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Is there any connection between category-theoretic term 'limit' (=universal cone) over diagram, and topological term 'limit point' of a sequence, function, net...?

To be more precise, is there a category-theoretic setting of some non-trivial topological space such that these different concepts of term 'limit' somehow relate?

This question came to me after I saw ( http://www.youtube.com/watch?v=be7rx29eMr4 ) a surprising fact that generalised metric spaces can be seen as categories enriched over preorder $([0,\infty],\leq)$.

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    $\begingroup$ Similar question on MO (without satisfactory answer). With some fleshing out, this answer in a linked question might lead to something. $\endgroup$
    – t.b.
    Aug 29, 2011 at 22:51
  • $\begingroup$ I seems that suggestion wouldn't work. If my objects are open sets and morphisms inclusions, then limit of any family of objects is just interior of intersection of that family. $\endgroup$
    – Rafael Mrden
    Aug 29, 2011 at 23:10
  • $\begingroup$ It struck me as naive as well... However I think that what one really should consider would be filters modulo some equivalence relation. A filter then should have a limit point $p$ if and only if it is equivalent to the associated principal ultrafilter (the neighborhood filter of $p$). $\endgroup$
    – t.b.
    Aug 29, 2011 at 23:15

2 Answers 2

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The connection is well-known (in particular I'm claiming no originality; I don't recall where I found this, though !): Let $(X,\mathcal O)$ be a topological space, $\mathcal F(X)$ the poset of filters on $X$ with respect to inclusions, considered as a (small, thin) 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 $E$. Kind regards - Stephan F. Kroneck.

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  • $\begingroup$ Ah, yes, thanks for spelling it out! It is essentially what I thought it should be (I don't know why I mentioned an equivalence relation in my comment above) only excuse: it was late :) $\endgroup$
    – t.b.
    Sep 8, 2011 at 11:22
  • $\begingroup$ Tnx, I'll accept now and check later :) don't know much about filters yet. $\endgroup$
    – Rafael Mrden
    Sep 8, 2011 at 18:36
  • $\begingroup$ This construction works out perfectly. Tnx again. :) $\endgroup$
    – Rafael Mrden
    Sep 10, 2011 at 20:20
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    $\begingroup$ Let me think. So $\lambda$ is a limit of $E$ iff $F$ is the intersection of all filters $G$ with $F \cup U_X(x) \subseteq G$ iff $U_X(x) \subseteq F$ iff $F$ converges to $x$. So this is really just a trivial reformulation. But quite amusing. $\endgroup$
    – Martin Brandenburg
    Jan 29, 2013 at 0:10
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    $\begingroup$ Isn't is a bit dissatisfying that the functor characterizing the convergence depends on the limit? IMO, an ideal answer to the question would associate to any filter on $X$ a functor such that the filter has a limit if and only if the functor does. $\endgroup$
    – Jochen
    Sep 28, 2020 at 11:17
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Let $\rm X$ and $\rm Y$ be $\rm T_1$ topological spaces. Let $f : \rm X \to Y$ be any function and let $x \in \rm X$.

Then because $\rm X$ is a $\rm T_1$ topological space, if $\mathcal V_x$ is the filter of neighborhoods of $x$, we have $$ \lim_{\mathcal V_x} \mathrm V = \bigcap_{\mathrm V \in \mathcal V_x} \mathrm V = \{x\}.$$

Now suppose $f$ is continuous at $x$. That means that the filter $f(\mathcal V_x)$ is finer than $\mathcal V_{f(x)}$. This implies that $$ \{f(x)\} \subset \lim_{\mathcal V_x} f(\mathrm V) \subset \bigcap_{\mathrm W \in \mathcal V_{f(x)}} \mathrm W = \{f(x)\}$$

Conclusion : if $f$ is continous at $x$ then $$\lim_{\mathcal V_x} f(\mathrm V) = f(\lim_{\mathcal V_x}\mathrm V).$$

More generally, if $\mathfrak F$ is any ultrafilter converging to $x$ :

  • if $\bigcap \mathfrak F = \emptyset$, then $\bigcap f(\mathfrak F) = \emptyset$ ;

  • or $\bigcap \mathfrak F = \{x\}$ and because $\mathcal V_{f(x)} \subset f(\mathcal V_x) \subset f(\mathfrak F)$, we have $\bigcap f(\mathfrak F) = \{f(x)\}$.

So in both cases, $$\lim_{\mathfrak F}f(\mathrm V) = f(\lim_{\mathfrak F} \mathrm V).$$

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    $\begingroup$ Yeah this actually answers the question, the currently accepted answer just explains your first equation, and in particular doesn't say anything about limits of functions. $\endgroup$
    – zzz
    Jan 31, 2018 at 2:53

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