Minkowski's Theorem - Convex Set interpretation

Question

I am studying an undergraduate course in number theory. In order to prove Minkowski's Theorem, we define convex subsets of $\mathbb{R}^2$:

A subset $X \in \mathbb{R}^2$ is convex if, for any $p, q \in X$, $\lambda p + (1 - \lambda) q \in X$ for all $\lambda \in [0; 1]$.

As it's $\mathbb{R}^2$, it seems like we can visualize this by imagining any subset of $\mathbb{R}^2$ that doesn't somehow "go back in on itself", and that's what convex is. Basically, the classical notion of convex.

Are there any exceptions to this? Is it safe, if I visualize any subset of $\mathbb{R}^2$, and it "looks" convex in this sense, to assume that it is - without an analytical proof?

Out of curiosity - if we parameterised a closed curve $\gamma: I \rightarrow \mathbb{R}^2$ with no self-intersections, and considered our subset of $\mathbb{R}^2$ to be the shape it encloses, would we be able to say something about the subset it encloses (whether or not it is convex) based on the oriented curvature of the curve?

Answer 1

Sure, that's more or less the right intuition. It basically says "If $P$ and $Q$ are in $X$, then so is the line segment joining them."

As for the second question: sure. Because knowing the signed curvature (for a unit-speed curve, just to keep things simple) lets you integrate from any starting point and direction you like as a way to recover the entire curve. Once you've done that, you can compute the area, etc. (using Green's theorem, for instance). The value you get doesn't depend on the initial point or direction, because the curves you get for different starting points/directions are all congruent.