I am a self-teaching beginner and unfamiliar with proofs - in particular I find proposals that are intuitively true harder to prove as my brain assumes too much or skips steps.
I'd like help with my solution to the following (second part of) exercise 3.5.2 from Tao Analysis I 4th edition:
Show that if $(X_i)_{1 \leq i \leq n}$ are an ordered n-tuple of sets, then the Cartesian product, as defined in Definition 3.5.6, is indeed a set. (Hint: use Exercise 3.4.7 and the axiom of specification.)
The relevant part of Definition 3.5.6 is:
$$ \prod_{1 \leq i \leq n} X_i := \{ (x_i)_{1 \leq i \leq n} : x_i \in X_i \text{ for all } 1 \leq i \leq n \} $$
Exercise 3.4.7 proves that: The collection of all partial functions from $X$ to $Y$ is itself a set.
Other Questions
I am aware of another question (link) but the proposed solution looks too complicated to me, and the answers and comments are not helpful at my beginner level.
Exploration
I wanted to be clear in my mind what "an ordered $n$-tuple of sets" was an how it relates to functions as hinted at by the question.
Tao defines an $n$-tuple to be a function as:
Suppose we define an ordered $n$-tuple to be a surjective function $x : \{i \in \mathbb{N} : 1 \leq i \leq n\} \to X$ whose codomain is some arbitrary set $X$ (so different ordered n-tuples are allowed to have different ranges); we then write $x_i$ for $x(i)$ and also write $x$ as $(x_i)_{1 \leq i \leq n}$.
So if we have an $3$-tuple $(a,b,c)$, we can say this is a function $f$ with domain $\{1,2,3\}$ and codomain $\{a,b,c\}$, which maps as $f(1)=a, f(2)=b, f(3)=c$. The ordering is required so we know which element of the domain maps to an element of the co-domain.
The co-domain doesn't have to be a set of primitive objects, it can be a set of sets. So a $3$-tuple, $(X_1, X_2, X_3)$ corresponds to a function which maps as $f(1)=X_1, f(2)=X_2, f(3)=X_3$.
My Solution Attempt
We are given an n-tuple of sets, $(X_i)_{1 \leq i \leq n} \equiv \{X_1, X_2, \ldots, X_n\}$.
We can only apply the axiom of specification to a single set, so this suggests we form a set which contains every element of every $X_i$. This is the union, which exists by axiom 3.12 (existence).
$$ Y = \bigcup X_i $$
However, the elements of $Y$ are not elements mapped to by a function $(X_i)_{1 \leq i \leq n}$. Instead, it is the elements of the powerset of $Y$, $P(Y)$, that are mapped to by a function $(X_i)_{1 \leq i \leq n}$. The powerset $P(Y)$ is a set by Axiom 3.11 (existence).
So a function $F\equiv (X_i)_{1 \leq i \leq n}$ maps from a domain $\{1 \ldots n\}$ to a codomain which is a subset of $P(Y)$.
Such a function $F$ is a partial function, because its domain is a subset of $\mathbb{N}$, and its codomain is a subset of $P(Y)$.
Exercise 3.4.7 tells us that "the collection of all partial functions from $X$ to $Y$ is itself a set".
So the collection of all partial functions from $\mathbb{N}$ to $P(Y)$ is a set (existence). Let's call this collection (a set), $G$.
Now, a key realisation is the cartesian product is itself a partial function in $G$. It is a function that maps from $\{1 \ldots n\}$ to a single specific element of $P(Y)$. That single specific element is a set that conforms to the definition of the cartesian product of an ordered $n$-tuple of sets $X_i$ which are in $P(Y)$:
$$ \prod_{1 \leq i \leq n} X_i := \{ (x_i)_{1 \leq i \leq n} : x_i \in X_i \text{ for all } 1 \leq i \leq n \} $$
We can now apply the (existence) Axiom 3.6 of Specification to $G$ with a property statement $S(f)$ pertaining to each element $f \in G$:
$$S(f) \text{ is true if } f:\{1 \ldots n\} \to \{ (x_i)_{1 \leq i \leq n} : x_i \in X_i \text{ for all } 1 \leq i \leq n \} $$
Therefore the cartesian product of an $n$-tuple of sets is itself a set. $\square$
My Thoughts
I am unsure of two points:
The requirement to use the powerset of the union, not just the union.
The cartesian product is one single partial function in the collection of partial functions.
There may be other areas I am wrong about but unaware.
