Questions on inverse limits

Question

Let $(S_n,\pi_n)$ be an inverse sequence, where $S_n=\mathbb N$ for each $n$. If each $\pi_n$ is identity, then the inverse limit, $\varprojlim S_n$, is (bijective to) $\mathbb N$.

My first question is: how $\varprojlim S_n\simeq\mathbb N$ ($\simeq$ means is bijective). My approach, let $s\in\varprojlim S_n$. Now $s=(s_0,s_1,s_2,...)$, and $s_1=s_0, s_2=s_1, ...$. Therefore $\forall n\in\mathbb N, s_0=s_n$. This means that all $s\in\varprojlim S_n$ are constant sequences. If I am correct, how this can be identified with $\mathbb N$. IF NOT, please provide some steps lead to the solution.

My second question is: Consider this example given here. What will be the $\varprojlim S_n$ if each $X_n=\mathbb Z$??

all ideas are welcome.

Answer 1

I assume you are referring to the classical explicit set-theoretic description of the inverse limit.

• About the first question. You argument is correct: as you noted the inverse limit $\varprojlim S_n$ is the sub-set of $\prod S_n$ made of all the sequences which are constant. Then the $0$-th canonical projection $\pi_0 \colon \prod S_n \to S_0=\mathbb N$ is a bijection, whose inverse is given by the mapping $$\pi_0^{-1} \colon \mathbb N \to \prod S_n$$ that sends every $s \in \mathbb N$ into the corresponding constant sequence.
• About the second question. If I've understood correctly, your considering the inverse diagram $$\mathbb Z \stackrel{-+1}{\longleftarrow}\dots \longleftarrow \mathbb Z \stackrel{-+1}{\longleftarrow} \mathbb Z \stackrel{-+1}{\longleftarrow} \mathbb Z \longleftarrow \dots $$ whose inverse limit is the set $$\varprojlim S_n=\{(x_n)_{n \in \mathbb N} \in \prod_{n \in \mathbb N}\mathbb Z \colon x_n = x_{n+1}+1\}$$ which can be rewritten as $$\{(x_n)_{n \in \mathbb N} \in \prod_{n \in \mathbb N}\mathbb Z \colon x_{n+1}=x_n - 1 \}\ .$$ Even in this case the projection $\pi_0 \colon \varprojlim S_n \to \mathbb Z$ defined by the equation $\pi_0((x_n))=x_0$ is a bijection. The inverse is given by the mapping $$\pi_0^{-1} \colon \mathbb Z \to \varprojlim S_n $$ $$\pi_0^{-1}(s)=(s-n)_{n \in \mathbb N}$$ indeed is not hard to see that $(x_n)_n=(s-n)_{n \in \mathbb N}$ is an element of $\varprojlim S_n$, by trivial calculations we can see that $$x_{n+1}=s-(n+1)=(s-n)-1=x_n-1\ ,$$ and again by calculations one can see that $\pi_0^{-1}\circ\pi_0=\text{id}_{\varprojlim S_n}$ and $\pi_0 \circ \pi_0^{-1}=\text{id}_{\mathbb Z}$.

Answer 2

For the first question, the bijection will be $$ (s, s, s, \dots) \mapsto s. $$ For the second one, I think the answer should be that $\varprojlim S_n \cong \mathbb Z$. To see this, note that any compatible sequence is determined by its first element, and this can be an arbitrary element of $\mathbb Z$. So in this case the bijection (in the other direction) will be $$ z \mapsto (z, z-1, z-2, \dots). $$