Convexity of a set C through a continuous function implies closedness? We define a function $f:$ $l^{1}$ $\rightarrow$ $\mathbb{R}$ by : 
$f(x)= \sum_{n=1}^\infty (1 - 1/n)x_n$ .
We consider the set $C= ${ $x\in l^{1} : f(x) = 1$}.
Show that $C$ is a closed convex set.
I had no problem to show that $C$ is convex, but I have no clue how to show $C$ is closed. I'm guessing I have to take a convergent sequence, but I don't know how to use the informations I have to show it does converge within $C$. 
I am tempted to start by saying that because $f(x) = 1$, every $(x_n)_{\mathbb{n \in N}}$ $\in$ $C$ is convergent, but I'm not sure it is true.
Can someone help ?

HINT:
Preimages of closed sets under continous maps are closed.
