TITLE: Ring of real-valued convergent sequences
QUESTION [12 upvotes]: Here is a fun and challenging problem:

Let $R$ denote the ring of real-valued convergent sequences and let
  $S$ denote the ring of real-valued sequences. Prove or disprove that
  $S\cong R$.

The cardinality of these two rings are the same (see Asaf's answer here), but I somewhat doubt the existence of a bijection that would preserve the ring structure.
I would appreciate some hints. 
Source: This is the last problem in this homework sheet (as you see, the deadline is long past).

REPLY [13 votes]: Recall that a ring element $r$ is idempotent if $r^2=r$.
The ring $R$ of convergent sequences has only countably many idempotents. For a sequence $(x_n)$ is idempotent precisely if any $x_i$ is $0$ or $1$. If the idempotent sequence $(x_n)$ converges, all but finitely many of the $x_i$ are $0$, or all but finitely many are $1$.
The ring $S$ of all sequences has uncountably many idempotents.
Thus $R$ and $S$ cannot be isomorphic.
Remark: The property of having only countably many idempotents is not expressible in the first-order language of rings. It would be interesting to know whether $R$ and $S$ are elementarily equivalent.