[proofplan]
We verify that $\varprojlim R_n$ is a topological ring by showing that addition and multiplication are continuous with respect to the inverse limit topology. The strategy is to use the universal property of the product topology: a map into a product is continuous if and only if each coordinate projection is continuous. We factor the ring operations on $\varprojlim R_n$ through the product $\prod R_n$, then observe that the $m$-th coordinate of the sum (or product) depends only on the $m$-th coordinates of the inputs, so continuity reduces to continuity of the ring operations in each $R_m$.
[/proofplan]
[step:Establish the inverse limit as a subspace of the product]
Recall that the inverse limit is defined as the subset
\begin{align*}
\varprojlim R_n = \bigl\{(x_n)_{n \geq 0} \in \prod_{n \geq 0} R_n : \varphi_{n+1}(x_{n+1}) = x_n \text{ for all } n \geq 0\bigr\},
\end{align*}
where $\varphi_{n+1}: R_{n+1} \to R_n$ are the transition maps, and the inverse limit topology is the subspace topology inherited from the product topology on $\prod_{n \geq 0} R_n$. The projection maps $\pi_m: \varprojlim R_n \to R_m$ sending $(x_n) \mapsto x_m$ are continuous by definition of the subspace and product topologies.
[/step]
[step:Verify that $\varprojlim R_n$ is a ring]
The ring operations on $\varprojlim R_n$ are defined coordinatewise: for $(x_n), (y_n) \in \varprojlim R_n$, set
\begin{align*}
(x_n) + (y_n) &:= (x_n + y_n), \\
(x_n) \cdot (y_n) &:= (x_n \cdot y_n).
\end{align*}
These are well-defined because the transition maps $\varphi_{n+1}$ are ring homomorphisms: $\varphi_{n+1}(x_{n+1} + y_{n+1}) = \varphi_{n+1}(x_{n+1}) + \varphi_{n+1}(y_{n+1}) = x_n + y_n$, so the sequence $(x_n + y_n)$ is again coherent. The same argument applies to multiplication. The zero element is $(0_n)$ and the identity is $(1_n)$, both coherent since $\varphi_{n+1}(0_{n+1}) = 0_n$ and $\varphi_{n+1}(1_{n+1}) = 1_n$. The ring axioms hold coordinatewise since each $R_n$ is a ring.
[/step]
[step:Factor the ring operations through the product and apply the universal property of the product topology]
Let $\odot$ denote either addition or multiplication. We must show that the map
\begin{align*}
\odot: \varprojlim R_n \times \varprojlim R_n &\to \varprojlim R_n \\
\bigl((x_n), (y_n)\bigr) &\mapsto (x_n \odot y_n)
\end{align*}
is continuous with respect to the product topology on the domain (product of two copies of $\varprojlim R_n$, each with the inverse limit topology) and the inverse limit topology on the codomain.
By the universal property of the product topology, this map is continuous if and only if for every $m \geq 0$, the composition
\begin{align*}
\pi_m \circ \odot: \varprojlim R_n \times \varprojlim R_n \to R_m
\end{align*}
is continuous, where $\pi_m$ is the $m$-th projection. This composition sends $\bigl((x_n), (y_n)\bigr)$ to $x_m \odot y_m$. We factor this as
\begin{align*}
\varprojlim R_n \times \varprojlim R_n \xrightarrow{\pi_m \times \pi_m} R_m \times R_m \xrightarrow{\odot_m} R_m,
\end{align*}
where $\odot_m$ is the ring operation in $R_m$. The map $\pi_m \times \pi_m$ is continuous because each factor $\pi_m$ is continuous (being the restriction of a product projection) and the product of continuous maps is continuous. The map $\odot_m$ is continuous because each $R_m$ is a topological ring by hypothesis. A composition of continuous maps is continuous, so $\pi_m \circ \odot$ is continuous for every $m$.
[guided]
We want to show that addition and multiplication on $\varprojlim R_n$ are continuous. The inverse limit topology is defined as the subspace topology inherited from $\prod R_n$, and the product topology is characterised by the universal property: a map $f: X \to \prod R_n$ is continuous if and only if $\pi_m \circ f: X \to R_m$ is continuous for every index $m$.
This universal property is the key tool. Rather than chasing open sets through the inverse limit directly, we reduce the problem to checking continuity of the individual coordinate maps. Fix an index $m \geq 0$. The $m$-th coordinate of $(x_n) \odot (y_n) = (x_n \odot y_n)$ is simply $x_m \odot y_m$. So the map $\pi_m \circ \odot$ factors as:
1. Project both inputs to $R_m$: the map $\pi_m \times \pi_m: \varprojlim R_n \times \varprojlim R_n \to R_m \times R_m$ sends $\bigl((x_n), (y_n)\bigr) \mapsto (x_m, y_m)$. Each factor $\pi_m: \varprojlim R_n \to R_m$ is continuous by definition of the subspace/product topology, so $\pi_m \times \pi_m$ is continuous.
2. Apply the ring operation in $R_m$: the map $\odot_m: R_m \times R_m \to R_m$ is continuous because $R_m$ is a topological ring by hypothesis.
The composition of these two continuous maps is continuous. Since this holds for every $m$, the universal property of the product topology gives continuity of $\odot: \varprojlim R_n \times \varprojlim R_n \to \prod R_n$. The image lies in $\varprojlim R_n$ (by the coherence verification above), and the inverse limit carries the subspace topology, so the map $\odot: \varprojlim R_n \times \varprojlim R_n \to \varprojlim R_n$ is continuous.
Why does this argument work so cleanly? Because the ring operations are defined coordinatewise and the inverse limit topology is defined via projections. The universal property translates a single continuity question about the product into infinitely many simpler questions about the individual factors $R_m$, each of which is answered by the hypothesis that $R_m$ is a topological ring.
[/guided]
[/step]
[step:Conclude that $\varprojlim R_n$ is a topological ring]
Since $\varprojlim R_n$ is a ring (with coordinatewise operations) and both addition and multiplication are continuous with respect to the inverse limit topology, $\varprojlim R_n$ is a topological ring. Additionally, the additive inverse map $(x_n) \mapsto (-x_n)$ is continuous by the same universal property argument: its $m$-th coordinate is $-x_m$, which is the composition of $\pi_m$ with the continuous negation map $R_m \to R_m$ (continuous since each $R_m$ is a topological ring).
[/step]
