are highlighting the distinction between $\forall n, \vdash P(n)$ versus $\vdash\forall n, P(n)$, where $\vdash P(n)$ means "there is a proof of $P(n)$". While this distinction is very important, it's not entirely applicable in this context.

It turns out that the axioms 3.4 and 3.5 (together with Extensionality and Empty Set) are sufficient to define an infinite wellorder, and to prove the singular statement encoding the fact that we have arbitrarily large collections. The wellordering property is substantially weaker without the principle of Specification, but it's still clear that we prove the existence of arbitrarily large sets.

It's entirely possible that Tao didn't know about the construction I referenced, as the details are rather subtle. It's more likely, however, that he was simply saying that the notions of "iteration" and "natural number" were not properly defined at that particular point in his formalization. His statements were very vague, so it's hard to nail down exactly what he was trying to say, so it's really not your fault for being confused by it.

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Answer 6

The principal of mathematical induction Axiom 2.5 is in the previous chapter, and Tao does not introduce this to his set theory until Axiom 3.8 (Axiom 3.7 in my edition). He does not need this for specific large unions (e.g. those made with a million unions) but he does for arbitrarily large finite numbers of unions.

The same axiom gives him an infinite set $\mathbf N$, but does not yet imply the possibility of a infinite number of unions. Axiom 3.12 (3.11 in my edition) gives this and so arbitrarily large numbers of unions more generally.

Answer 7

Some other previous answers wrongly claim that Tao's excerpt implies that we can prove "there is a set with 2 elements". No. $\{a,b\}$ may not have 2 elements, unless you can prove $a ≠ b$. It's unclear whether Tao himself made this mistake, since his excerpt is actually a bit ambiguous.

Nevertheless, the point is that even if you can write down a property $Q$ and prove that it holds for every explicit natural number (i.e. "$0$" or of the form "$1+\cdots+1$"), it does not imply that you can prove "$∀k{∈}ℕ \ ( \ Q(k) \ )$"!

That is, it is possible that you can prove "$Q(0)$" and "$Q(1)$" and "$Q(1+1)$" and "$Q(1+1+1)$" and so on, but cannot prove "$∀k{∈}ℕ \ ( \ Q(k) \ )$"! This is not due to any defect in logical reasoning but instead is a fundamental phenomenon. Given any foundational system $F$ for mathematics, which we use for all our proofs, if $F$ can prove all the basic properties of natural numbers (i.e. that $⟨ℕ,0,1,+,·,<⟩$ forms a discrete ordered semiring), then we can in fact write down an explicit property $C$ and prove (within $F$) that $F$ proves "$C(0)$" and "$C(1)$" and "$C(1+1)$" and "$C(1+1+1)$" and so on and either $F$ proves "$0 = 1$" or $F$ cannot prove "$∀k{∈}ℕ \ ( \ C(k) \ )$"! This is a direct consequence of the Gödel-Rosser incompleteness theorem (for which you can read a relatively simple proof here).

By the way, relying on "common sense" is often foolish in mathematics; unless you can perform exceptionally clear logical thinking, your "common sense" is likely to be flawed, in which case the easiest way to avoid fallacies is by using rigorous logical reasoning and not intuition.

Answer 8

1. "Any object" does not include any object we can think of (what if it's a paradoxical object?) It includes any object we can show exists by using the axioms we have available.

2. Axiom 3.12 implies Axiom 3.5 (if we have the pairing axiom) because if we let $A = \{B, C \}$, then $\cup A = B \cup C$. Axiom 3.5 does not imply Axiom 3.12 but says that if we have two sets their union is a set; Axiom 3.12 says that if we have a set $A$, the set made by taking the union of all the elements of $A$ is a set.

Axiom 3.12 gives us that if $\{a,b,c, \ldots \}$ is a set, then $a \cup b \cup c \ldots$ is a set.

The set $A$ may be infinite, and no finite amount of applications of Axiom 3.5 can produce the repeated union of all the elements of $A$.

Meta-mathematically, we can construct sets to be arbitrarily large finite cardinalities.

If we need a set with $50$ elements, we know how to construct such a set. And we can intuit that such a process can work for any finite cardinality.

But can we prove, on the basis of the axioms, that there exists a set with arbitrary amounts of elements?

Furthermore, how do we reason about infinite sets? Does our intuition work with infinite sets? Don't we need some rigorous method for dealing with infinite sets/constructions, where our intuitions don't extend to?

Does the Axiom of Union + Axiom of Set Existence + Axiom of Pairing prove that there exist sets of arbitrary size?

Sure, we can 1-by-1 construct increasingly bigger sets.

But, can you write out a proof of the existence of a set of cardinality ${{10^{\text{google}}}^{\text{google}}}^{\text{google}}$? - Even if you could write on individual atoms, there aren't enough in the Observable Universe to produce such a proof.

For that, we need some kind of Recursion Theorem, which is the point of the additional axioms.

Also, sure, you can construct a set of some cardinality $n$, but can you construct a set of arbitrary finite cardinality?

However big you construct, you can never cover all possible sizes in a finitely long proof by taking one union pair at a time, and so you cannot prove that such an object exists.