Is the definition of a Sigma Algebra not implied by the Power Set? Given a *subset of the* power set of $X$, called $A$, $A$ is a Sigma algebra of $X$ if:


*

*$X$ is an element of $A$

*The complement of a set $B$, element of $A$, with reference to $X$, is also in $A$.

*A countable collection of sets in $A$ has a union which is also in $A$.


This is the setup as I understand it. What I am curious about is the seeming redundancy between these conditions, and the characteristics a power set will have implicitly. I cannot think of a power-set which would violate any of the conditions of a sigma-algebra. Am I mistaken? If so, what are/is the counter-example(s)?
*correction
 A: The power set is a subset of itself, and it is a sigma algebra, but there may be others.  
In practice, for example in measure theory on $\mathbb{R}$, we don't take the power-set because it is too complicated.  Instead, we work with a smaller sigma-algebra, like the Borel sigma-algebra on $\mathbb{R}$, which is generated by open intervals.  
This ultimately allows for the consistent theory of Lebesgue Integration which extends the Riemann integral.  Why do we want to extend the Riemann Integral?  That's another story.  (It has to do with completeness of the function spaces $L^p(\mathbb{R})$.)
A: A $\sigma$-algebra on $X$ is a collection of subsets of $X$. As such, it is a subset of the power set of $X$, but it need not be all of the power set of $X$.
Now, each of the conditions you give are such that, if you start with subsets of $X$, you will end up with subsets of $X$. That guarantees that the conditions do not require a $\sigma$-algebra to “go outside” the power set of $X$; that is, every element of $A$ is a subset of $X$, and you don’t need to consider things that are not subsets of $X$. So, for example, $X$ is a subset of $X$, so asking that $X\in A$ does not require your $A$ to have things that are not subsets of $X$. If $B\in A$, then $B$ is a subset of $X$, and hence so is $X\setminus B$. And a union of subset of $X$ is also a subset of $X$.
However, not every collection of subsets of $X$ is a $\sigma$-algebra. For example, if $X=\{a,b\}$, and $A=\{\varnothing,\{a\}, X\}$, then $A$ is a subset of the power set of $X$, but is not a $\sigma$-algebra on $X$, because it fails to satisfy condition 2: evern though $\{a\}$ is an element of $A$, its complement relative to $X$, $X\setminus\{a\}=\{b\}$ is not an element of $A$.
On the other hand, both $\{\varnothing,X\}$ and $\{\varnothing, \{a\}, \{b\}, X\}$ are $\sigma$-algebras on this $X$, as you should verify.
For any set $X$, the following two collections are always $\sigma$-algebras on $X$:


*

*The trivial $\sigma$-algebra $A=\{\varnothing, X\}$;

*The total $\sigma$-algebra, $A=\mathcal{P}(X)$, the collection of all subsets of $X$.


I believe this is where you got tripped: yes, the whole power set is always a $\sigma$-algebra; but it’s not the only $\sigma$-algebra. If you want a more interesting example, let $X$ be the set of real numbers, and let $A$ contain every set that is either countable or whose complement is countable (these sets are called “co-countable”). It is a good exercise to verify that this collection satisfies your conditions 1, 2,  and 3, and so is a $\sigma$-algebra.
A: Given a set $X$, the power set of $X$, denoted as $P(X)$, is the set of all the subsets of $X$. (It does not make sense to say "a power set [of $X$]" as (originally) mentioned in your post.) 
For any given set $Y$, observe that $P(Y)$ is always a $\sigma$-algebra on $Y$, no exception. 
On the other hand, there are of course examples of collections of subsets that are not $\sigma$-algebras of a given set. Consider for instance $X=\{1,2\}$ and $\mathcal{A}=\{\emptyset,\{1\}\}$. The set $\mathcal{A}$ is not a $\sigma$-algebra of $X$ since $X$ is not in $\mathcal{A}$, which violates the first axiom in your list.  
