Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $\mathbb{R}$.

share|cite|improve this question
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
up vote 4 down vote accepted

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 $\mathfrak c$.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.