Space of compactly supported function is NOT complete 
Claim: The space $C_c$ of real valued continuous, compactly supported functions is not complete

Solution:
I don't really know how to prove this claim. Nonetheless, I had some Ideas: I was able to prove that $C_c$ is not closed. Also we know that every complete subspace of a metric space is closed. But I cannot conclude. Can someone help me?
Thank you
 A: You are missing something in the statement you propose to use. The complete subspace should namely be a subspace of a complete metric space. Thus: 

A subspace of a complete metric space is closed if, and only if, it is complete. 

I guess you consider $C_c$ as a subspace of the space of continuous functions? If so, then this space is complete with respect to the supremum norm. You can finish your proof then as follows:

Suppose towards a contradiction that $C_c$ is complete. Then, according to the above statement, the space $C_c$ is closed in the space of continuous functions. A contradiction, which implies that $C_c$ cannot be complete.

A: You don't mention the norm for completion? but with the sup norm you may take a sequence  $f_n\in C_c$ converging to e.g. $1/(1+x^2)$ uniformly  (just truncate in a continuous way, not by indicator function). Then $f_n$ is Cauchy but does not converge in $C_c$.
A: @ H. H. Rugh has the answer. The bell-shaped function $f(x)=\frac{1}{1+|x|^2}$ he mentioned asymptotically vanishes at infinity. Then pick a large $[-n,+n], \ n \in \mathbb{N}$ and via a short line segment, continuously join the ends of the function to the $x$-axis, so that you get a continuous, and now compactly supported function.
But if $n$ is larger and larger, then your truncations will infinitely closely approximate your original function. They themselves will form a Cauchy sequence. Since $f$ is not compactly supported, this sequence negates completeness of the space.
