The set of singular points of the boundary of a convex body is meagre.

Question

I get stuck with the following proof. Specifically the exposition of claim (5) below doesn't make much sense to me. How should I visualize the set $G$ and the ball $B$ touching a point $x \in G$?

Answer 1

$\DeclareMathOperator{\bd}{bd}$You should include some definitions. I assume "singular points" means "points with more than one supporting hyperplane", and that $\mathcal{C}_p$ is the set of convex bodies (though I don't know what that $p$ is for). I'm not sure what "angle at least $\frac{1}{n}$" means (angle to what? in radians? as a fraction of a circle?), or why $S_n$ is introduced (I guess because each $S_n$ is relatively closed in $\bd C$, but the set of all singular points need not be?)

$G$ is a hypothetical open patch of the boundary of $C$. If a closed set is not meagre in $\bd C$, then it contains a relatively open subset of $\bd C$. A small ball contained in $C$ can then be placed to touch $G$ in just one point.

The open patch gives you "wiggle room" to make sure such a ball exists. For instance, you can find an open ball $\mathcal{B}$ such that $\mathcal{B} \cap \bd C \subset G$, then take a little ball inside $\mathcal{B} \cap C$ which meets $\bd C$ in a single point. This wouldn't work without a relatively open patch of $\bd C$ (which is why you cannot, in fact, find any open balls contained in $C$ which meet $\bd C$ at singular points.)