Countability of a set: only 2 options? So I know sets can be countable (bijection between set and $\mathbb{N}$, finite) or uncountable. Is there another option or are all sets either or?
 A: Once you accept the law of excluded middle, and $\varphi$ is a property which you can express (e.g. countability) every element in the universe either satisfies $\varphi$ or its negation.
Recall the definitions:


*

*$A$ is countable if there is an injection from $A$ into $\Bbb N$.

*$A$ is uncountable if it is not countable.


So every set $A$ either satisfies the property "being countable", in which case it is countable; or it doesn't in which case... it is uncountable.
Of course, much like "finite" doesn't tell you much about the cardinality of a set, just that it is finite, being uncountable doesn't tell you much about the cardinality of a set except that it is not countable. There are many uncountable sets of different cardinalities, but they are all uncountable nonetheless (much like there are many irrational numbers with varying algebraic and algebraic-like properties, but they are all irrational numbers nonetheless).
A: There are varying degrees of uncountability, but since "uncountable" means "not countable," a set is either countable or not, hence either countable or uncountable. 
As for the varying degrees, in general $2^A$, the set of all subsets of a set $A$, has a different cardinality than does $A$. So $U = 2^\mathbb N$ is bigger than $\mathbb N$, and $2^U$ is even bigger than $U$, and so on. 
A: Cantor's theorem says that the cardinality of the power set $2^A$ (the power set is the set of all subsets of a set) of $A$ is larger than the cardinality of $A$. This means that $2^{\Bbb{R}}$ has a larger cardinality than $\Bbb{R}$. But if you take the definition of uncountable to be anything which isn't finite and can't be put into bijection with $\Bbb{N}$ then yes, everything is either at most countable or uncountable. There are an infinite number of different infinite cardinalities though.
We can know surprisingly little about these cardinalities beyond this. It turns out that the continuum hypothesis, which states there are no sets with cardinality between that of the intergers and that of the real numbers (i.e. the continuum) cannot be proven or disproven using ZFC, which is the "standard model" of mathematics.
It can be proven if you assume the Axiom of Choice that the cardinality of the integers is the smallest non-finite cardinal. This is called $\aleph_0$.
It also turns out that if you assume the Axiom of Choice, you know that there is  a smallest cardinality, or cardinal number, which is larger than any given one. The smallest one larger than $\aleph_n$ is called $\aleph_{n+1}$ and then you have a linear ordering on all cardinals. Without AC things are more challenging.
