Arbitrary Unions and Intersections I'm currently reading James Munkres Topology and I just have a few questions (two of them will be very quick, the other, you will have to read my "understanding"):
As said by Dr. Munkres,

$ $When we have a set whose elements are sets, we shall often refer to it as a collection of sets and denote it by the script letter $\mathcal{A}$ or $\mathcal{B} $

Question 1: So does this mean that whenever a set is thought of as a collection of sets, the elements of this set are themselves strictly sets?
To illustrate my question, suppose we had the following set:

$\mathcal{A}=\{\{1,2,3,4\},\{1,2,6,7,8\},\{6,7,8\},\{8\},9,8\}$

Would this be unacceptable given that 9 and 8 are both objects and not sets?
Also stated by Dr. Munkres,

$ $Given a collection of $\mathcal{A},$ the union of the elements of $\mathcal{A}$ is defined by the equation: $ $
$$\bigcup_{A\in \mathcal{A}}A=\{x\mid x\in A\,\exists A\in\mathcal{A}\}$$ (1)
$ $The intersection of the elements of $\mathcal{A}$ is defined by the equation: $ $
$$\bigcap_{A\in \mathcal{A}}A=\{x\mid x\in A\,\forall A\in\mathcal{A}\}$$
(2)

Note: the set notation used by the author in (1) was actually $\{x\mid x\in A$ for at least one $A\in\mathcal{A}\}$ and $\{x\mid x\in A$ for every $A\in\mathcal{A}\}$ for (2).
Question 2: To start off, to my understanding, the elements of the set $\mathcal{A}$ are denoted individually as $A,$ and given that each $A$ is a set itself (assuming question 1 is correct), each $A$ will contain objects as elements, and for the sake of how the author defined the above equations, we can say that individually, the objects of $A$ are denoted as $x$. To my understanding, the union of the elements of $\mathcal{A}$ will contain every object which is in each element of $\mathcal{A}$ (given that each element is in fact itself a set made up of objects), and if one particular object (for example) is found within all of the elements that make up $\mathcal{A},$ the union of the elements of $\mathcal{A}$ will also contain this object. On the other hand, the intersection of the elements of $\mathcal{A}$ will be made up of all the objects that are common to each and every element of $\mathcal{A}$ and if no such objects are common to every element of $\mathcal{A},$ well then the intersection of the elements of $\mathcal{A}$ will be the empty set. Is this the correct way of thinking about it or am I way off?
Final question: if you'd like to "describe" each element of $\mathcal{A},$ we could allow each individual element of $\mathcal{A},$ which we denote as $A,$ to take on a consecutive  index from the natural numbers (ie. $A_{i\in\mathbb{N}^+}).$ But how can we do this? I thought sets had no order?
 A: Question 2: Yes, this is exactly it: an element belongs to the union of some sets if it belongs to at least one of them, and the intersection contains only the common elements for all given sets.
Question 1: Yes, basically correct, probably something like that was the intention of the author. 
However, using the most widely accepted axiom system ZFC of set theory, we find that every object is a set, meaning that all of our everyday mathematical objects (such as numbers, points on plane, etc.) are represented somehow as sets. 
So in this sense, it's rather just a notation that suggests that elements of $\mathcal A$ will be handled as sets.
Final Question: Well, one of the benefits of these definitions of union and intersection is that they are not defined as extensions of the binary $\cup$ and $\cap$ operations, like we could inductively define
$$A_1\cup A_2\dots\cup A_n:=(A_1\cup\dots A_{n-1})\cup A_n$$
but we can do it at once for all $A\in\mathcal A$, without assuming any ordering on $\mathcal A$. 
However, if you insist, using the axiom of choice (letter 'C' in ZFC) one can put a well-order to any set, in particular, to $\mathcal A$. (A linear order is well-order if every subset contains a unique minimal element, like in $\Bbb N$).
A: Q1: No ,even if you are permitting non set individuals (as you seem to regard "9" and "8") you still have the empty set $\emptyset$ so if you replace "8" and "9" by $\emptyset$ your collection can still be regarded as a collection of sets. Most treatments of set theory as the foundations for mathematics do not have individuals (i.e. non sets) but do need an empty set . Of course you are much of the time not interested in the members of say the number 8 ,but note that in the Dedekind cut construction of mathematics the real numbers are infinite sets but are in most uses in  mathematics are regarded as objects . How you wish to interpret you objects (set or non set) does not effect the logic of doing Mathematics rigorously .
