[Math] Axiom of union; union of natural and real numbers

elementary-set-theory

In set theory we assume the axiom of union to be true for all universes, more formally $\forall x \exists y \forall [z \in y \Leftrightarrow \exists t (t \in x \land z \in t)]$. We call the set $y$ the union of $x$.

This is intuitively understood as the set consisting of the elements of the elements of $x$, but I have trouble understanding the scenario where the elements of $x$ aren't generally considered to be sets.

Take the natural numbers $\mathbb{N}$ as an example. After the axiom $\cup \mathbb{N}$ exists and it consists of the elements of the natural numbers; however, numbers aren't intrinsically sets. As everything has to be a set in set theory many different set theoretic constructions of our number systems have been devised, such as the Von Neumann construction of the natural numbers.

What I'm wondering is what $\cup \mathbb{N}$ or $\cup \mathbb{R}$ in fact are, especially in the usual set theoretic constructions. Do they have some kind of a deeper property or are they arbitrarily determined by the universe $\mathscr{U}$ and the construction of $\mathbb{N}$ and $\mathbb{R}$ in $\mathscr{U}$?

Best Answer

In the usual set-theoretic construction of $\mathbb{N}$, in which $0 = \{\}$ and $n+1 = n \cup \{n\}$, you have $$ \mathbb{N}=\{0,1,2,3,\ldots\}=\{\{\},\{0\},\{0,1\},\{0,1,2\},\ldots\}, $$ and so $$ \bigcup\mathbb{N}=\{0,1,2,3,\ldots\}=\mathbb{N}. $$ There's not a comparably simple construction of $\mathbb{R}$ in ZFC, so I think $\bigcup\mathbb{R}$, while it must exist, will depend heavily on the details. For instance, there are at least two different constructions in which the elements of $\mathbb{R}$ are equivalence classes (subsets) of another set: the set of Cauchy sequences in one case, and the set of Dedekind cuts in the other. Using these constructions, $\bigcup\mathbb{R}$ is either the set of the Cauchy sequences or the set of Dedekind cuts, which are clearly quite different.

Indeed, even $\bigcup\mathbb{N}=\mathbb{N}$, while elegant, isn't necessary. Other constructions of $\mathbb{N}$ include representing its elements as collections of finite subsets of some other infinite set $A$. (I.e., "3" is the set of all 3-element subsets of $A$, "4" is the set of all 4-element subsets, etc.) In this case, $\bigcup\mathbb{N}$ would be the set of all finite subsets of $A$.

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