# Equivalent Definitions for Finite Tree

## Theorem

Let $T$ be a finite tree of order $n$.

The following statements are equivalent:

$(1): \quad T$ is connected and has no circuits.
$(2): \quad T$ has $n-1$ edges and has no circuits.
$(3): \quad T$ is connected and has $n-1$ edges.
$(4): \quad$ Any two vertices of $T$ are connected by exactly one path.
$(5): \quad T$ has no circuits, but adding one edge creates a cycle.

## Proof

Statement $1$ is the usual definition of a tree.

### 1 implies 2

• The fact that $T$ has no circuits is part of statement $1$.

$\Box$

### 2 implies 3

• The fact that $T$ has $n-1$ edges is part of statement $2$.

$\Box$

### 3 implies 1

• The fact that $T$ is connected is part of statement $3$.

$\Box$

### 1 implies and is implied by 4

This is proved by Path in Tree is Unique.

$\Box$

### 1 implies 5

Suppose $T = \struct {V, E}$ is connected and has no circuits.

Let $u, v \in V$ be any two vertices of $T$.

Let $P = \tuple {u, u_1, u_2, \ldots, u_{n - 1}, v}$ be a path from $u$ to $v$.

Let a new edges $\set {u, v}$ be added.

Then $\tuple {u, u_1, u_2, \ldots, u_{n - 1}, v, u}$ is now a cycle, which is by definition also a circuit, in $T$.

Note that this applies even when $P = \tuple {u, v}$: $\tuple {u, v, u}$ is still a cycle in $T$, but now $T$ is a multigraph.

$\Box$

### 5 implies 1

Suppose $T$ has no circuits, but adding one edge creates a cycle, which is by definition also a circuit.

If $T$ were disconnected, then it would be possible to add an edge $e$ to connect two components of $T$.

By definition, $e$ would be a bridge.

From Condition for Edge to be Bridge, it follows that $e$ does not lie on a circuit.

So, if the only way to add an edge to $T$ forms a cycle, it follows that $T$ must be connected.

So $T$ is connected and has no circuits.

$\Box$

Thus, all the above can be used as a definition for a finite tree.

$\blacksquare$