The theorem I am reading is as follows:
Suppose $G$ is a connected planar graph. Let $T$ be the set of edges in $G$ which form a spanning tree and let $T^*$ be the set of all edges which are duals of the edges not in $T$. Then the edges of $T^*$ form a spanning tree of $G^*$.
The proof that I want to understand (from this book) proceeds by showing that $T^*$ is acyclic and then connected. What I have a doubt is in that $T^*$ is connected. The argument (slightly changed from the book) is as follows:
We show that $T^*$ is connected. To prove this let $T^*=T_1\sqcup T_2$. Let $U$ be the union of regions in $G$ with capital in $T^*$. $U$ is not the whole plane (no point of $T_2$ belongs to it). Hence the boundary $B$ of $U$ is non-empty. Now $B$ consists of certain edges of $T$, having a region of $U$ on one side and a region not in $U$ on the other side. Therefore there is no vertex of degree $1$ on the boundary, which is a contradiction as that would mean that the boundary yields a cycle in $T$. So $T^*$ is a spanning tree.
I don't understand what is $U$ and why $B$ will be the way he describes. What does the author mean by capital?
The proof is fairly simple but badly put. We first assume $T^*$ is disconnected. The author denotes this by $T^*=T_1\sqcup T_2$, but should probably be better written as $T^*=\bigsqcup_i T_i$.
The next line is very badly worded, I think the author means that $U$ is the union of regions of $G$ that contain a particular $T_i$. This follows from the next line, which states that $U$ contains no regions which 'have capital' in some other $T_j$.
The remaining argument states that $U$ is not the whole plane, and so has either an inner boundary, or an outer boundary, or possibly both.
And because this boundary splits $U$ from $U'$ it must be a cycle, which is a contradiction with the premise that the original $T$ is a tree.