# Totally ordered set with greater cardinality than the continuum

Does there exist a totally ordered set $S$ with cardinality greater than that of the real numbers? Sequences are continuous functions with domain $\mathbb{N}$ and paths are continuous functions with domain $\mathbb{R}$; both of them are very important "traversals" of the points inside a space. Would there be another interesting "traversal", a continuous function with domain $S$? In a sense, $S$ would have to be "denser" than $\mathbb{R}$.

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What about ordinal numbers and transfinite sequences? Transfinite sequences were mentioned at MSE a few times. – Martin Sleziak Aug 15 '12 at 13:45
Depending on the axioms of set theory you want to use, every set can be well-ordered, thus totally ordered. – Hagen Knaf Aug 15 '12 at 14:09
I was hoping for a familiar example like $\mathbb{N}$ or $\mathbb{R}$... but then again I'm actually looking more for a "denser traversal" with notable properties, so if one existed it should be in the literature. – Herng Yi Aug 15 '12 at 14:14
Well, perhaps the main problem is that if you take a set as big as the powerset of $\mathbb{R}$, you're not going to get injective functions from it into familiar spaces. I'd suggest Conway's surreal numbers, the maximal totally ordered field, except that's in fact so big it's not even a set but a proper class! – Kevin Carlson Aug 15 '12 at 14:16
It sounds like what is really desired is a nicely describable total order on a set larger than $\mathfrak c$. – Ross Millikan Aug 15 '12 at 14:31

As mentioned in the comments, (assuming axiom of choice) there are many examples of linearly ordered sets of cardinality greater than $\mathfrak c$, for example the cardinal $\mathfrak c^+$ (successor of $\mathfrak c$).
Perhaps more interestingly, you can indeed find linear orders that are "denser" than $\mathbf R$ in the intuitive sense. If you take the real interval $[0,1]$, then you can extend it to a dense linear order of arbitrary cardinality while preserving endpoints: you can choose some arbitrary cardinal $\kappa\geq \mathfrak c$ and find $T\supseteq [0,1]$ with linear ordering $\leq$ which agrees with the ordering on $[0,1]$, and has the property that $0$ is the least element of $T$, $1$ is the greatest, and between each pair of distinct elements of $T$ (including any pair of real numbers) there are at least $\kappa$ other elements (e.g. if we choose $T$ to be a $\kappa$-saturated elementary extension of $[0,1]$ in the language $\{0,1,\leq\}$, if you're into model theory).
Of course, finding interesting "traversals" in your sense from such a $T$ for $\kappa>\mathfrak c$ will be quite troublesome, since most commonly used (outside set-theoretical topology) spaces have cardinality at most $\mathfrak c$.
re tomasz: I find your post very interesting, but I have a problem, is $T$ a field? – ᴊ ᴀ s ᴏ ɴ Dec 2 '15 at 6:15
@ᴊᴀsᴏɴ: It can't be a field, simply because it's contained in the interval $[0,1]$ (so there's no $2$ within it, for instance). You can't have a field which is linearly ordered with endpoints (or even a group). On the other hand, if you take a model of real closed fields of some very large cardinality $\kappa$, then this model will have many similar properties. It will also have a natural (order) topology and a give rise to somewhat well-behaved notions of path-connectedness, smooth manifolds etc. – tomasz Dec 2 '15 at 22:26