Tree has Center or Bicenter

Theorem
Every tree has either:
 * $(1): \quad$ Exactly one center

or:
 * $(2): \quad$ Exactly one bicenter,

but never both.

That is, every tree is either central or bicentral.

Proof
A tree whose order is $1$ or $2$ is already trivially central or bicentral.

Let $T$ be a tree of order at least $3$.

First we establish that the construction of a center or bicenter actually works.

From Finite Tree has Leaf Nodes, there are always at least two nodes of degree $1$ to be removed.0

By the Handshake Lemma, it is clear that $T$ must also have at least one node whose degree is greater than $1$:


 * $\ds \sum_{i \mathop = 1}^n \map {\deg_G} {v_i} = 2 q$

where $q$ is the number of edges in $T$.

But $q = n-1$ from Size of Tree is One Less than Order.

So if each node has degree $1$, then $n = 2 \paren {n - 1}$ and so $n = 2$.

Therefore if $n > 2$ there must be at least one node in $T$ of degree is greater than $1$.

Next, from Connected Subgraph of Tree is Tree, after having removed those nodes, what is left is still a tree.

Therefore the construction is valid.

We need to show the following:


 * $(1): \quad T$ has only one center or bicenter


 * $(2): \quad $T$ has either a center or a bicenter.

Suppose $T$ has more than one center or bicenter.

It would follow that at least one of the iterations constructing the center or bicenter disconnects $T$ into more than one component.

That could only happen if we were to remove an edge between two nodes of degree greater than $1$.

Hence $T$ has at most one center or bicenter.

Now to show that $T$ has at least one center or bicenter.

The proof works by the Principle of Complete Induction.

We know that a tree whose order is $1$ or $2$ is already trivially central or bicentral. This is our base case.

Suppose that all tree whose order is $n$ have at most one center or bicenter. This is our induction hypothesis.

Take a tree $T$ whose order is $n+1$ where $n > 2$.

Let $T$ have $k$ nodes of degree $1$.

We remove all these $k$ nodes.

This leaves us with a tree with $n + 1 - k$ nodes.

As we have seen that $T$ has at least one node whose degree is greater than $1$, $n + 1 - k \ge 1$.

As there are always at least two nodes of degree $1$, $n+1-k \le n-1$.

So after the first iteration, we are left with a tree whose order is between $1$ and $n-1$ inclusive.

By the induction hypothesis, this tree has either a center or bicenter.

The result follows by the Principle of Complete Induction.