Recession cone of a set in $l_2$ Let $\mathbb{C}=\\{(x_n)_{n\geq1}\in l_2/|x_n|\leq n, n=1,2,3...\\}$. Prove that this set is convex, closed and not bounded but $recc(A)=\\{(0,0,...,0,...)\\}$. I am using the first definition of recession cone found here Recession cone And $l_2$ is this space: $l_2=\\{(x_n)_{n\geq1}\in \mathbb{R^\mathbb{N}}:\sum_{n=1}^\infty|x_n|^2<\infty\\}$ Convexity is easy to prove, but I'm stuck in the rest.

The set is closed since the mapping $x\mapsto x_n$ is continuous on $l_2$. The set is unbounded as $ne_n\in C$.

Assume $d\in recc(A)$. Then $0+\lambda d \in C$ for all $\lambda\ge0$, i.e., $\lambda |d_n|\le n$ for all $\lambda\ge0$. This implies $d_n=0$ for all $n$, and $recc(C)=\\{0\\}$.