# 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.

• I think you can show that the limit is path connected. This follows from the fact that every $\pi_{m,n}$ is a covering map, right? Thus if you have a path $[0,1]\to\mathbb{R}/\mathbb{Z}$ then it can be lifted to a path $[0,1]\to\mathbb{R}/n\mathbb{Z}$. Thus a path defined on the first coordinate of the limit (as a subset of the Cartesian product) can be extended to every other coordinate. Some calculations have to be made to ensure that everything agrees though. Commented Jan 10, 2018 at 15:01
• Also every inverse limit of Hausdorff compact spaces is Hausdorff compact. And if the spaces are additionally connected and the invserse system is directed, then the limit is also connected. The point is that the limit is a nice intersection of closed subsets of the product. Commented Jan 10, 2018 at 18:53

## 2 Answers

The connectedness can be proved as follows. By a continuum we mean a connected compact Hausdorff space.

1. 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.
2. 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.

• Could you please explain how we use the existence of a cofinal subsequence? In particular, which subsequence do you have in mind? Commented Jan 10, 2018 at 22:24
• @Algebrus For example the sequence induced by numbers $2, (2 ⋅ 3)^2, (2 ⋅ 3 ⋅ 5)^3, …, (2 ⋅ 3 ⋅ 5 ⋅ 7 ⋅ 11 ⋅ 13 ⋅ 17)^7, …$. Also, existence follows just from the fact that the system is directed and countable. Commented Jan 11, 2018 at 0:16

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.

• What quotient map? $\mathbb{R}$ is not even a part of the system. Plus by the definition the projection of inverse limit is the other way around: from the limit to elements. Commented Jan 10, 2018 at 14:19
• How? Give me the formula. Commented Jan 10, 2018 at 14:21
• Again: the projections $p_n$ are from the limit to members, not the other way around. Commented Jan 10, 2018 at 14:36
• Yes, @freakish is right, the projections are maps $p_k \colon \varprojlim \mathbb{R}/n\mathbb{Z} \to \mathbb{R}/k \mathbb{Z}$. Commented Jan 10, 2018 at 14:48
• How is $C × ℝ$ connected? Commented Jan 11, 2018 at 0:31