Compactly Supported Function's Inductive Limit Topology is Independent of Inductive System 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 $C_n$?
 A: 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
