Is it possible to regard limits in analysis (say, of real sequences or more generally nets in topological spaces) as limits in category theory? Is there some formal connection?
Edit ('13): Perhaps it is more interesting to ask whether limits in category theory can be seen as special limits of ultrafilters or nets.
I have asked this question on math.stackexchange last year, and got satisfying answer. (So this construction did not come from me.)
Let $(X,\mathcal O)$ be a topological space, $\mathcal F(X)$ the partialy ordered set of filters on $X$ with respect to inclusions, considered as a small category in the usual way. Given $x\in X$ and $F\in\mathcal F(X)$ let $\mathcal U_X(x)$ denote the neighbourhood filter of $x$ in $(X,\mathcal O)$ and $\mathcal F_{x,F}(X)$ the full subcategory of $\mathcal F(X)$ generated by $\{G\in\mathcal F(X):F\cup\mathcal U_X(x)\subseteq G\}$, let $E:\mathcal F_{x,F}\hookrightarrow\mathcal F(X)$ be the obvious (embedding) diagram, $\Delta$ the usual diagonal functor and $\lambda:\Delta(F)\rightarrow E$ the natural transformation where $\lambda(G):F\hookrightarrow G$ is the inclusion for each $G\in\mathcal F_{x,F}$. It is not hard to see that $F$ tends to $x$ in $(X,\mathcal O)$ iff $\lambda$ is a limit of $E$.
In very special cases, the notions coincide. Let $R$ be the category (poset) whose objects are the real numbers and in which $Hom(x, y)$ has a single element if $x \leq y$ and is empty otherwise. Then for a nonincreasing sequence of real numbers, its limit in the classical sense (if not $-\infty$) is also its limit in the categorical sense (if it exists).
I have always justified this to my self by thinking:
But to make the first into an instance of the second, one would need a category representing a topological space where points are objects. And I can't think of one right now.
I agree with Tom Leinster's answer to the previous question.
To this I would add that I believe that the general usage of "limit" in category theory, ie including binary products and pullbacks, is due to Peter Freyd (in his thesis), whereas previously "projective" "inductive limits" had been indexed by N or ordinals. This extension of the usage is another example of the over-stretching of language that Tom mentioned.
On the other hand, I also strongly agree with Martin that this answer is unsatifactory, but this does not mean that I think that any satisfactory answer can be given by referring to a single (contrived) example.
This is the kind of question that those (like me) who are interested in both category theory and analysis should come back to from time to time and reconsider.
I am not completely satisfied by the accepted answer because the functor which characterizes the convergence of a filter depends on the limit. I therefore add another quite simple answer (written for sequences but this easily generalizes to filters and nets) to this old post.
The definition of a limit as a universal cone of a functor resembles the infimum (greatest lower bound) of a set in a very transparent way: Considering a partially ordered set $(X,\le)$ as a category with only one morphism from $x$ to $y$ if $x\le y$ and none otherwise, a subset $A$ of $X$ has an infimum if and only if the inclusion functor $A\hookrightarrow X$ has a limit. In particular, if the power set $\mathscr P(X)$ is ordered by inclusion the intersection of any subfamily $\mathscr A$ is a limit.
Let now $(x_n)_{n\in\mathbb N}$ be a sequence in some topological space $X$. Then the limit of the contravariant functor $F:\mathbb N\to \mathscr P(X)$ assigning to $n$ the set $F(n)=\overline{\{x_k:k\ge n\}}$ is the set of all limit points of the sequence.
I think that this is a strong relation between analytical and functorial limits although it does not yet characterize convergence of sequences. At least, if either $X$ is a compact Hausdorff space or $(x_n)_{n\in\mathbb N}$ is a Cauchy sequence in a Hausdorff uniform space, the sequence converges if and only if the set of limit points is a singleton.
I think this doesn't quite work:
Let $\mathcal{C}$ be the category whose objects are the point of $X$, and define $$ \mathrm{mor}_\mathcal{C}(x,y) = \{ \mbox{closed sets containing both $x$ and $y$} \}. $$ Composition is union.
Now (for example) a sequence $\{ x_n\}$ in $X$ defines a functor $F: \mathbb{N} \to \mathcal{C}$ and a cone from $F$ to $y$ is essentially a single closed set containing the entire sequence and $y$. Since this set must contain the topological limit $x$ of the sequence, this means that the cone factors through the same closed set viewed as a morphism $x\to y$, so $x$ is the categorical colimit of $F$.
And since the morphism sets are symmetrical, the sequence $\{ x_n\}$ can be viewed as a contravariant functor $G: \mathbb{N}\to \mathcal{C}$, and the topological limit $x$ is the categorical limit of $G$.
PROBLEM: the factorization is not unique!
A more abstract version of the answer https://mathoverflow.net/a/120183 :
Recall that a filter $F$ has limit $x$ if all open sets containing $x$ are in $F$, equivalently, if
$$U_x\subseteq F$$
where $U_x$ is the filter of neighbourhoods of $x$.
Set up a diagram that contains all supersets of $F\cup U_x$. The limit of the diagram is the intersection of all these supersets, that is, $F\cup U_x$ itself.
It follows that $F$ is the limit of the diagam iff $F=F\cup U_x$, that is, iff $U_x\subseteq F$, that is, iff $F$ has limit $x$.
