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.

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    $\begingroup$ This was previously addressed here: mathoverflow.net/questions/6554/terminology-in-category-theory $\endgroup$ Dec 28, 2009 at 12:55
  • $\begingroup$ Twice, in fact. $\endgroup$ Dec 28, 2009 at 13:39
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    $\begingroup$ While not the answer to your question you might be interested in considering the natural ordering of the L^p spaces on measurable eu sets, now take the direct limit. Notice that depending on the category in which you take the spaces to be objects, you may, or may not get L^/infty. I found this excercise very amusing when I first considered it. $\endgroup$
    – B. Bischof
    Dec 28, 2009 at 17:37
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    $\begingroup$ @Buschi: I don't understand what you're saying. $\endgroup$ Feb 23, 2012 at 14:29
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    $\begingroup$ I have asked this qouestion on math.stackexchange last year, and got sasisfactory answer, at least for me :) ( math.stackexchange.com/questions/60590/… ) $\endgroup$ Jan 28, 2013 at 20:30

6 Answers 6


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).

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    $\begingroup$ Much more generally, metric spaces can be identified with certain categories enriched over [0,∞] with the opposite of the usual order, as realized by Lawvere in his paper "Metric spaces, generalized logic, and closed categories." At least some limits of sequences or nets in metric spaces can then be identified with limits in the categorical sense. $\endgroup$ Dec 28, 2009 at 18:07
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    $\begingroup$ I hope you don't mind a dumb question on an 8 year old post, but how should I think of a sequcence of numbers as being a diagram in category theory? (I hope that's the right question to ask. My confusion is that in category theory we speak of the limit of a diagram, so in order for a sequence to have a limit in the categorical sense, it seems like the sequence would have to correspond to a diagram.) $\endgroup$
    – N. Virgo
    Dec 5, 2017 at 3:50
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    $\begingroup$ @Nathaniel Consider the category $J$ whose objects are natural numbers and there is a single morphism in $\mathop{\mathrm{Hom}}_J(x,y)$ whenever $x \le y$ and empty otherwise. Given a sequence of real numbers, $(a_n)$, there is a functor $F\colon J \to R$ assigning $F(n) = a_n$. The functor $F$ is the diagram in this example. A cone to $F$ is a real number which is a lower bound for the sequence $(a_n)$, so the limit of $F$ is the least upper bound, i.e. $\lim_{n\to\infty} a_n$ in the classical sense. $\endgroup$
    – Tyler
    Feb 14, 2020 at 23:15
  • $\begingroup$ So, @Tyler, would the actual limit of a sequence be a zero object in the category of cones? I mean, if the limit of $F$ and the colimit are the same? $\endgroup$ May 16, 2022 at 17:18

I have always justified this to my self by thinking:

  • A limit of a sequence is the "best approximation" of the sequence by a single point.
  • A limit of a diagram is the "best approximation" of the diagram by a single object.

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.

  • $\begingroup$ Why does it not yet characterize convergence of sequences? Would you know if this has been worked out somewhere? $\endgroup$ May 16, 2022 at 17:29
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    $\begingroup$ The sequence $x_{2n}=n$ and $x_{2n+1}=0$ satisfies $\bigcap_{n\in\mathbb N}\overline{\{x_k:k\ge n\}}=\{0\}$ but it diverges. I don't know whether this simple (and maybe superficial) functorial interpretation of convergence appears somewhere in the literature. $\endgroup$ May 17, 2022 at 7:15

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!


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