Proof that some inverse limit is connected Consider the inverse limit
$$\varprojlim \mathbb{R}/n\mathbb{Z}$$
with respect to the system $(\mathbb{R}/n\mathbb{Z})_{n \geq 1}$ of additive groups togehter with the projections $\pi_{m,n} \colon \mathbb{R}/m\mathbb{Z} \to \mathbb{R}/n\mathbb{Z}$ whenever $n \mid m$.

We equip each $\mathbb{R}/n\mathbb{Z}$ with the quotient topology
 and $\varprojlim \mathbb{R}/n\mathbb{Z}$ with the induced topology from 
$\prod_{n \geq 1} \mathbb{R}/n\mathbb{Z}.$

My question: How can one prove that $\varprojlim \mathbb{R}/n\mathbb{Z}$ is a connected topological space?

I see that each $\mathbb{R}/n \mathbb{Z}$ is connected, hence the product $\prod_{n \geq 1} \mathbb{R}/n\mathbb{Z}$ is connected. But I do not know how to proceed from here.
The connectedness can be proved as follows. By a continuum we mean a connected compact Hausdorff space.


Every intersection of a downwards directed family of continua is a continuum. For simplicity, imagine just a sequence of continua $X_0 ⊇ X_1 ⊇ \cdots$ and put $X := ⋂_{n ∈ ω} X_n$. Clearly, $X$ is Hausdorff compact. It is also connected. Otherwise $X = A ∪ B$ for some nonempty disjoint subsets $A$, $B$ that are clopen in $X$ and so closed in $X_0$. Since $A$ and $B$ are compact and $X_0$ is Hausdorff, there are disjoint open sets $U, V ⊆ X_0$ such that $A ⊆ U$ and $B ⊆ V$. We have that $(X_n)_n$ is a decreasing sequence of compacta such that $⋂_n X_n ⊆ U ∪ V$. Hence, for some $n$ we have $X_n ⊆ U ∪ V$. Since $X_n ∩ U ⊇ A ≠ ∅$ and similarly $X_n ∩ V ≠ ∅$, we have that $X_n$ is disconnected.
Every limit of a directed inverse system is homeomoprhic to the intersection of subspaces of the product, and the subspaces may be taken homeomorphic to products of some of the original spaces. Again, imagine a sequence as before. We have $\varprojlim (X_n, f_{n,m}) = \{x ∈ ∏_n X_n: x_n = f_{n,m}(x_m)$ for every $n ≤ m\} = ⋂_k Y_k$ where $Y_k= \{x ∈ ∏_n X_n: x_n = f_{n,m}(x_m)$ for every $n ≤ m ≤ k\} \cong ∏_{n ≥ k} X_k$.


By combining the two facts we obtain the connectedness since our spaces are circles. Also note that we don't need a general proof for directed systems since our system has a cofinal subsequence.
Consider $C$ the subset of $\mathbb{N}^{\mathbb{N}}$ the set of maps $\mathbb{N}\rightarrow \mathbb{N}$ such that for every $u\in C$, $u(n)= u(m)$ mod $n$ if $n$ divides $m$ endowed with the discrete topology,  there exists a continuous map $f:C\times \mathbb{R}\rightarrow \mathbb{R}^{\mathbb{N}}$ defined by $f(u,x)=(x+u(1),..,x+u(n),..)$ the image of the composition of $f$ with $\mathbb{R}^{\mathbb{N}}\rightarrow \Pi_n\mathbb{R}/n\mathbb{Z}$ is $\varprojlim\mathbb{R}/n\mathbb{Z}$. We deduce that $\varprojlim\mathbb{R}/n\mathbb{Z}$ is connected since $C\times \mathbb{R}$ is connected.