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 $\mathbb{R}$.
Totally ordered set with greater cardinality than the continuum
-
0Although no Axiom of Choice [AC] is needed if you use the theory of transfinite ordinals, let me give a "simple" example (simple in the sense that you do not need to know ordinals) using Zorn's Lemma (which is known to be equivalent to [AC]). First consider the partial $( \mathcal{P}(\mathbb{R}), \subseteq )$, and then use Zorn's Lemma to prove that every partial order can be extended to a linear order (look at http://en.wikipedia.org/wiki/Linear_extension and http://en.wikipedia.org/wiki/Szpilrajn_extension_theorem ) – 2012-08-16
1 Answers
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 $\mathfrak c$.
-
0@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. – 2015-12-02