# Category-theoretic limit related to topological limit?

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 $([0,\infty],\leq)$.

– t.b.
Aug 29 '11 at 22:51
• 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. Aug 29 '11 at 23:10
• 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 $p$).
– t.b.
Aug 29 '11 at 23:15

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.

• 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 :)
– t.b.
Sep 8 '11 at 11:22
• Tnx, I'll accept now and check later :) don't know much about filters yet. Sep 8 '11 at 18:36
• This construction works out perfectly. Tnx again. :) Sep 10 '11 at 20:20
• 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. Jan 29 '13 at 0:10
• 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. Sep 28 '20 at 11:17

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, $$\lim_{\mathfrak F}f(\mathrm V) = f(\lim_{\mathfrak F} \mathrm V).$$

• 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.
– zzz
Jan 31 '18 at 2:53