Conceptual question about set theory fundamentals: Tao's Analysis I Chapter 3

Question

In Tao's Analysis I, Chapter $3$ introduces the fundamentals of set theory. In particular, the following lemma and axiom are provided (in this order...i.e. lemma precedes the axiom in the chapter layout):

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Lemma (Single Choice): Let $A$ be a non-empty set. Then there exists an object $x$ such that $x \in A$

There is an accompanying mini-proof by contradiction that establishes this.

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Axiom (Singleton sets and pair sets): 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 $y=a$; we refer to $\{a\}$ as the singleton set whose element is $a$. Furthermore, if $a$ and $b$ are objects, then there exists a set $\{a,b\}$ whose only elements are $a$ and $b$; i.e. for every object $y$, we have $y \in \{a,b\}$ if and only if $y=a$ or $y=b$; we refer to this set as the pair formed by $a$ and $b$.

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The aim of this question is not meant to be pedantic; I want to make sure I am understanding the difference between the two statements...because the order of presentation has me slightly confused.

It seems to me that it would have made more sense to present the Axiom (Singleton sets and pair sets) first because this is effectively establishing the existence of sets beyond just the empty set (which was stated one page earlier)...i.e. this establishes that "non-empty sets" exist (albeit, only non-empty sets containing one or two elements).

The lemma seems like it is already baked into this axiom...for example, the singleton set $\{a\}$ is literally defined by $a\in\{a\}$ and the pair set is literally defined by $a\in \{a,b\}$. And this is sort of why I am confused.

Why is there a need for this lemma when the axiom essentially already states this? Moreover, I am anticipating that all sets are effectively "derived" from the union of singleton and pair sets...and therefore, once again, the axiom already encodes the lemma, even for non-empty sets greater than size $1$ and $2$.

Any clarity would be greatly appreciated!

Answer 1

In Tao's system there are more than one "sort" of mathematical objects; numbers, sets, functions.

For simplicity, consider natural numbers: we have numbers: $0,1,2,\ldots$ and sets of numbers: $\{ 0 \}, \{ 1,3,5,\ldots \}$, etc.

The author assumes the existence of sets: at least one, the empty set (Axiom 3.2).

For every non-empty set $A$ (if any) we can always pick one object $x$ belonging to $A$: $x \in A$.

The proof is trivial, using the definition of empty set: $\forall x \ \lnot (x \in \emptyset)$, and the (stated see (why?)) fact that the empty set is unique.

Assume that the lemma to be proved is false, i.e. that $\lnot \exists x (x \in A)$; but this is the definition of empty set. Thus, $A= \emptyset$, contradicting the assumption that $A$ is not empty.

Having assumed (Axiom 3.2) that at least one set exists is not enough for our purpose; thus, in order to develop mathematics, we have to "produce" more sets from the existing stuff (up to now: the empty set and numbers).

How to do this ? With further axioms that we can classify as:

• "conditional existence" axioms, i.e. axioms producing new sets from existing ones:

Axiom 3.3 (Singleton sets and pair sets), as well as Axiom 3.5 (Axiom of specification).

With them we can generate e.g. $\{ \emptyset \}$, i.e. a set with one element, and then apply the trivial Lemma above to conclude that: $\emptyset \in \{ \emptyset \}$;

• "unconditional existence" axioms, i.e. axioms asserting the existence of brand new sets:

Axiom 3.7 (Infinity): There exists a set $\mathbb N$, whose elements are the natural numbers.

This is a much more interesting axiom for mathematics: we have that $0 \in \mathbb N, 1 \in \mathbb N$, etc. In addition, using Specification, we may prove that the set $\{ 0,1 \}$ exists (it is a subset of $\mathbb N$), and much more using other "conditional" axioms.

See e.g. Set Theory and Zermelo-Fraenkel Set Theory.