Say T is a tree and e is an edge in T. H is a subgraph of T obtained by removing edge e in T. Can there be a walk in H that connects to T?
Edit: I've been trying to work it out, and what I have is that there cannot be walk that connects the two, since graph T is made of both graphs connected by e, and by removing edge e, subgraph H is formed, so both subgraph H and a walk connecting them can't exist.
Is this right? Is there a better way to say this if so?
One characterization of trees is that there is a unique $xy$-path between any two vertices $x$ and $y$. In your example, the edge $vw$ is the unique $vw$-path in $T$. Having removed the edge $vw$, there is now no other $vw$-path. That is to say, $v$ and $w$ belong to separate connected components. (This leads us to a second characterization of trees. They are minimally connected, meaning that the removal of any edge disconnects the tree.) Thus, a slightly stronger claim could be made: Let $x$ be any vertex in the connected component containing $v$ and $y$ be any vertex in the connected component containing $w$. There is no $xy$-path in $H$.