Naive set theory, empty set as subset of mathematical objects

Question

Hello I am self teaching foundational math, I am thinking about union of a set. It definition assumes the set to have only elements that are also sets, else it would break. But I started thinking is the empty set infact an element of any mathematical object, like say the integer n? In that case taking the union of the set U {1,{2,3}} would yeild {2,3} Do I understand this correctly? Also since the integers can be constructed as sets of empty sets etc, the empty set would be an element of any integer. But does this extend to any mathematical object?

My question boils down to do union of a set imply that the set is composed of sets, or do we allow the empty set to be the result when asking for the element if any mathematical object that is not an explicit set.

Answer 1

In modern treatments of set theory, for the most part everything is a set.

Naive set theory is—well—naive. Since it is not axiomatized or formalized in full, this question is a bit awkward to answer. But the general wisdom is that if you write $x\in y$, then $y$ is a set. Namely, only sets can have elements. This is also the case in set theories with urelements, or atoms, or objects which are not sets.

In particular, this means that if we define $\bigcup\cal A$ to be $\{a\mid\exists A\in\mathcal A:a\in A\}$, then all the non-sets objects which might exist in $\cal A$ sort of disappear. In other words, if $\bf Set$ is the collection of all sets, then $\bigcup\mathcal A=\bigcup(\mathcal A\cap\mathbf{Set})$.

The point here is that mathematics is written by humans and for humans. It is us who decide what to do with the end cases, or whether we ignore the exceptions or type errors, or however else we handle them. This is why mathematical texts have definitions. To formalize this sort of thing. And once you start formalizing naive set theory... you end up with a more modern treatment of it.

Answer 2

In a naive set theory, you typically just start with binary unions: $\{a, b, c\} \cup \{c, d, e\} = \{a, b, c, d, e\}$. The resulting set contains those (and only those) elements which appear in at least one of the input sets.

You can then extend this principle to arbitrary unions, where if you say $A_i$ is a set for all $i \in I$, then you can form $\bigcup_{i \in I} A_i$, which contains those (and only those) elements which appear in at least one of the $A_i$.

If you collect these $A_i$ into a single set themselves, for instance $\mathcal A = \{A_0, A_1, A_2, A_3, \ldots\}$, then you can write the "union of $\mathcal A$", usually denoted $\bigcup \mathcal A$, for what we called above the union of the elements of $\mathcal A$. This is typically, as far as I know, the way the union is defined/worked with in axiomatic set theory, where everything is always a set, and thus in particular it makes sense to define the union on "the set of its operands".

To get back to your question: yes, the union operation only really makes sense when applied to (a collection of) sets. It's unclear what the union of a set and a non-set (or of two non-sets) should be. Of course, if you wanted to, you could define this to be whatever you want (including "skipping over" non-sets) but there doesn't appear to be a particular reason why you would want that definition.

But I started thinking is the empty set infact an element of any mathematical object

No, the empty set is a subset of any set. If all your mathematical objects are sets, then the empty set is a subset of every object. However, the empty set is certainly not an element of every object: it's not an element of itself!