A totally ordered finite set $\quad \mathcal P_0 \varsubsetneq \mathcal P_1\varsubsetneq \dots \mathcal \varsubsetneq \mathcal P_n \quad$ of prime ideals of a ring $A$ is said to be a chain of length $n$. As is well known, the supremum of the lengths of such chains is called the Krull dimension $\dim(A)$ of the ring $A$.

If the lengths of these chains are not bounded, the ring is said to be infinite dimensional: $\dim(A)=\infty$.This can happen, surprisingly, even for a Noetherian ring $A$.

But in the infinite dimensional case we could consider arbitrary totally ordered subsets $\Pi \subset Spec(A)$ of prime ideals, their cardinality $card(\Pi)$ and then take the sup of all those cardinals. Let us call this sup the cardinal Krull dimension of the ring $A$.

An equality $\dim(A)=\aleph$ would then be a more quantitative measure of the infinite dimensionality of $A$ than just $\dim(A)=\infty$

My question is whether results are known related to that cardinal Krull dimension. For example: for X a topological space, has the cardinal Krull dimension of $\mathcal C(X)$ (the ring of continuous functions on $X$) been calculated? I don't find this trivial, even for $X=\mathbb R$. There are obvious variants of this question concerning rings of differentiable functions on manifolds, etc. Thanks in advance for any information on this topic.