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Let $\{C_n\}$ be an increasing sequence of compact subsets of $\mathbb{R}^n$ with non-empty interiors. Define the inductive limit topology on $C_c(\mathbb{R}^m)$ using the inductive system $(C_n,i_{n,m})$ where $i_{n,m}$ is the obvious inclusion map.

Why is the topology on $C_c(\mathbb{R}^m)$ independent of the $C_n$?

ABIM
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You can consider $C_c(\mathbb{R}^n)$ by instead thinking of the directed system $\{C_i\}_{i \in I}$, which consists of all compact subsets of $\mathbb{R}^n$ partially ordered by inclusion. Then, you can define the direct limit topology exactly as you did. In this case, the direct limit is $C_c(\mathbb{R}^n)$.

The question is, why does your $\{C_n\}$ of increasing compact subsets of $\mathbb{R}^n$ (that in the limit covers $\mathbb{R}^n$ I assume, else this does not hold), also yield the same direct limit? It's because the $\{C_n\}$ is cofinal in $\{C_i\}$ i.e. for any element $a \in \{C_i\}$, there is an element $b \in \{C_n\}$ such that $a \leq b$.

It is a fact (whose proof you can look up) that the direct limit over a cofinal subset is isomorphic to the direct limit over the entire directed set. This is why the choice of the cofinal subset does not matter.

Edit: You can find a proof of my claim in the last $2$ pages here http://www-users.math.umn.edu/~garrett/m/fun/Notes/06_categories.pdf

Osama Ghani
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  • Interesting. From a conceptual standpoint. What does it mean to converge in $C_c(\mathbb{R}^n)$? – ABIM May 29 '20 at 18:52
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    I’m honestly not familiar at all with the topology on $C_c(K)$ with $K$ compact. I feel like this is the topology of uniform convergence (you can correct me otherwise). Regardless, given a sequence of functions in $C_c(\mathbb{R^n})$, there is probably a way (that I can’t think of right now) to find some compact $K$ so that the supports of all the $f_n$ lie in $K$. Thus convergence in $C_c(\mathbb{R}^n)$ boils down to convergence in $C_c(K)$. – Osama Ghani May 29 '20 at 19:05
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    This is sort of the essence of direct limits. In fact the reason why the direct limit of the cofinal and overall system agree is precisely because any elements I want to work with, I could find somewhere in the cofinal subset to work with them (in a very handwavey sort of sense). – Osama Ghani May 29 '20 at 19:06
  • This is very interesting. Do you happen to have a reference, besdies the notes, that does in more detail. Like a book with a descent chapter/section on this sort of thing? – ABIM May 29 '20 at 19:09
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    I actually had a hard time finding references for this statement (and was able to find MSE answers about inverse limits instead of direct ones). Eventually I found one in Categories for the Working Mathematician but it's well into abstract nonsense. It uses the concept of final categories (which generalize cofinal subsets), but does prove the statement in generality. In particular, chapter IX (Special Limits) section 3 (Final Functors) has the statement and proof. – Osama Ghani May 29 '20 at 19:28
  • https://math.stackexchange.com/questions/334523/bijection-between-direct-limits Actually I just found this here that's way more tractable. – Osama Ghani May 29 '20 at 19:33
  • Just one thing, what does $a\leq b$ mean in this case? – ABIM May 30 '20 at 09:20
  • You mean $i\leq n$ in the poset I? – ABIM May 30 '20 at 10:53
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    $i$ and $n$ are indices from two different index sets so you can’t really compare them. If you compare $a$ and $b$ in $I$, then $a \leq b$ i.e. $a$ includes into $b$. If you look at this as a poset category, than there is a morphism $a$ to $b$. – Osama Ghani May 30 '20 at 13:39
  • Sorry I got confused for a second because I thought a and b were functions. But yes, in this case all is clear :) – ABIM May 30 '20 at 13:56