I have been told to think of inverse limits as "Cauchy Completions" under some metric, for instance through the construction of the p-adic numbers. This got me thinking, though, and I wonder if the following is a valid construction of the completion of a metric space.
Consider the category of metric spaces with suitable maps (maybe continuous, or Lipschitz). Take a metric space $(X,d)$. Then, for every open set $U\subset X$, $(U,d)$ is another metric space, and if $V\subset U$, we have the natural inclusion $V\hookrightarrow U$. We can create a diagram in the following way. Let the nodes be the set of nonempty open sets $N = \{U\subset X| U\text{ is open}, U\ne \emptyset\}$, and let the arrows be the natural inclusion maps $A = \{V\hookrightarrow U| U,V\in N\text{ and } V\subset U\}$.
This diagram may be uncountable. I don't know if this is a problem or not, but to get rid of this for separable spaces, maybe we could use balls of rational radii centered on points from our countable dense subset.
My question is, is the inverse limit of this diagram the Cauchy Completion of the metric space $X$? For that matter, if all these inclusions are forced to be strict (so $X\notin N$), is the direct limit of the diagram $X$ itself?
First of all, you only use the topology of your metric spaces, so that you don't have to compute anything in order to see that the limit cannot be the completion of the metric space (which heavily depends on the metric). Secondly, the category of metric spaces doesn't have all limits. Therefore it is unclear at all if the limit of all non-empty open subsets exists. But here, since every map is an inclusion, the limit is just the intersection of all non-empty open subsets. This is empty (if the space isn't just a singleton), because metrizable spaces are Hausdorff. In particular, it is not the completion, in fact nothing interesting at all.