If $d(X)$ denotes the minimal cardinality of a dense set in a Banach space $X$, do we have $|X|=d(X)^{\aleph_0}$?

Question

Let $X$ be a Banach space and let $d(X)$ be the minimal cardinality of a dense set in $X$. Is it then true that $|X|=d(X)^{\aleph_0}$ ?

Answer 1

Yes, this is true and it is due to I. Juhász and Z. Szentmiklóssy.

The inequality $|X|\leqslant d(X)^\omega$ is easy as every point is $X$ is a limit of a sequence from a set of cardinality $d(X)$.

As for the converse, by Bing's metrisation theorem, every metric space contains a family of pairwise disjoint open sets of cardinality equal to its density. As we are in a Banach space, we may take such a family whose members are additionally bounded. We iterate this process in every member of this family. We then end up with a tree of countable height and $d(X)^\omega$ many branches. As the diameters of sets in every branch go to 0, these are Cauchy sequences hence convergent. The function that assings the limit to a branch of this tree is an injection, hence $d(X)^\omega \leqslant |X|$.