Let $X$ be a non-empty subset of $\mathbb{R}^n$, and call the maximally convex subsets of $X$ (ordered by inclusion $\subseteq$) the convex components of $X$.
I was wondering if, in possible analogy to the case of connected components, any such set $X$ is necessarily the union of its convex components?
Best Answer
Of course it is a union of its convex components. That's because any point $x\in X$ is convex, and thus contained in some convex component.
The interesting question would be: is it necessarilly the disjoint union of its convex components? The answer is no. Here's a simple counterexample: consider $\mathbb{R}^2$ with:
$$C_1=[0,1]\times\{0\}$$ $$C_2=\{0\}\times[0,1]$$ $$X=C_1\cup C_2$$
Note that both $C_1$ and $C_2$ are convex components of $X$ (and these are all there are) by your definition. However $C_1\cap C_2=\{(0,0)\}$ is nonempty. So $X$ is not a disjoint union of its convex components.