I don't see a whole bunch of use for it. However if the empty graph is a connected graph and every connected graph has a spanning tree then...
It all depends what you want to do with it. I thought of generating functions. The first article in a Google search says "It will turn out to work better if we ... do not count the empty tree as a rooted tree." So that indicates that "who knows, it could have been better the other way but it isn't." It also seems to say "it is a tree but not a rooted one." Then again, the author does not need to take a position since rooted trees are what is needed.
Would it be so perverse to say that a rooted tree is precisely a finite partial order $(S,\prec)$ such that
- for all $u \in S$ the set $\{x \mid x \preceq u\}$ is totally ordered by $\prec$
- for all $u,v \in S$ there is a common lower bound.
If we can get away with that then the empty order is an order...
For the purposes of this question let me stipulate a structure called an Rooted Oriented Trinary Tree (a ROTT.) This is simply a rooted tree such that each node may have a left, and/or middle, and/or right child. So, it might have only a middle and a right child.
It might be convenient to inductively define
- The empty graph is a ROTT
- any ordered tripple $[T_{\ell},T_m,T_r]$ of ROTTs is a ROTT
- That is all ROTTs
Sure you could avoid it, but in that case It will turn out to work better if we "count the empty tree as a rooted tree."
I can't remember the venerable book I first learned the theory of convex sets from, but I recall that the introduction said something like "to save ink we will not mention the word "nonempty" in the statement of theorems." The point being that we want the intersection of convex sets to be convex and there is an operation which sends $A,B$ to the convex hull of their union and the empty set is the identity for this operation. But we ignore it when it suits us.