What are the union and intersection class? From what I know, the union and intersection class of any class $A$ is defined as the following;
$$\bigcup A = \{x:x \in y \text{ for some } y \in A\}$$
$$\bigcap A = \{x:x \in y \text{ for every } y \in A\}$$
I am new to set notation so I am having trouble figuring out what it is. If I have a class $A = \{1,2,3\}$ what will its union and intersection class be?
I worked it out and from my understanding, the former is any subset of $A$ and the latter is simply the set of all subsets of $A$, but I don't feel that's right; it seems unnecessary since they are just repeating the definitions of a subset and a powerset.
 A: When we speak of $\bigcup A$ and $\bigcap A$, we require all the elements of $A$ to themselves be sets.
Depending on the precise foundational theory you're working in, one of a few things can happen.

*

*You're working in some variation of ZF set theory without atoms/urelements - that is you're working in a 1-sorted set theory - which does not support classes.

A 1-sorted theory is, as the name suggests, a logical theory in which there's only 1 sort of thing. In ZF, the only sort of thing that exists is sets. All the "things" discussed in ZF are sets.
In this case, all elements of $A$ are automatically sets, so the notation always makes sense.
In ZF, one typically defines $0 = \{\}$, $1 = \{0\}$, $2 = \{0, 1\}$, $3 = \{0, 1, 2\}$. In this case, we see that $\bigcup \{1, 2, 3\} = \{0, 1, 2\} = 3$. In fact, if $A$ is a finite set of natural numbers, $\bigcup A$ will be the largest element of $A$ and $\bigcap A$ will be the smallest element.
However, there is a small wrinkle with ZF-like theories. The wrinkle is that $\bigcap A$ is not defined when $A = \emptyset$. This is because according to the definition of $\bigcap \emptyset$, we would expect absolutely anything to be an element of $\bigcap \emptyset$ - that is, $\bigcap \emptyset$ would be the set of all sets. But we know there is no set of all sets, since if there were such a thing we would have Russell's Paradox.


*You're working in some variation of NBG set theory without atoms/urelements.

In NBG set theory, the things one quantifies over are "classes". There is, for example, a class of all sets. A "set" is a special kind of class which is "small enough" in some sense.
This allows a different approach to axioms than ZF. In ZF, there is an axiom saying that for all $a$, there is a set $P(a)$ of all subsets of $a$. In NBG, given a set $a$, you can immediately define the class $P(a) = \{b \mid b \subseteq a\}$; the axiom says that $P(a)$ is a set.
In NBG set theory, the intersection $\bigcap A$ is always defined. But if $A$ is empty, the intersection will not be a set. It will instead be a "proper class" - that is, a class which is not a set.


*You work in some variation of NBG or ZF with atoms/urelements.

In this version of events, the "things" one quantifies over might be "atoms". An atom is something which is not a set/class and which is thus not defined solely in terms of its elements (atoms are generally taken to have no elements at all). In such a set theory, we would only discuss $\bigcap A$ and $\bigcup A$ when all elements of $A$ are sets (not atoms), though (assuming we adopt the convention that atoms have no elements) the notation $\bigcap A$ and $\bigcup A$ makes sense even when $A$ has atoms - in this case, $\bigcup A = \bigcup \{x \in A \mid x$ a set$\}$ and $\bigcup A = \emptyset$.


*You work in some version of type theory (including, for the purposes of this discussion, topos theory).

In type theory, everything has a "type". It does not make sense in type theory to take the union or intersection of types.
However, given a type $T$, one can (sometimes) form the type $P(T)$, which is (roughly) the type of all subsets of $A$. Given some $A : P(P(T))$, one can define $\bigcup A = \{x : T \mid \exists y \in A (x \in y)\}$ and $\bigcap A = \{x : T \mid \forall y \in A (x \in y)\} : P(T)$.
Here, $t : T$ means "$t$ has type $T$".
In this account, $\bigcap A$ is actually well-defined even when $A = \emptyset$.
In type theory, trying to take the union $\bigcup \{1, 2, 3\}$ would be a "type error". One cannot even discuss it. This is because $1, 2, 3$ are not sets.
The type theoretic/category theoretic account of intersections and unions is a bit more subtle when one does not have "power set" types. I will not go into that here.
A: Let me bring general formal definition of union with respect to  indexed family:
Suppose we have set $X$, called indices set, and for each $\alpha \in X$ is defined some $U_\alpha$ set. Then, by definition, set
$$\bigcup\limits_{\alpha \in X}U_\alpha=\{x\colon \exists \alpha \in X, x\in U_\alpha\}$$
is called union of indexed family $\{U_\alpha\}_{\alpha \in X}$ with respect to indices set $X$. If, for example, we take $X=\{1,2, 3\}$, then we obtain  union of tree sets asked. For $X=\{1, \cdots, n\}$ it is union of $n$ sets etc.
For intersection existential quantifier is changed to universal.
Very interesting is example where $X= \emptyset$.
A: You can simply interpret $\bigcap$ and $\bigcup$ as follows: If $A$ is finite and $A=\{a_0,\cdots,a_{n-1}\}$, then
\begin{align*}
\bigcap A&=a_0\cap a_1\cap\cdots\cap a_{n-1},\\
\bigcup A&=a_0\cup a_1\cup\cdots\cup a_{n-1}.
\end{align*}
But when $A$ is infinite, for example $A=\{a_n\mid n\in\mathbb{N}\}$, we can't write them as follows:
\begin{align*}
\bigcap A&=a_0\cap a_1\cap\cdots,\\
\bigcup A&=a_0\cup a_1\cup\cdots,
\end{align*}
since $a_0\cap a_1\cap\cdots$ and $a_0\cup a_1\cup\cdots$ can't be well defined （i.e., $\cap$ and $\cup$ can't be used over infinite sets） while in fact you can still interpret $\bigcap$ and $\bigcup$ as above. And this is also why we introduce $\bigcap$ and $\bigcup$.
Now for $A=\{1,2,3\}$, it's trivial that $\bigcap A=1\cap 2\cap 3=1$ and $\bigcup A=1\cup 2\cup 3=3$.
Hope this can help you.
