Let $C\_c(\mathbb{R})$ be the set of compactly-supported continuous functions on $\mathbb{R}$. We can view this with a number of different topologies but I have my eye on two in particular. Let $X$ be $C\_c(\mathbb{R})$ equipped with the inductive limit topology and let $Y$ be the same set with the compact-open topology.


What is an example of a convergent sequence in $Y$ which fails to converge in $X$?  

Intuitively, such a sequence must exist since the map $x\to x$ is not continuous from $Y$ to $X$.



Convergence in $C_c$ but not in $C$
Let $\phi \in C\_c$ be nonzero (say for simplicity that $\phi$ is supported in $(0,1)$) and let $f\_n(x) = \phi(x-n)$ be its translation to the left by $n$.


This sequence converges to 0 uniformly on compact sets (indeed, on any compact set it is eventually equal to 0). But if we think of the inductive limit topology as coming from the inclusion of the spaces $X\_k = C\_c((-k,k))$ into $C\_c(\mathbb{R})$, then we see that $f\_n$ does not converge in this topology because there is no single $X\_k$ which contains all the $f\_n$.


As Jochen says in the comment, it is a general fact about strict inductive limits that any convergent sequence must be contained in one of the $X\_k$ (where technically to make this a strict inductive limit I ought to replace the $X\_k$ with their closures). Apparently it's due to Dieudonné and Schwartz but I think the proof is a fairly simple exercise once you see how to do it.


But we can also prove it more directly in the current case. Consider the set $U = \{ f : |f(x)| \le 1/|x|\}$. Clearly $U \cap X\_k$ is a neighborhood of 0 in $X\_k$ (since it contains the uniform ball of radius $1/k$ centered at 0), so by definition of the inductive limit, $U$ is a neighborhood of 0 in $X$. But clearly there are infinitely many $f\_n$ that are not in $U$, so $f\_n$ does not converge to 0 in $X$. You can similarly show that the sequence can't converge to any other $f \in C\_c$ either.


