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Is it possible to construct a category $\mathcal{C}$ with $\mathrm{Ob}\,\mathcal{C}=\mathbb{R}$ and for every diagram of the from $$a_0\leftarrow a_1\leftarrow\cdots a_n\leftarrow\cdots$$ the inverse limit exists if and only if the corresponding sequence $(a_n)$ converges and in this case the limits coincide?

One should probably give some restrictions so that there are enough morphisms. To throw away stupid examples like $\mathrm{Mor}(x,y)=\varnothing$ for $x\neq y$ and $\mathrm{Mor}(x,x)=\{\mathrm{Id}_x\}$ let's assume that for every $x,y$ at least one of the sets $\mathrm{Mor}(x,y)$ and $\mathrm{Mor}(x,y)$ is nonempty(maybe someone will suggest more adequate restrictions)

Probably this question doesn't make any sense, though taking category structure from the structure of ordered set makes $\varprojlim a_i=\sup a_i$ which is close to the desired.

Any results for other metric spaces are also interesting.

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First example. Let $\mathrm{Mor}(x,y)$ be empty for $x<y$, and consists of one element if $x\ge y$. So, if you can write such a diagram, $(a_n)$ is non-decreasing sequence. Therefore this diagram has the inverse limit iff $(a_n)$ bounded above, and inverse limit equals to supremum.

Second example. Let $\mathrm{Mor}(x,y)$ be the set of all paths connecting $x$ with $y$ (we need to consider this paths up to reparametrization, for composition law will be associative). This example works for all converging sequences, when $a_n\to a_{n-1}$ is "straight path", as it easy to check.

(it looks like a fundamental $\infty$-groupoid of a topological space, but i'm not shure that $\infty$-constructions will work as you want)

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  • $\begingroup$ actually, it works for all (locally) path-connected spaces $\endgroup$ – Andrey Ryabichev Aug 26 '15 at 15:01

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