Is it possible to construct (without using Axoim of Choice) a totally ordered set S with cardinality larger than $\mathbb{R}$?


**Motivation:** A total ordering is often called a “linear ordering”. I have heard the following explanation: “If you have a total ordering on a set S, you can plot the set on the real line such that elements to the right are greater than elements to the left”. Formally this means that there exist a function $\phi:S\rightarrow \mathbb{R}$ such that for all $ a$ ,$b\in S$, x < y $ \Leftrightarrow \phi(x)$ < $\phi(y)$. This is of course correct if the set is finite or countable (and it gives a good intuition on what a total ordering is), but obviously not if $|S|>|\mathbb{R}|$, and using the axiom of choice it is easy to “construct” a total ordering on, say, the power set of $\mathbb{R}$. But I would prefer to have a more concrete counterexample, and this is why I asked myself this question.


Later I realized that is was possible to construct a total ordering on a set $|S|=|\mathbb{R}|$, such that no such function $\phi$ exist, but I still think that the above question is interesting.



Explicit ordering on set with larger cardinality than R
Yes. By [Hartogs' theorem](http://en.wikipedia.org/wiki/Hartogs_number), there is an ordinal that has no injection into $R$. The minimal such ordinal is the smallest well-ordered cardinal not injecting into $R$. It is naturally well-ordered by the usual order on ordinals. None of this needs AC. 


One can think very concretely about the order as follows: Consider all subsets of $R$ that are well-orderable. By the Axiom of Replacement, each well-order is isomorphic to a unique ordinal. Let $\kappa$ be the set of all ordinals that inject into $R$ in this way. One can show that $\kappa$ itself does not inject into $R$, and this is the Hartogs number for the reals. 


More generally, of course, there is no end to the ordinals, and they are all canonically well-ordered, without any need for AC. 


But in terms of the remarks in your "motivation" paragraph, there are linear orders that do not map order-preservingly into $R$ that are not larger than $R$ in cardinality. For example, the ordinal $\omega\_1$ cannot map order-preservingly into $R$, since if it did so, then there would be an uncountable family of disjoint intervals (the spaces between the successive ordinals below $\omega\_1$), but every such family is countable by considering that the rationals numbers are dense. Another way to see this is to observe that the real line has countable cofinality for every cut, but $\omega\_1$ has uncountable cofinality.


Lastly, there is a subtle issue about your request that the order by "larger than $R$". The examples I give above via Hartog's theorem are not technically "larger than $R$", although they are not less than $R$ in size. The difficulty is that without AC, the cardinals are not linearly ordered, and so these two concepts are not the same. But you can turn the Hartogs argument into a strict example of what you requested by using the lexical order on $R\times\kappa$, where $\kappa$ is the Hartog number of $R$. This order is strictly larger than $R$ in size, and it is canonically linearly ordered by the lexical order.


