Size of a set of Well Orders Given a set $X$ of size $\kappa$, is there any way to work out the number of well-orders on $X$?
It seems like it should be possible but I can't see how to do it.
Surely if $X$ is infinite then $X \times X$ is equinumerous with $X$ and so the set of well-orders on $X$, which is a set of subsets of $X \times X$ has size no larger than $\mathcal{P}(X \times X)$, which is equinumerous with $\mathcal{P}(X)$. I know that $\mathcal{P}(X) = 2^{\kappa}$, so the set is of size at most $2^{\kappa}$?
Is this right, and how cn I finish it off? What about the finite case?
 A: If you want the number of well-orders up to isomorphism then the answer is $\kappa^+$, since every ordinal of size $\kappa$ represents a well-ordering of $X$, and up to isomorphism, every such well-ordering is an ordinal of size $\kappa$. And in the case of a finite $X$, of course, the answer is $1$.
If you want the full number, then you can notice that if you fix just one well-ordering, every permutation of $X$ defines another. How many permutations does $X$ have?
Conclude from this that $\kappa^+|\cdot X!|$ is the number of well-orders that $X$ has. Once you have calculated $|X!|$, it will be easy to see what the answer is.
A: The number of relations on $X$ is $2^{\kappa}$ as you remark. It now suffices to show that there are  $2^{\kappa}$ many well orders. Let $<$ be any fixed linear order, Then for each bijection $f:X \rightarrow X$ define 
$$x<_f y \leftrightarrow f(x)<f(y)$$  is is easy to check that $<_f=<_g$ then $f=g$ since there is only one possible isomorphism between well orders.
Now since we know that there are $2^{\kappa}$ many bijections we have that many distinct well orders.
A: Your question cannot be answered assuming only basic axioms of set theory. Given a fixed well-ordering on X, every bijection gives rise to another well-ordering on X, by transfer of structure. However, this does not give rise to every well ordering on X. For instance, consider the natural order on the natural numbers $N$. Every bijection will give rise to another ordering, but in each well-ordering there is no maximum. On the other hand, the extended natural numbers, which includes a number $\omega$, does have a maximum, and is a well-ordering of $N$. 
So the real answer is $|X|!$ times the number of distinct well-orderings of X. For finite X, there is only one well-ordering up to isomorphism. However, for X = N, the number of distinct well-orderings is called $\aleph_1$. The continuum hypothesis states that $\aleph_1$ = $2^N$. Since the continuum hypothesis is independent of the other axioms of set theory, this means that the answer to your question depends on which model of sets you are working with (a model where the continuum hypothesis is true, or a model where it is false).
