# Simple questions about a finite tree [closed]

Let $$X$$ be a finite tree (a contractible graph) which has at least one edge.

1. There is a vertex of $$X$$ that meets only one edge of $$X$$.

2. If we exclude the edge (and the vertex) in 1 from $$X$$, then $$X$$ is still a tree.

These two statements are intuitively clear, but I can't think of a way to prove these. Am I missing something?

• Exactly what definition of a tree are you using here? – Somos Mar 16 at 12:21
• @Somos It is a $1$-dimensional CW complex which is contractible. – user302934 Mar 16 at 12:24
• By "CW complex" I take it that you are using a topological space. So you are using a finite topological space? What do you mean by "finite" in this context? – Somos Mar 16 at 12:29
• What if $X$ consists of a single vertex (and no edges)? How do the numbered items apply in this case? – paw88789 Mar 16 at 12:38
• @Somos Here, finite tree means a tree with finitely many vertices and edges. – user302934 Mar 16 at 12:46

1. Assume it is not true. Start at any vertex $$v_0$$ (with an edge to it). We can go over the edge to another vertex $$v_1$$ and from that to a vertex $$v_2$$, using another edge since $$v_1$$ doesn't have only one edge by our assumption. By repeating this process we get a sequence of vertices $$(v_0,v_1,v_2,v_3,\ldots)$$ with the property that for any $$n\in\mathbb{N}_0$$ we have $$v_n\neq v_{n+1}$$. Now two things can happen. The first is that all vertices are mutually distinct, i.e. $$v_i\neq v_j$$ for $$i\neq j$$, but then we would have infinite vertices in the graph, a contradiction. The other thing that can happen is that there is a minimal $$n\in\mathbb{N}_0$$ and $$k\in\mathbb{N}$$ such that $$v_n=v_{n+k}$$ (why?). But then the sequence $$(v_n,v_{n+1},\ldots,v_{n+k})$$ constitutes a cycle, which I suppose is a contradiction to the contractability of the graph. Hence our assumption was false and the statement 1. follows.