Are limits categorical limits? [duplicate]

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

Solution 1:

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.

Solution 2:

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