What is the largest, well-ordered set? Intuitively it seems to me that the largest set (up to bijection) must be the set of real numbers, but am I right? I keep feeling the proof must be straight forward, but I just can't see it. Can anybody provide some insight?
 A: Under the Axiom of Choice, any set can be given a well-ordering - so your question is really just whether $\mathbb{R}$ is the largest set. The answer is no, in more or less the most extreme way possible - there are more differently-sized sets larger than $\mathbb{R}$ than there are elements of $\mathbb{R}$. To see an example of one of them, recall that Cantor's theorem states that for any set $A$, $\mathscr{P}(A)$ (the set of all subsets of $A$) is strictly larger than $A$. So, for example, the set of all subsets of $\mathbb{R}$ is strictly larger than $\mathbb{R}$ - and the set of subsets of that is even larger, and so on.
If you want a more natural example, the set of functions $f:\mathbb{R} \to \mathbb{R}$ (not necessarily continuous) is strictly larger than $\mathbb{R}$.
A: The usual order on the reals is not a well order as there are infinite descending chains like $\frac 1n, n\in \Bbb N$.  Under the axiom of choice every set has a well order, so $2^{\mathfrak c}$, which is larger than $\Bbb R$, can be well ordered.  As there is no largest set, there is no answer to the question as asked.  If you do not assume the axiom of choice, it is consistent that the reals cannot be well ordered.
A: $\mathbb{R}$ isn't actually a well-ordered set; a well-ordered set is a totally ordered set in which every subset has a minimum element, which does not apply to $\mathbb{R}$ under its usual ordering. 
As for your question, there is no largest well-ordered set. For any well-ordered set, there is a larger well-ordered set. For consider an ordinal $\alpha$. ZF guaruntees the existence of the Hartogs number of $\alpha$, denoted $\aleph(\alpha)$, which is the smallest ordinal not in bijection with $\alpha$. 
If you use the axiom of choice, then for every set there exists a relation on that set which is a well-ordering. So, given any set $S$, $S$ has a well order, and the power set of $S$ is strictly larger than $S$ (due to Cantor's theorem) and also has a well-ordering. 
