Let $C$ be a set that is not necessarily convex. The recession cone of $C$ is defined as $$\{y\in C \,\vert\, x + ty \in C, \forall x\in C \text{ and } t\ge 0\}.$$ It seems to me that the recession cone is always convex, even if $C$ is not. Here's why I think that:
Let $y_1,y_2$ be in the recession cone, and $\lambda \in [0,1]$. We want to show that $\lambda y_1 + (1-\lambda)y_2$ is also in the recession cone. Let $x\in C$ and $t\ge 0$, then $$x + t(\lambda y_1 + (1-\lambda)y_2) = x + t\lambda y_1 + t(1-\lambda)y_2.$$
Since $y_1$ is in the recession cone, we have $x + t\lambda y_1\in C$. And since $y_2$ is also in the recession cone, we must have $$x + t\lambda y_1 + t(1-\lambda)y_2 \in C.$$
I have seen references, such as Wikipedia, that suggest that $C$ needs to to be convex in order for the recession cone to be convex. That does not seem to be the case, so I feel like I'm missing something obvious.