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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?

flag	asked Dec 28 2009 at 12:50 Martin Brandenburg 22.5k●1●25●96

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This was previously addressed here: mathoverflow.net/questions/6554/… – Kevin Lin Dec 28 2009 at 12:55

Twice, in fact. – Harry Gindi Dec 28 2009 at 13:39

I'm not really satisfied with the answers. if we consider the partial order of open subsets of X, a diagram in this category has nothing to do with a net/sequence. – Martin Brandenburg Dec 28 2009 at 14:47

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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 2009 at 17:37

I think that obstacle is that give a topology viewed as a category objects are open sets, and points (concept inherent to analysis limit concept) isn't descrivible in simple categorical way (a ultrafilter is topology is T1). THen need enrich the base, and considering the limits on a locales relative to its "point", or more generally in a topos on Set (where a point is a geometric morphism from Set). But this requires a more accurate exploration... – Buschi Sergio Feb 23 at 12:22

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Håkon Gylterud · Answer 1 · 2009-12-28 15:52:17Z UTC

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.

David White · Answer 2 · 2012-02-23 22:51:15Z UTC

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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).

edited Feb 23 at 22:51

David White

answered Dec 28 2009 at 16:25

Reid Barton
13.8k●1●25●79

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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 2009 at 18:07

Reid Barton · Answer 3 · 2009-12-29 23:20:09Z UTC

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.

Jeff Strom · Answer 4 · 2012-02-23 15:03:33Z UTC

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!

Strongart · Answer 5 · 2012-02-23 11:13:16Z UTC

Please look at my seminar video:A Story of Limite(From calculus to Category),which discuss the limit from calculus to topology,to set,then to category,I think that will match you question well.

http://you.video.sina.com.cn/api/sinawebApi/outplayrefer.php/vid=71512309_1215048895/s.swf

limits in category theory and analysis

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