Inverse limit of (sub)sets Let $(X_i)_{i\in I}$ be a family of subsets $X_i\subset X$ partially ordered by inclusion. If $X_j\subseteq X_i$ let $\iota_{ij}\colon X_j\hookrightarrow X_i$ be the inclusion and write $i\le j$. This gives rise to a projective system $(X_i,\iota_{ij})_{i,j\in I,i\le j}$ as clearly

*

*the self inclusion $\iota_{ii}\colon X_i\hookrightarrow X_i$ is in fact just the identity on $X_i$ and therefore $\iota_{ii}=\operatorname{id}_{X_i}$

*if we have $i\le j\le k$ the composition $\iota_{ij}\circ\iota_{jk}\colon X_k\hookrightarrow X_j\hookrightarrow X_i$ corresponds to the iterated inclusion $X_k\subseteq X_j\subseteq X_i$ which is precisely given by $\iota_{ik}\colon X_k\hookrightarrow X_i$ and hence $\iota_{ik}=\iota_{ij}\circ\iota_{jk}$
We consider the intersection $\bigcap_{i\in I}X_i$ with (canoncial) inclusions $\iota_j\colon\bigcap_{i\in I}X_i\hookrightarrow X_j$. These are compatible with the maps $\iota_{ij}$ as $\iota_{ij}\circ\iota_j$ corresponds, again, to an iterated inclusion and hence $\iota_i=\iota_{ij}\circ\iota_j$.

In fact, we have $\lim\limits_{\substack{\longleftarrow\\\small i\in I}}X_i=\bigcap\limits_{i\in I}X_i$.

It remains to verify the universal property of the inverse limit. For this purpose let $\mathfrak X$ be a set accompanied by maps $f_i\colon\mathfrak X\to X_i$ which are compatible with the maps $\iota_{ij}$ as usual. Consider $x\in\mathfrak X$ and its image(s) $f_j(x)\in X_j$. By the compatibility property we have $f_i(x)=(\iota_{ij}\circ f_j)(x)=f_j(x)$ for all $i\le j$ since the $\iota_{ij}$ are inclusion maps. In particular, there is some index $k\in I$ such that $i,j\le k$ and hence $X_k\subseteq X_i\cap X_j$. But then also $f_k(x)=f_i(x)=f_j(x)$ and hence there is a well-defined element $\overline x\in\bigcap_{i\in I}X_i$ such that we can set $\rho(x)=\overline x$ which is compatible with the maps $\iota_i,f_i$. This guarentees the existence of a map $\rho\colon\mathfrak X\to\bigcap_{i\in I}X_i$. For the uniqueness part consider a map $\rho\colon\mathfrak X\to\bigcap_{i\in I}X_i$ such that $f_i=\iota_i\circ\rho$ for all $i\in I$. An element $x\in\mathfrak X$ determines an element $f_j(x)$ for all $j\in I$ which in turn uniquely determines this element for all indices $i$ such that $i\le j$. This ensures the uniqueness of $\rho(x)$.

I am not really confident about this proof/construction. While the existence parts seems relatively rigorous to me the uniqueness part appears completely handwavy. I am trying to get a better grasp on inverse limits and the simple example in $\sf Set$ should help for intuition.


So..., here are the main questions: Is the given proof correct? Can it be improved; if so, how? I would like to see a precise verification of $\lim\limits_{\substack{\longleftarrow\\\small i\in I}}X_i=\bigcap\limits_{i\in I}X_i$ as I was not able to locate a reference for this fact (as simple as it might be) and the more involved examples (say, in $\sf Grp$ or ${\sf R}$-${\sf Mod}$) did not really help  me to understand this particular scenario.


Bonus: Explain how this case relates to the dual situation of a inductive system where the direct limit is $\lim\limits_{\substack{\longrightarrow\\\small i\in I}}X_i=\bigcup\limits_{i\in I}X_i$ and hence the role of intersection(s) is replaced by the union(s).

Thanks in advance!
 A: I think the idea is good; let me comment on how I would think about it:

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*In order for the inverse limit $\varprojlim_{i \in I} X_i$ to be defined, you need $I$ to have the structure of a category, and you need $i \mapsto X_i$ to have the structure of a functor. You do this by making $I$ into a preorder when you define the "$\leq"$ relation, but to be super-pedantic, you didn't explicitly check that the relation $\leq$ is reflexive and transitive. Then you do check that $i \mapsto X_i$ is a functor with the action $(i\leq j) \mapsto \iota_{ij}$ on morphisms. To be super-pedantic again, you never explicitly said that $X$ was to be a functor to the category $Set$ (rather than to some other category); I suppose this was understood, but see the second bullet below.


*You then define a cone on the functor $X: I \to Set$, with vertex $\cap_{i \in I} X_i$ and legs $\iota_i$, and verify that this is indeed a cone. Great.


*Next you verify the universal property, which I think looks good modulo the one significant issue of your post which I come to next.


*The only real issue is the following. In the hypotheses, you never stipulated that the family of sets $(X_i)_{i \in I}$ be downward directed in the sense that for every $i,j \in I$ there exists $k$ with $X_k \subseteq X_i \cap X_j$. But you use this assumption in your proof of the universal property, when you produce the index $k$.
So what you've shown is that if $(X_i)_{i \in I}$ is a family of subsets of $X$ which is downward-directed, then $\varprojlim_{i \in I} X_i = \cap_{i \in I} X_i$. This is good. Without the hypothesis that $(X_i)_{i \in I}$ be downward-directed, the statement would be false in general. For example, suppose that $I = \{1,2\}$ and neither $X_1$ nor $X_2$ is contained in the other. Then $I$ is a discrete poset with the $\leq$ order you define, and $\varprojlim_{i \in I} = X_1 \times X_2$, which of course is typically different from $X_1 \cap X_2$.
Some related notes:

*

*In the usage I'm familiar with, the terms "inverse limit" and "projective limit" are ambiguous. Sometimes they mean (1) "limit indexed by a downward-directed poset" (or more generally, "limit indexed by a co-filtered category"), but sometimes they mean more generally (2) "limit indexed by an arbitrary category". I think that at least nowadays most people tend to just say "limit" when they mean (2), so that the usage (1) is more common, but it's something to watch out for.


*In general the limit $\varprojlim_{i \in I} X_i$ can change depending what category you regard the assignment $i \mapsto X_i$ as taking values in. Let's go back to the example where $I = \{1,2\}$ is discrete. If you regard the assignment $i \mapsto X_i$ as taking values in the category $Set$, then as remarked above, you get $\varprojlim_{i \in I} X_i = X_1 \times X_2$, the cartesian product. But instead, you might regard $i \mapsto X_i$ as a functor $I \to Set/ X$, where $Set / X$ is the slice category, i.e. the category where an object is a set $Y$ equipped with a map $Y \to X$, and a morphism is a map $Y \to Y'$ making the obvious diagram commute. If you do this, then you do end up getting $\varprojlim_{i \in I} X_i = X_1 \cap X_2$. Similarly, if you regard $i \mapsto X_i$ as taking values in the poset $Sub(X)$ of subsets of $X$ and inclusions (note that this is a full subcategory of $Set/X$), then you again get $\varprojlim_{i \in I} X_i = X_1 \cap X_2$.


*Similarly, the limit can change depending on what we regard the morphisms of $I$ as being, so it's good that you were explicit about this. An exception is when we compute limits in a category which is a preorder. In this case, $\varprojlim_{i \in I} X_i$ is the greatest lower bound of the $X_i$'s (either existing if the other does), and it doesn't matter what category structure you regard $I$ as having for this to be true.
