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 ?

To show $f(x)$ is continuous I started by saying:
$f(x)$ is continuous if: $\forall \epsilon \gt 0, \exists \delta = (\delta_n) \gt 0$, such that $(\; \mid x_n - y_n \mid \lt \delta_n, \forall n \;) \Rightarrow (\mid f(x) - f(y) \mid \lt \epsilon)$.
Then, I got to the point:
$ \mid f(x) - f(y) \mid \; \le \; \sum_{n=1}^\infty (1 - 1/n)\mid x_n - y_n \mid  \; \le \;\sum_{n=1}^\infty \mid x_n - y_n \mid $
Now, if we take $ \epsilon \gt \sum_{n=0}^\infty \delta_n$, we indeed would have shown then that: $(\; \mid x_n - y_n \mid \lt \delta, \forall n \;) \Rightarrow  (\mid f(x) - f(y) \mid \lt \epsilon)$.
Hence, $f(x)$ is continuous.
Is this correct ?
