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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    This was previously addressed here: – Kevin H. Lin Dec 28 '09 at 12:55
  • Twice, in fact. – Harry Gindi Dec 28 '09 at 13:39
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    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. – B. Bischof Dec 28 '09 at 17:37
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    @Buschi: I don't understand what you're saying. – Martin Brandenburg Feb 23 '12 at 14:29
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    I have asked this qouestion on math.stackexchange last year, and got sasisfactory answer, at least for me :) (… ) – Rafael Mrđen Jan 28 '13 at 20:30
up vote 19 down vote accepted

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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    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. – Mike Shulman Dec 28 '09 at 18:07
  • 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.) – Nathaniel Dec 5 '17 at 3:50

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 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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