Look at the domain of a function $y=x-2$ where $x\in\mathbb{R_+}$.
Then, the triangle produced by x and y-intercepts is bounded and closed. So it is compact. Suppose it is also nonempty.
Does this imply the set is convex?
Graphically, the answer is YES. And using the definition of a set being convex, YES (e.g. for any $\alpha\in[0,1]$...).
But my question is can we make a direct implication between compact set and convex set?
Thanks!
2026-04-24 22:14:25.1777068865
Does being Nonempty Compact Set on $\mathbb{R^+_2}$ imply being Convex set?
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No, of course not. Any finite set with more than one point is compact and not convex.