Let $(X,d)$ be a metric space where $X$ has cardinality $2^{\aleph_0}$. Does it follow that the continuous functions $C(X)$ with values in $\mathbb{C}$, considered as an abelian group, has rank at most $2^{\aleph_0}$?
I thought I had read something to this effect in a paper, but I may not be interpreting it right, since if $X = pt$ then $C(X) = \mathbb{C}$, which already has uncountable rank as an abelian group, right?
If $X$ is separable (contains a dense countable subset) then this rank is at most the cardinality ${\mathfrak c}$ of continuum since $C(X)$ has the cardinality of continuum. (Actually, in this case, rank is exactly ${\mathfrak c}$.) Most likely, the source that you remember assumed separability.
On the other hand, if you take, say, a discrete metric space $(X,d)$ of cardinality ${\mathfrak c}$, then $C(X)=2^{{\mathfrak c}}$ and, hence, will have rank $2^{{\mathfrak c}}$.