I am looking at a proof of the following property: "If graph T had order n, T is a tree if and only if it contains no cycles, and has n-1 edges."
The proof of <= is as follows: Suppose T is a graph with n vertices, n-1 edges and no cycles. Since there are no cycles, there exists no more than one path between any two vertices. Now if T is disconnected, suppose it is made up of k connected subgraphs, k>1, none of which contains a cycle. Therefore, T is made up of k components, each of which is a tree. But a tree with m vertices has m-1 edges, so for k disconnected trees with a total of n vertices, the total number of edges is n-k. Hence, k=1, which is a contradiction. Therefore, T must be connected, and since it contains no cycles it is a tree.
Now I understand the proof except for the part which states that k=1. I do not see a reason for k to have to equal 1. Why is it so?
We obtain $k=1$ because we assumed that $T$ has $n-1$ edges and showed that $T$ has $n-k$ edges.