Find an uncountable number of subsets of $l_n^p(\mathbb{R})$ and $l_n^p(\mathbb{C})$ which is neither open nor closed . Find an uncountable number of subsets of $l_n^p(\mathbb{R})$ and $l_n^p(\mathbb{C})$ which is neither open nor closed .
I think if we define for positive $a\in\mathbb{R}-\mathbb{Q}$ the subset $A_a=[0,a)\times\mathbb{R}^{n-1}$ .Number of $A_a$ is uncountable .
Each $A_a$ is neither open or closed 
Reason : For each $a$ we have $\mathbb{R^n}-A_a=\{(- \infty,a)\cup[a,\infty)\}\times\mathbb{R^{n-1}}$ which is neither open nor closed . 
So the collection of subsets  $A_a$ of $\mathbb{R^n}$ satisfies the requirements . 
In a similar way if defines for each $a\in \mathbb{R}$ the subset $B_a=\mathbb{Z}\times\mathbb{C}^{n-1}$ satisfies the requirement of being in uncountable in number and neither closed nor open .  
I would just like to know if my proof is alright . If you any other subsets satisfying the requirements please provide . Thank you .  
 A: Thanks for the clarification. Your answer works for $l_n^p(\mathbb{R})$, but not for $l_n^p(\mathbb{C})$, because $B_a$ is closed. But even your construction and arguments for $l_n^p(\mathbb{R})$ are a bit sloppy. 
For example, why do you choose $a \in \mathbb{R} - \mathbb{Q}$? If we are dealing with real numbers, the irrationals are nothing special and nowhere in your argument is it used or needed that $a$ is irrational. To me it looks like you remember some argument about intervals being open/closed when dealing with $\mathbb{Q}$ embedded in $\mathbb{R}$, but that has no relevance here.
Then you continue with discussing why $A_a$ is not open/closed. $A_a$ is defined as a cross product between a finite interval and several copies of $\mathbb{R}$. Your argument goes that the complement of $A_a$, which is a slighter more complicated cross product, is not open/closed. But any argument you could do to actually prove that would most probably be applicable to $A_a$ itself.
For $A_a$, you correctly found a neither open nor closed set in $\mathbb{R}$ and could extend this to correctly create a neither open nor closed set in $\mathbb{R}^n$. If you can similarly construct a neither open nor closed set in $\mathbb{C}$, you can also extend  this to $\mathbb{C}^n$.
A: For $r\in \mathbb R$ and $m\in \mathbb N$ , let $v(r,m)$ be the vector whose  $j$-th co-ordinates are all $0$ for $j>1$ and whose first co-ordinate is $r-2^{-m}$. Then $S(r)=\{v(r,m):m\in \mathbb N\}$ is not open and not closed.
For $0<s$ let $T(s)=[0,s)\times \{0\}^{n-1}$ if the dimension $n$ is greater than $1$, or $T(s)=[0,s)$ if $n=1$. Then $T(s)$ is not open and not closed.
Let $\bar 0$ denote the $0$-vector. When $v$ is a vector with $d_p(v,\bar 0)\geq 1$ then $U(v)=\{v\}\cup B_p(\bar 0,1)=\{v\}\cup \{u: d_p(u,\bar 0)<1\}$ is not open and not closed.
