A set $Y \subset \mathbb{R}^{p}$ is said to be $C$-compact with respect to a cone $C \subset \mathbb{R}^{p}$, if for all $y \in Y$, then $(y-\text{cl }C)\cap Y$ is compact.
For $p = 2$, can anyone see a set $Y$ that is compact but not $C$-compact? I do not have an intuitive sense for how this is possible to destroy the strong conditions of compactness in $\mathbb{R}^{2}$!
Edit: A cone $C \subset \mathbb{R}^{p}$ is a set such that $x \in C \implies \alpha x \in C$ for all $\alpha \geq 0$.
Also, $(y-\text{cl }C)\cap Y$ is equivalent to $(Y + (-\text{cl }C))\cap Y$.
$y - \mbox{cl}(C)$ is the image of the closed set $\mbox{cl}(C)$ under the isometry $x \mapsto y - x$ and hence is closed. So if $Y$ is compact $(y - \mbox{cl}(C)) \cap Y$ is a closed subset of the compact set $Y$ and hence is compact. So compact sets are $C$-compact.