I am currently reading Tao's Analysis I, specifically, the section about set theory and I got stuck with one exercise which consists of deducing some basic axioms of set theory assuming the axiom of universal specification.

Axiom (Universal Specification) Suppose for every object $x$ we have a property $P(x)$ pertaining to $x$ (so that for every $x$, $P(x)$ is either a true statement or a false statement.) Then there exists a set $\{x : P(x) \text{is true}\}$ such that for every object $y$, $$y \in \{x : P(x) \text{is true}\} \text{if and only if} \space P(y) \space \text{is true}$$

The exercise consists of deducing the following axioms from the universal specification:

Axiom 1

There exists a set $\emptyset$, known as the empty set, which contains no elements, i.e., for every object $x$ we have $x \not \in \emptyset$.

Axiom 2 If $a$ is an object, then there exists a set $\{a\}$ whose only element is $a$. Furthermore, if $a$ and $b$ are objects, then there exists a set $\{a,b\}$ whose only elements are $a$ and $b$.

Axiom 3 (Pairwise union)

Given any two sets $A,B$ there exists a set $A \cup B$, called the union $A \cup B$ of $A$ and $B$, whose elements consists of all elements which belong to $A$ or $B$ or both.

Axiom 4 (axiom of specification)

Let $A$ be a set, and for each $x \in A$, let $P(x)$ be a property pertaining to $x$. Then there exists a set, called $\{x \in A: P(x) \text{is true}\}$ whose elements are precisely the elements $x$ in $A$ for which $P(x)$ is true.

First of all I am having some difficulty understanding the universal specification axiom, isn't exactly by definition of the set $\{x : P(x) \text{is true}\}$ that $y$ is in this set if and only if $P(y)$ is true? So, why do we need an axiom to be able to say this?

So, now suppose we take the universal specification as an axiom, then 4 it's immediate since $\{x \in A: P(x) \text{is true}\}=A \cap \{x \in A: P(x) \text{is true}\}$.

I don't see how can I show the remaining three axioms assuming universal specification, any hints that could suggest me how should I prove the implications would be greatly appreciated.

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    $\begingroup$ Start by deriving a contradiction using Russell's paradox (namely, $\{x:x\notin x\}\in\{x:x\notin x\}$ and $\{x:x\notin x\}\notin\{x:x\notin x\}$ must both be true). Since everything follows from a contradiction, you will in particular get the axioms you list. $\endgroup$ – hmakholm left over Monica Dec 31 '15 at 4:09
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    $\begingroup$ To elaborate a bit on @HenningMakholm 's comment above, this particular Axiom of Universal Specification would seem to be equivalent to the axiom of unrestricted comprehension (from Frege's early attempt to axiomatize set theory?): For any unary predicate P, there exists a set $\{x: P(x)\}$. If we have $P(x)\equiv x\notin x$ then we can obtain a contradiction, namely Russell's Paradox. $\endgroup$ – Dan Christensen Dec 31 '15 at 4:34
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    $\begingroup$ Perhaps more in the spirit meant by the exercise, to prove the existence of the empty set, you could use $P(x)\equiv x\neq x$. $\endgroup$ – Dan Christensen Dec 31 '15 at 4:58
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    $\begingroup$ The axiom of universal specification wasn't meant to define sets, it was to give the existence of them, so only with it can you say such a set as described exists. $\endgroup$ – Sean English Dec 31 '15 at 17:16

Due to the fact that the "wrong" Axiom of Universal Specification [page 52] is introduced after the "right" axioms and in the context of the discussion of Russell's paradox, I think that Exercise 3.2.1 [page 54] asks for a "correct" derivation of the requested axioms, avoiding to draw upon the Principle of explosion (or ex contradictione (sequitur) quodlibet, "from contradiction, anything follows").

Having said that, from the above Axiom :

For every property $P(x)$ there exists a set $\{ x : P(x) \text { is true } \}$ such that for every object $y$,

$y∈ \{ x : P(x) \text { is true } \}$ if and only if $P(y)$ is true

we can derive Axiom 3.2 (Empty set) [page 40] using $x \ne x$ as $P(x)$ :

$y∈ \{ x : x \ne x \text { is true } \}$ if and only if $y \ne y$ is true.

But for no $y$ we have that $y \ne y$ is true, and thus for no $y$ we have : $y∈ \{ x : x \ne x \text { is true } \}$.

Thus, the set $\{ x : x \ne x \text { is true } \}$ has no elements, i.e. it is empty.

By Definition 3.1.4 (Equality of sets) [page 39] the empty set is unique: if there are two, having the same elements - i.e. none - they are equal.

For Axiom 3.3 (Singleton sets and pair sets) [page 41] we have tu use the formula :


Thus : $y∈ \{ x : x = a \text { is true } \}$ if and only if $y = a$ is true.

For Axiom 3.4 (Pairwise union) [page 42] we have to use as $P(x)$ the formula :

$x \in A \lor x \in B$.

Thus : $y∈ \{ x : (x \in A \lor x \in B) \text { is true } \}$ if and only if $(y \in A$ or $y \in B)$ is true.

For Axiom 3.5 (Axiom of specification) [page 45] the condition is trivially:

$x \in A \land P(x)$.

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