about set of real sequence with meter sup ... Let  
$$ X=\{ \{x_n \}_{n=1} ^{ \infty } \mid {x_n  } \to 0\ \} $$
where ${x_n  }$ is a  real sequence  and put the metric $ d(x_n ,y_n)= \sup  _{n \in \Bbb {N}} |x_n -y_n|$ on $X$.
Let $ E=\{ \{x_n \}_{n=1} ^{ \infty } \mid {x_n  } \in X, \forall n \in \Bbb {N} \ , x_{n} \geq 0 \} $  now  which of following options is true?


*

*$E$ is not connected.

*$E ^\circ = \emptyset $  .($E ^\circ $ is interior points of E)

*$E ^\prime \neq E $  .($E  ^\prime $ is limit points of E)

*$E$ is not closed.
I think that "2" is true because for every $a\in E , r>0 $ we have $B_r(a) \nsubseteq E$ because if we let $a= (a_1 ,a_2 ,a_3 ,...)$  there exists at least one $ t <0 , |a_1 -t| < r$ , $b= (t ,a_2 ,a_3 ,...) \in B_r(a) $ . note that  $B_r(a)= \{ x \in X | d(x,a)< r  \}$ .
 A: Your claim that for every $a\in X $ and $r>0$ there exists a $t<0$ such that $|a_1 -t|<r$ is not true. For example, take $a_n=\frac{1}{n}$ and $r=\frac{1}{2}$. This clearly does not allow any such $t$ as you have described.
This, however, is pretty close to the right solution. Take $a\in E$ and $r>0$. Then since $\lim a_n =0$, we have that for some $N$, $a_N <\frac{r}{3}$. Let $a_t\in X$ be such that $(a_t)_n=a_n$ for $n\neq N$, and $(a_t)_N = -\frac{r}{3}$. From here the result follows. Good intuition.
1 is not true since for any two points in our space $a,b \in E$, we may construct a continuous path between them by $f(t)= ta + (1-t)b$ with component wise addition and multiplication.
3 is false since $E$ is connected and hence has no isolated points, and is closed because 4 is true.
4 is false since for any $a\in X\setminus E$, there exists at least one $a_n<0$. Hence we may let $r=|\frac{a_n}{2}|$, and then it follows that $B_r(a)\subseteq X\setminus E$.
A: Here are some hints and ideas.
No. 2 is true, 3 and 4 are false.
2:
This is because if $x_{n} \to 0$, then it becomes smaller than any fixed $\epsilon$. Let $m$ be such that $x_{m} < \epsilon$. Then let $y$ be a sequence which is same as $(x_{n})$ except at $m$ where it is $y_{m} = x_{m} - \epsilon$. Then $(y_{n}) \in B((x_{n}), \epsilon)$ but is not in $E$.
3:
Let $x \in E$. Then $x = (x_{1}, x_{2}, \ldots )$ with $x_{i} \geq 0$ and the sequence goes to $0$. We will construct a sequence in $E$ going to $x$.
Let $y_{1}$ be the sequence $(x_{1} + 1, x_{2}, \ldots)$ and $y_{2} = (x_{1}, x_{2} + 1/2, x_{3}, \ldots)$ and so on. Then each sequence $y_{j}$ is an element of $E$ and $d(x, y_{j}) = 1/j$ and hence $(y_{j}) \to x$. Thus, $E \subset E^{\prime}$.
Conversely, let $x = (x_{1}, x_{2}, \ldots) \in X$ be such that $x_{n} < 0$ for some $n$. Then one can see relatively easily that no sequence from $E$ can converge to $x$ (since the distance has to be smaller than the value $|x_{n}|$ from some point on). Thus, $E^{\prime} \subset E$.
4:
This follows from 3. Set of limit points of a set is closed and $E$ is equal to the set of its limit points.
