Find frontier of c0 in the metric space (X,d) Find $Fr(c_0)$ in the space (c,$d_\infty$)
where $FrA := \overline{A} \cap \overline{A'}$, $\overline{A}:= closure\ of\ A$ and, $A':= complement\ of\ A$
$c$: consisting of all convergent sequences
$c_0$: the space of all sequences that converge to zero
$c_0\subset c$
$d_\infty$: $\sup_k|a_k-b_k| $
(c, $d_\infty$): metric space
 A: I will write $d$ for $d_{\infty}.$ You should say $d((a_k)_k,(b_k)_k)=\sup_k|a_k-b_k|$, not $\max_k|a_k-b_k|,$ because the set $\{|a_k-b_k|:k\in \Bbb N\}$ may fail to have a largest member. 
(i). For $a=(a_k)_k\in c$ \ $c_0$ let $L_a=\lim_{k\to \infty}a_k$. The open ball $B_d(a,L_a)$ is disjoint from $c_0$ because for any  $b=(b_k)_k\in c_0$ we have $$d(a,b)=\sup_k|a_k-b_k|\geq \lim_{k\to \infty}|a_k-b_k|= L_a.$$ Therefore $c\setminus c_0$ is open. So $c_0$ is closed.   
(ii). $c$ \ $c_0$ is dense in $c.$ To show this it suffices to show that $\overline {c\setminus c_0}\supset c_0,$ as follows:  For $a=(a_k)_k\in c_0$ and for  $r>0$ take $b^{(r)}=(b^{(r)}_k)_k$ where $b^{(r)}_k=a_k$ if $|a_k|\geq r/4$ and $b^{(r)}_k=r/4$  if $|a_k|<r/4.$   Then $\lim_{k\to \infty} b^{(r)}_k=r/4\ne 0$ and $$d(a,b^{(r)})=\sup \{|a_k-b^{(r)}_k|: r/4>|a_k|\}\leq \sup |\{|a_k|+r/4: r/4>|a_k|\}\leq r/2<r.$$  So $b^{(r)}\in B_d(a, r) \setminus c_0. $
(iii). Since $c_0$ is closed and $c$ \ $c_0$ is dense in $c$ we have Fr$(c_0)=\overline {c_0}\cap \overline {(c \setminus c_0)}=c_0\cap c=c_0.$
Remark. For $a= (a_k)_k$ and $a'=(a'_k)_k$ in $c$ and $r\in \Bbb R$ let $a+a'=(a_k+a'_k)_k$ and $ra=(ra_k)_k,$ and $-a=(-1)a.$ Let $\|a-a'\|=d(a,a').$ Then $c$ is a  normed vector space over $\Bbb R,$ and $c_0$ is a vector sub-space of $c.$.... In any normed vector space $v,$ if $v_0$ is a $closed$ vector subspace of $v,$ and $v_0\ne v,$ then $v_0$ has empty interior and Fr$(v_0)=v_0.$
