A closed set in vector space of limited sequences $l^{\infty}$ Consider $l^{\infty}$ the vector space of limited sequences $(x_n)$ of real numbers with norm $||(x_n)|| = \sup |x_n|$. For $n \in \mathbb{N}$, let $e_n = (0,...,0,1,0,...)$ the sequence whose n-th coordinate is $1$. Let $F=\{(1+\frac{1}{n})e_n | n \in \mathbb{N} \}$. Show that $F$ is closed, $d(0,F)  =1$, but $\nexists f \in F$ such that $d(0,f)=1$.
I tried solving this exercise, but I don't think I'm on the right way.
Here is my attempt:
Since it is a normed space, it is $T_1$, then $x \in \bar{F} \iff \exists x_n \in F$ st. $x_n \to x$. Then for $F$ to be closed I need to find a sequence in F converging to a given point of $F$. Perhaps that's useful, but I can't think on a way to start a proof with this.
Then I tried using this theorem:
$x \in \bar{F} \iff d(x,F) = 0$
Let $(x_n) \in \bar{F}$. 
$d((x_n),F) = \inf \{ d((x_n),(f_n)) | (f_n) \in F \} = \inf \{ ||(x_n)-(f_n)|| | (f_n) \in F \} = \inf \{ \sup|x_n-f_n|; (f_n) \in F \} = 0$ 
How can I conclude $(x_n) \in F$ ?.
$d(0,F) = \inf \{d(0,(f_n)) ; (f_n) \in F \} = \inf \{||(f_n)|| ; (f_n) \in F \} = \inf \{\sup |f_n|; (f_n) \in F \}$ . I can "see" that $n \to \infty$ there will be some $f_\infty$ whose non-zero coordinate will be $1$, then this value will be the infimum, but how can I write this?
 A: The idea is $F$ is discrete.  Every point in $F$ is at least 1 away from every other point.  So any sequence in $F$ converging in $l^\infty$ has to eventually be constant.
You can find the distance to $F$ by computing the distance of each element to zero (which is just the infinity norm), then take the inf.  $\inf 1+\frac{1}{n}=1.$
Finally, the distance of each element to zero is $1+\frac{1}{n}\neq 1$.
A: HINT: If $x\notin F$, there are two possibilities.


*

*There are distinct $m,n\in\Bbb Z^+$ such that $x_m\ne 0\ne x_n$. Let $\epsilon=\min\{|x_m|,|x_n|\}$, and show that the open $\epsilon$-ball centred at $x$ is disjoint from $F$.  

*There is a unique $n\in\Bbb Z^+$ such that $x_n\ne 0$. Clearly $x_n\ne 1+\frac1n$; use this last inequality to find an $\epsilon>0$ such that the open $\epsilon$-ball centred at $x$ is disjoint from $F$.


For the rest, you know that 
$$d(0,F)=\inf\left\{\left\|\left(1+\frac1n\right)e_n\right\|:n\in\Bbb Z^+\right\}\;;$$
and 
$$\left\|\left(1+\frac1n\right)e_n\right\|=1+\frac1n\;,$$
so
$$d(0,F)=\inf\left\{1+\frac1n:n\in\Bbb Z^+\right\}=1\;.$$
However, for each $n\in\Bbb Z^+$ we have
$$d\left(0,\left(1+\frac1n\right)e_n\right)=1+\frac1n\ne 1\;.$$
