Tree has Center or Bicenter

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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 [[Definition:Center of Tree|center]] or a [[Definition:Bicenter of Tree|bicenter]]. Suppose $T$ has more than one [[Definition:Center of Tree|center]] or [[Definition:Bicenter of Tree|bicenter]]. It would follow that at least one of the iterations constructing the [[Definition:Center of Tree|center]] or [[Definition:Bicenter of Tree|bicenter]] disconnects $T$ into more than one [[Definition:Component (Graph Theory)|component]]. That could only happen if we were to remove an [[Definition:Edge (Graph Theory)|edge]] between two [[Definition:Node (Graph Theory)|nodes]] of [[Definition:Degree (Vertex)|degree]] greater than $1$. Hence $T$ has at most one [[Definition:Center of Tree|center]] or [[Definition:Bicenter of Tree|bicenter]]. Now to show that $T$ has at least one [[Definition:Center of Tree|center]] or [[Definition:Bicenter of Tree|bicenter]]. The proof works by the [[Principle of Complete Induction]]. We know that a [[Definition:Tree (Graph Theory)|tree]] whose [[Definition:Order of Graph|order]] is $1$ or $2$ is already trivially [[Definition:Central Tree|central]] or [[Definition:Bicentral Tree|bicentral]]. This is our base case. Suppose that all [[Definition:Tree (Graph Theory)|tree]] whose [[Definition:Order of Graph|order]] is $n$ have at most one [[Definition:Center of Tree|center]] or [[Definition:Bicenter of Tree|bicenter]]. This is our induction hypothesis. Take a [[Definition:Tree (Graph Theory)|tree]] $T$ whose [[Definition:Order of Graph|order]] is $n+1$ where $n > 2$. Let $T$ have $k$ [[Definition:Node of Graph|nodes]] of [[Definition:Degree of Vertex|degree $1$]]. We remove all these $k$ [[Definition:Node of Graph|nodes]]. This leaves us with a tree with $n + 1 - k$ [[Definition:Node of Graph|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 [[Definition:Node of Graph|nodes]] of [[Definition:Degree of Vertex|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 [[Definition:Center of Tree|center]] or [[Definition:Bicenter of Tree|bicenter]]. The result follows by the [[Principle of Complete Induction]]. $\blacksquare$