Understanding mathematical terminology in Aluffi I'm reading the first section of Aluffi's "Algebra: Chapter 0" and don't fully understand this below passage concerning naive set theory.

But note that while it is clear from the definitions that, for example,
$$S_1 \cup S_2 \cup S_3 = (S_1 \cup S_2) \cup S_3 = S_2 \cup (S_2 \cup S_3),$$
it is not so clear in what sense the sets
$$S_1 \times S_2 \times S_3, \; (S_1 \times S_2) \times S_3, \ S_1 \times (S_2 \times S_3) $$
should be `identified' (where we can define the leftmost set as the set of 'ordered triples' of elements of $S_1, S_2, S_3$, by analogy with the definition for two sets.) In fact, again, we can really make sense of such statements only after we acquire the language of functions. However, all such statements do turn out to be true, as the reader probably expects; by virtue of this fortunate circumstances, we can be somewhat cavalier and gloss over such subtleties.

My biggest confusion is the use of the word 'identified.' This is the second time the author uses this term with respect to a set. I'm also lost as to what the author means by "all such statements turn out to be true."
 A: For example, the statement $$(X):\quad\{1,2\}\times \{3,4\}\times \{5,6\}\mbox{ has $8$ elements}$$ is meaningless, technically: since $\times$ is not associative, an expression like "$\{1,2\}\times \{3,4\}\times \{5,6\}$" is ambiguous (compare "$2+3+5$" and "$2-3-5$"). The non-associativity of $\times$ can be seen by carefully thinking about the definition of ordered pairs: we will generally have $((a,b),c)$ and $(a,(b,c))$ not be literally the same object.
However, statement $(X)$ above is "morally correct:" obviously no matter how we make sense of the ambiguous expression $\{1,2\}\times \{3,4\}\times \{5,6\}$, the resulting set does indeed have $8$ elements. There's a general principle that as long as we don't ask too detailed questions about the sets we build using $\times$, we don't need to be entirely precise with our parentheses.
Now once you understand that $\times$ is indeed non-associative, the previous paragraph should worry you a lot at first; it should feel like exactly the sort of slippery "reasoning" that mathematics is all about not allowing. And at the moment it is! But with some care, and a clever application of functions (as the text suggests), we can in fact turn that vague idea into a perfectly precise result.
A: If you want to be really really really precise, $S_1\times(S_2\times S_3)=\{ (a, (b, c))\ |\ a\in S_1, b\in S_2, c\in S_3\}$ and $(S_1\times S_2)\times S_3 =\{ ((a, b), c)\ |\ a\in S_1, b\in S_2, c\in S_3\}$. Since $((a, b), c)\neq(a, (b, c))$, those two sets are not equal. However, one can just use them as if they were the same set, a set of triples $(a, b, c)$. This is what he means when he "identifies" those two sets.
