Does any connected graph G have a spanning tree T with the same domination number?

Question

Let $G$ be a simple graph. A spanning tree of a connected graph $G$ is an acyclic connected subgraph $T$ of $G$ such that $V_T = V_G$.

A dominating set of $G$ is a subset $W$ of $V_G$ such that every vertex in $V_G\setminus W$ is adjacent to some vertex in $W$. The domination number of $G$, $\gamma(G)$, is the minimum on the cardinalities of the dominating sets of $G$.

Evidently, for any spanning subgraph $H$ of $G$, the domination number of $H$ is lower-bounded by the domination number of $G$, i.e. $\gamma(G) \le \gamma(H)$. Particularly, for every spanning tree $T$ of a connected graph $G$, $\gamma(T) \ge \gamma(G)$.

Does every connected graph $G$ have a spanning tree $T$ such that $\gamma(G)=\gamma(T)$?

Answer 1

Yes, and what's more, for any dominating set $W$ of $G$, there is a spanning tree $T$ for which $W$ is also a dominating set.

Begin choosing $T$ by going through all vertices $v \notin W$, and adding an edge from $v$ to some vertex $w \in W \cap N(v)$. This gives us a star forest in which $W$ is a dominating set. It can be extended to a spanning tree however you like.

Answer 2

Assume $G$ is a graph and $X$ is a set such that every vertex is in $X$ or adjacent to a vertex in $X$.

We prove there is a spanning tree $T$ of $G$ such that every vertex is in $X$ or adjacent to a vertex in $X$. For each vertex $v$ not in $X$ we add exactly one edge from $v$ to $X$ that is in $G$. After doing this the graph has no cycles, because it is bipartite, and all the vertices in $G\setminus X$ have degree $1$. We can make this graph into a tree by adding edges in $G$.