Inverse limit and direct limit of an infinite sequence of sets

Question

Let $S_1 \subset \cdots \subset S_n \subset \cdots$ be an infinite sequence of finite sets. We can define both inverse limit and direct limit of this sequence. What is the difference between them? Thank you very much.

Edit: Let $\iota_{i,j}: S_i \to S_j$ be the map sending $a$ to $a$, for all $i<j$. Then we have a direct system. Then we can take direct limit $\lim_{\rightarrow} S_i = \sqcup_{i} S_i/\sim$, where $\sim$ is defined as: for $x_i \in S_i, x_j \in S_j$, $x_i \sim x_j$ if $x_i = x_j$.

Assume further that every $S_i$ has some element $0$. Let $\pi_{i,j}: S_j \to S_i$ be the map sending $a$ to $a$ if $a$ is in $S_i$ and sending $a'$ to $0$ if $a'$ is not in $S_i$, for all $i<j$. Then we have an inverse system. Then we can take inverse limit $\lim_{\leftarrow} S_i =\{ s \in \prod _{i} S_{i} : s_{i}=\pi_{i,j}(s_{j}), \forall i\leq j \}$.

I am asking this question because I have the following situation:

I have an infinite sequence of polytopes: $P_1, P_2, \ldots$ (every $P_i$ has finitely many vertices and faces), where for all $i$, there is a map sending all facets (codimension 1) of $P_i$ to facets to $P_{i+1}$ ($P_{i+1}$ has more vertices and has more facets). I want to define some polytope $P$ which is the limit of $P_1, P_2, \ldots$ in some sense ($P$ is the largest polytope in this sequence which possibly has infinite many vertices and facets). Should I define $P$ as direct limit or inverse limit? Thank you very much.

Edit: More precisely, the polytopes $P_i$'s are defined as follows. Let $A_1 \subset A_2 \subset \cdots$ be a sequence of sets of polynomials. Define $P_i$ to be the Newton polytope of the polynomial which is the product of the polynomials in $A_i$, see the webpage.

Answer 1

The colimit (better avoid the term "direct limit" since it is utterly confusing) of the sequence $S_0 \subseteq S_1 \subseteq S_2 \subseteq \cdots$ is the union set $\bigcup_{n \geq 0} S_n$. In fact, it is easily verified that this union, equipped with the inclusion maps $S_k \hookrightarrow \bigcup_{n \geq 0} S_n$, satisfies the universal property of the colimit.

However, the limit (better avoid the term "inverse limit") of $S_0 \subseteq S_1 \subseteq S_2 \subseteq \cdots$ is just $S_0$. This is trivial to verify. More interesting limits can be computed for sequence of maps in the other directions, e.g. $\cdots \twoheadrightarrow S_2 \twoheadrightarrow S_1 \twoheadrightarrow S_0$.

More generally, if $D : \mathcal{I} \to \mathcal{C}$ is any functor and $\mathcal{I}$ has an initial object $0$, then $\lim(D) = D(0)$. Dually, if $\mathcal{I}$ has a terminal object $1$, then $\mathrm{colim}(D) = D(1)$.

Answer 2

This is a supplement to Martin Brandenburg's answer.

Regarding $S_0 \subseteq S_1 \subseteq S_2 \subseteq \cdots$ as an inverse system with inclusions $\iota_{i,j} : S_i \to S_j$ as bonding maps requires that the index set $\mathbb N_0$ is ordered "reversely" by $\triangleleft$ with $i \triangleleft j$ iff $i \ge j$. This index set has $0$ as maximal element and thus the inverse limit is $S_0$.

But in the question we consider $\mathbb N_0$ with its standard order and for $i \le j$ we have the surjective functions $\pi_{i,j} : S_j \to S_i$ as bonding maps. We get $$\varprojlim (S_i,\pi_{i,j}) = \bigcup_{i=1}^\infty S_i ,$$ but this is a special feature coming from the definition of the $\pi_{i,j}$.

Note that together with the "inverse limit object" $\varprojlim (S_i,\pi_{i,j})$ we need "projections" $\pi_i : \varprojlim (S_i,\pi_{i,j}) \to S_i$ such that the characteristic universal property of the inverse limit is satisfied (see the general definition here). It is $(\varprojlim (S_i,\pi_{i,j}), (\pi_i)_{i \in \mathbb N_0})$ what constitutes the inverse limit of the inverse system $(S_i,\pi_{i,j})$. Without the projections $\pi_i$ the inverse limit object does not provide all necessary information.

We shall use the universal property to prove $(1)$. It is certainly also possible to examine $\{s \in \prod _{i} S_{i} \mid s_{i}=\pi_{i,j}(s_{j}) \phantom{x} \forall i\leq j \}$ as in Anne Bauval's comment, but a rigid proof based on this approach also requires some effort.

We define $$\pi_i : \bigcup_{i=1}^\infty S_i \to S_i, \pi_i(x) = \begin{cases} x & x \in S_i \\ 0 & x \notin S_i\end{cases} .$$ Clearly $\pi_i = \pi_{i,j} \circ \pi_j$ for $i \le j$.

Now let $f_i : T \to S_i$ be a collection of functions such that $f_i = \pi_{i,j} \circ f_j$ for $i \le j$. For $t \in T$ let $\nu(t) = \min\{ i \in \mathbb N_0 \mid f_i(t) \ne 0 \}$; if all $f_i(t) = 0$ we understand $\nu(t) = \infty$.

We have $f_i(t) = f_{\nu(t)}(t)$ for all $i \ge \nu(t)$:

It is impossible that $f_i(t) \notin S_{\nu(t)}$ because then $f_{\nu(t)}(t) = \pi_{\nu(t),i}(f_i(t)) = 0$ which contradicts the definition of $\nu(t)$. Hence $f_i(t) \in S_{\nu(t)} \subset S_i$ and by definition $f_{\nu(t)}(t) = \pi_{\nu(t),i}(f_i(t)) = f_i(t)$.

Define $$f : T \to \bigcup_{i=1}^\infty S_i, f(t) = \begin{cases} f_{\nu(t)}(t) \in S_{\nu(t)} & \nu(t) < \infty \\ 0 & \nu(t) = \infty \end{cases}.$$

We have $\pi_{i} \circ f = f_i$:

If $f(t) = 0$, then all $f_i(t) = 0$ and $\pi_i(f(t)) = \pi_i(0) = 0 = f_i(t)$. So let $f(t) \ne 0$. For $i \ge \nu(t)$ we get $f(t) = f_{\nu(t)}(t) = f_i(t)$, thus $\pi_i(f(t)) = \pi_i(f_i(t)) = f_i(t)$ since $f_i(t) \in S_i$. For $i < \nu(t)$ we have $\pi_{i,\nu(t)}(f_{\nu(t)}) = f_i(t) = 0$; we conclude $f_{\nu(t)} \notin S_i$. Hence $\pi_i(f(t)) = \pi_i(f_{\nu(t)}(t)) = 0 = f_i(t)$.

Next, let $f' : T \to \bigcup_{i=1}^\infty S_i$ be any function such that $\pi_{i} \circ f' = f_i$ for all $i$. We have to show that $f' = f$.

If $f'(t) = 0$, then all $f_i(t) = 0$ and thus $f'(t) = f(t)$. So let $f'(t) \ne 0$. Let $\nu'(t) = \min \{ i \in \mathbb N_0 \mid f'(t) \in S_i \}$. Then for $i \ge \nu'(t)$ we have $f'(t) \in S_{\nu'(t)} \subset S_i$, thus $f'(t) = \pi_i(f'(t)) = f_i(t)$. Taking $i = \max(\nu(t),\nu'(t))$ we get

$$f'(t) = f_i(t) = f{\nu(t)}(t) = f(t) .$$

Let us finally address your question wtether you should define $P$ as direct limit or inverse limit.

In your case the limit objects $\varinjlim (S_i,\iota_{i,j})$ and $\varprojlim (S_i,\pi_{i,j})$ agree. So the adequate perspective depends on what you want to do with $P$. In my opinion it is more natural to regard $P$ a direct limit.