Claim: $Q$ a topological space, $X$ the space of all complex-valued, continuous functions $f$ with compact support. The norm is $$ |f|_{\max }=\max _{Q}|f(q)| $$ This space is not complete unless $Q$ is compact.
Let $Q= R$(the set of real number). Can someone give me some example of $f_n$ converging to f to show that it is not complete?
On
Take $f$ to be any continuous function which is not compactly supported but decays to $0$ at $\infty$. Then $f\not\in X$ but it may be approximated uniformly by taking $f_n$ to be its restriction to $[-n,n]$ and then altering $f_n$ so that it is continuous at $-n$ and $n$ by simply making it go to $0$ on the interval $[n,n+1]$ and $[-n-1,-n]$.
For $n\in\Bbb Z^+$ let
$$f_n(x)=\begin{cases} 0,&\text{if }x<0\\ x,&\text{if }0\le x\le 1\\ \frac1x,&\text{if }1\le x\le n\\ \frac1n(n+1-x),&\text{if }n\le x\le n+1\\ 0,&\text{if }x>n+1 \end{cases}$$
It’s not hard to see that
$$|f_n-f_m|_{\text{max}}=\frac1{\max\{m,n\}}\,,$$
so $\langle f_n:n\in\Bbb Z^+\rangle$ is Cauchy. However, it clearly converges to the function
$$f(x)=\begin{cases} 0,&\text{if }x<0\\ x,&\text{if }0\le x\le 1\\ \frac1x,&\text{if }x>1\,, \end{cases}$$
which does not have compact support.