Let $p$ and $q$ be distinct primes. What is the projective limit $$\varprojlim \mathbb R^2 / (p^n \mathbb Z \times q^n \mathbb Z)?$$
That's an exercise from Robert's book A Course in p-adic Analysis.
Is it true that this limit is isomorphic to $\varprojlim (\mathbb R / p^n \mathbb Z) \times (\mathbb R / q^n \mathbb Z)$ which is just a product of projective limits of $\mathbb R / p^n \mathbb Z$ (and respective expression involving $q$), e.g. $\mathbb S_p \times \mathbb S_q$ where $\mathbb S$ denotes a solenoid?
Yes, your guess is correct. For each $n$, we have $\mathbb{R}^2/(p^n\mathbb{Z}\times q^n\mathbb{Z})\cong \mathbb{R}/(p^n\mathbb{Z})\times \mathbb{R}/(q^n\mathbb{Z})$ (just treat each coordinate separately), and these isomorphisms are compatible with the maps in the inverse system. If you have two inverse systems $(A_n)$ and $(B_n)$ and form the inverse system $(A_n\times B_n)$ with the maps just acting separately on each coodinate, then there is a natural isomorphism $\lim (A_n\times B_n)\to\lim A_n\times \lim B_n$ (you can construct the isomorphism quite explicitly in terms of elements, or more categorically, you can observe that both objects have the universal property that a map into them is the same as a family of compatible maps to both $A_n$ and $B_n$ for each $n$). Thus $\lim \mathbb{R}^2/(p^n\mathbb{Z}\times q^n\mathbb{Z})\cong\lim\mathbb{R}/(p^n\mathbb{Z})\times \lim\mathbb{R}/(q^n\mathbb{Z})=\mathbb{S}_p\times\mathbb{S}_q$.