I was looking proof of a result of Baer for his result: the infinite abelian group $\prod_{i=1}^{\infty} \mathbb{Z}$ is not free abelian group. I was looking proof in Rotman's Introduction to Homological Algebra ($1979$):
(1) To prove non-free ness of above abelian group, it suffices to produce a non-free subgroup of it.
(2) Take $S$ to be collection of those members $(m_1,m_2,m_3,\ldots,)$ in $G$ such that $p^k$ divides all except finitely entries of this infinite-tuple, for all $k\ge 1$.
(3) $|S|=2^{\aleph_0}$, so $S$ is uncountable subgroup.
(4) It then follows that $\dim (S/pS)$ over $\mathbb{Z}/p$ is uncountable (why?)
(5) We finish proof by showing that $\dim(S/pS)$ is countable; (this gives contradiction, so $S$ is non-free.)
Question: It is simply step (4), I am not getting. It may be trivial, but I was unable to see the reason behind it.