Axiom 3.1: (Sets are objects). If $A$ is a set, then $A$ is also an object. In particular, given two sets $A$ and $B$, it is meaningful to ask whether $A$ is also an element of $B$.

Axiom 3.3 (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 set formed by $a$ and $b$.

Axiom 3.4 (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 the elements which belong to $A$ or $B$ or both. In other words, for any object $x$, $$x \in A \cup B \iff (x \in A \text{or} x \in B).$$

Axiom 3.11 (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 $x$ $$ x \in \bigcup A \iff (x \in S \text{for some} S \in A).$$

Exercise 3.4.8. Show that Axiom 3.4 can be deduced from Axiom 3.1, Axiom 3.3 and Axiom 3.11.

I am struggling with this exercise in Analysis 1 by Tao. My attempt is as follows: Let $C = \{A , B\}$. Then, $x \in \bigcup C \iff x \in A \text{or} B$ by the axiom of the union, and $\bigcup C$ is a set. I think that $\bigcup C$ is equal to $A \cup B$, but I don't know how to complete the proof. I appreciate if you give some help.


Yes, $\bigcup C=A\cup B$, because $$x\in\bigcup C\iff (x\in A) \text{ or } (x\in B) \iff x\in A\cup B \,. $$

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  • $\begingroup$ Thanks, but I haven't used the first two axioms. Why do we need them for this exercise? $\endgroup$ – shk910 Feb 23 at 11:10
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    $\begingroup$ @shk910 to guarantee that $\{A,B\}$ is a set $\endgroup$ – Alessandro Codenotti Feb 23 at 11:55

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