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This is essentially the 2D version of this question. To start with, I apologise that, when I asked that question, I claimed "2D version is easy" and didn't give much serious thought on it. However, when I really settled down on writing a strict proof, it turned out none too trivial. Whatever I tried I couldn't help ending up with a mess.

So here goes the question:

In $\Bbb R^2$, there's a closed convex polygon $P$ and some point light source $s\notin P$. For each edge $E_i$ of $P$, the line passing through $E_i$ cuts the plane into two open regions, one of which doesn't intersect $P$ and we call it $R_i$. We say an edge $E_i$ of $P$ is lighted if $s\in R_i$. Collect all the lighted edges as $\mathcal F_s:=\{E_{k_{j}},j=1,\cdots,m\}$. Show that $\cup_j E_{k_j}$ is connected.

If this is proved, I think (again, without much serious thought, so don't bet on it!) that, by using Ethan Bolker's answer inductively, we can prove the general cases in $\Bbb R^n$.

But, for now, could anybody provide a rigorous proof for the 2D case? Thanks!

Vim
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Given lighted edges $E_i,E_j$, we want to show that $E_i$ and $E_j$ are connected by lighted edges. This is immediate if $E_i$ and $E_j$ are adjacent, so assume them to be non-adjacent and let $L_i,L_j$ be the lines passing through $E_i,E_j$. Because both edges are lighted we have $s\in R_i\cap R_j$, so $L_i$ and $L_j$ intersect at some point $p$. For any edge $E_k$ between $E_i$ and $E_j$ and on the $p$-side of $E_i,E_j$ we have $R_k \supset R_i\cap R_j$, so $s\in R_k$ and $E_k$ is lighted. This gives us a sequence of lighted edges connecting $E_i$ to $E_j$, meaning that the set of lighted edges is connected.

Edit: The line $L_k$ intersects both $L_i$ and $L_j$ on the boundary of the polygon $P$, and thus cannot intersect either of these lines anywhere else. Now $R_i\cap R_j$ is disconnected from $P$, and by construction $p\in R_k$. If there were some point of $R_i\cap R_j$ that $R_k$ failed to contain, then the boundaries of $R_k$ and $R_i\cap R_j$ would intersect somewhere, since $R_k$ contains at least one point of $R_i\cap R_j$. But this would require $L_k$ (the boundary of $R_k$) to intersect either $L_i$ or $L_j$ at a point not on the boundary of $P$, and we have prohibited this possibility. So $R_i\cap R_j\subset R_k$.

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