Skip to main content
3 of 6
added lemma
Sam Nead
  • 28.2k
  • 5
  • 72
  • 133

Note that a point is a tree, and a pair of points is a forest. Thus with the given definition the circle $S^1$ is a two-tree. (In fact, any finite graph is a two-tree with the given definition.) Thus all surfaces are three-trees, and more generally all $n$-manifolds are $(n+1)$-trees.

The examples given in the original post strongly suggest that this is not the desired outcome. Perhaps the following definition is closer to what you want.

  1. The zero-tree is the one-point space.
  2. A connected topological space $X$ is an $(n+1)$-tree if for any pair of distinct points $x, y \in X$ there is an $n$-tree $Y \subset X - \{x, y\}$ separating $x$ from $y$.

I think that life is nicer if we also add the assumption that $n$-trees are homeomorphic images of CW-complexes. With this additional assumption I believe we can prove the following.

Lemma: Suppose that $X$ is an $n$-tree. Then $X$ is simply connected.

Sam Nead
  • 28.2k
  • 5
  • 72
  • 133