Union of a set of (finite) sets

Question

Axiom of Union: Given a set of sets $A$ (all of whose elements are sets) there exists a set $\bigcup A$ such that for every object $x$, $$x \in \bigcup A \iff (\exists S\in A \quad x\in S) $$ (Modified form Tao's Analysis 1)

Given a set $I$ and sets $A_{i} \forall i \in I$ we have, $$\bigcup_{i \in I}A_i = \bigcup \{A_i\ : i \in I \}$$ How do I prove the existence of the set $\{A_i:i \in I\}$ given $A_i$ as above? I am hinted at the use of the axiom of replacement but can't think of a proper property (or formula) that defines the set.The axiom of replacement is stated as, (Copied)

(Replacement): Let $A$ be a set. For any object $x ∈ A$, and any object $y$, suppose we have a statement $P(x, y)$ pertaining to $x$ and $y$, such that for each $x ∈ A$ there is at most one $y$ for which $P(x, y)$ is true. Then there exists a set $\{y :\exists x∈ A\quad P(x, y)\}$, such that for any object $z$, $$z ∈ \{y :\exists x∈ A\quad P(x, y)\} \iff \exists x \in A \quad P(x,z).$$

Answer 1

The sentences

if one has some set $I$, and for every element $\alpha \in I$ we have some set $A_\alpha$

and

we often refer to $I$ as an index set, and the sets $A_\alpha$ are then called a family of sets, and are indexed by the labels $\alpha \in I$

are intuitively clear, but mathematically not very precise. Indeed, a few pages later Tao spends some time in introducing the concept of an ordered $n$-tuple which is nothing else than a family of objects $x_i$ indexed by the set $\{1,\ldots, n\}$. This shows that the general conecpt of a family is not that trivial as it looks at first glance. For $n$-tuples Tao goes even deeper in an exercise; see functional definition of tuples (Terrence Tao Analysis, execise 3.5.2)

Anyway, let us not further discuss this point.

Here is Tao's axiom of 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 $\{y : P(x, y) \text{ is true for some } x \in A\}$, such that for any object $z$, $$z \in \{y : P(x, y) \text{ is true for some } x \in A\} \\\Longleftrightarrow P(x, z) \text{ is true for some } x \in A.$$

In the original formulation of your question you wrote $\{y : P(x, y) , x \in A\}$ instead of $\{y : P(x, y) \text{ is true for some } x \in A\}$ which is too sloppy because you replaced the existential quantifier "for some" (cf. Tao's Appendix A.4) by a comma.

To apply the axiom of replacement, Tao works with the following statement $P(\alpha, y)$ applicable to any object $\alpha \in I$ and any object $y$ :

• $P(\alpha, y)$ holds true iff $y = A_\alpha$.

Now we can use the axiom of replacement for the set $I$ and the statement $P(\alpha, y)$ and get the set $$\{y : P(\alpha, y) \text{ is true for some } \alpha \in I \} = \{ y : y = A_\alpha \text{ is true for some } \alpha \in I\}$$ which is simply denoted by $$\{A_\alpha : \alpha \in I\} .$$

Note that the above approach works for any family of objects $y_\alpha$ indexed by $\alpha \in I$; we do not need to assume that the $y_\alpha$ are sets. We always get the set $$\{y_\alpha : \alpha \in I\} .$$

Answer 2

Let $F=(a_i)_{i\in I}$ be a family (of sets or whatsoever) indexed by a set $I$, i.e. $F$ is a set of ordered pairs such that for each $i∈I$, there is a unique ordered pair in $F$ whose first component is $i$ (the second one being then denoted by $a_i$).

Its range, $R=\{a_i:i\in I\}$, is defined by $$a\in R\iff\exists i\in I\quad(i,a)\in F,$$ i.e. by using the axiom of replacement: $$R=\{a:\exists i\in I\quad P(i,a)\},\text{ where }P(i,a)\text{ is }(i,a)\in F.$$