Totally ordered set with greater cardinality than the continuum

Question

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 $\mathbb{R}$.

Answer 1

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 $\mathfrak c$.