Generalized Set Operation I can't seem to understand Generalized Set Operations. Can someone explain it for me? For example, in Example 1.12, suppose k = 3 then set A(of k) = {3, 4, 5, 6 ...}, then how is {3,4,5 ...} = {1,2,3 ...}

 A: When you're learning set theory, the definitions of union and intersection you're first given are binary: they're defined on a pair of sets.
\begin{align}A\cup B&=\lbrace x:x\in A\text{ or }x\in B\rbrace\\
A\cap B&=\lbrace x:x\in A\text{ and }x\in B\rbrace\end{align}
Since they're also associative (the order you do a chain of consecutive unions or intersections in doesn't matter), you can define the union or intersection of any finite collection of sets.  This is the first pair of equations your book gives above.
However, it's always good to generalize your definitions!  In particular, those definitions don't allow us to take a union or intersection of an infinite collection of sets -- even though the operation makes intuitive sense to us!  (The intersection of sets is just the elements that are in all the sets, for example, and that should make just as much sense with an infinite number of sets as with a finite number of sets.)
So what we do is redefine union and intersection to apply to a single set -- this time, a set of sets.  This is $\mathcal{A}$ above, and the second pair of equations your book gives.  There are a couple different notations for this:


*
*$\bigcup\mathcal{A}$

*$\bigcup(A|A\in\mathcal{A})$ (the one your book uses)

*$\bigcup_{A\in\mathcal{A}}A$

*$\bigcup_{i\in I}A_i$, where $I$ is what's called an _indexing set_.  (For example, if $I=\mathbb{N}$, this is equal to $A_1\cup A_2\cup A_3\cup\dotsb$.

For me, the last notation is the most revealing.  I don't know if you're familiar with cardinality yet.  If not, it's a measure of the size of a set, and one of the most important theorems in set theory is that there is no greatest cardinality -- that is to say, there are multiple sizes of infinity. 
Now, in the indexing set notation, $I$ itself doesn't really matter as much as its cardinality.  If it is $\mathbb{N}$, you're taking the union of $A_1,A_2,\dotsc$, a countable number of sets.  But it could also be $\mathbb{R}$, in which case you're taking the union of the sets $A_x$ for every real number $x$.  Because $\mathbb{R}$ has a higher cardinality than $\mathbb{N}$, there's no way to write this out in a list!  So thinking of this as $$A_{x_1}\cup A_{x_2}\cup\dotsb$$ is just wrong.  You can't get this union from a generalization of the binary union.  Hopefully this shows how powerful this generalization is.
To conclude, I'll explain the example at the bottom of the page.  In this case, the set $A_n=\lbrace n,n+1,n+2,\dotsc\rbrace$: that is, all integers greater than or equal to $n$.  The union $\bigcup_{n\in\mathbb{N}}A_n$ is the set of integers that are in at least one set $A_n$.  But every $n\ge 1$ is in such a set -- for example, $n\in A_n$.  So this union is equal to $\mathbb{N}$.  On the other hand, no integer is in all sets $A_n$ -- for example, $n\not\in A_{n+1}$.  So $\bigcap_{n\in\mathbb{N}}A_n=\emptyset$.
I hope this helps.  Ask if you're still confused!
