$l_\infty$ is the space of bounded sequences and $c_0$ is the space of sequences converge to $0$, is a closed subspace of $l_\infty$. I am trying to prove that for any $x \in l_\infty$, $\;d(x,c_0) = \displaystyle \limsup_{n\to \infty} |x_n|$.
Progress: I started with $d(x,c_0) = \inf_{y \in c_0} {d_\infty (x,y)} = \inf_{y \in c_0} \sup_{i \in N} |x_i - y_i|$, then I got stuck. Here we have $\limsup y_n = 0$, but I dont know how can I make use of that
For any sequence $y\in c_0$ we have $$ \|x-y\|_\infty = \sup_n|x_n-y_n|\ge \limsup_{n\to\infty} |x_n-y_n|=\limsup_{n\to\infty} |x_n| $$ This gives a lower bound on the distance.
To get the matching upper bound, let $y_n=x_n$ when $n\le N$, and $y_n=0$ otherwise. This is an element of $c_0$, and $$ \|x-y\|_\infty = \sup_{n>N}|x_n| \xrightarrow[N\to\infty]{} \limsup_{n\to\infty} |x_n| $$ where the last step uses the characterization of $\limsup$ as the limit of suprema of tail sequences.