Let G be a simple graph. Prove that the connection relation in G is an equivalence...

Let G be a simple graph. Prove that the connection relation in G is an equivalence relation on V (G)

Expert Solution

Theorem: Let G = (V, E) be a graph. Then the connectivity relation
over V is an equivalence relation.
Proof: Consider an arbitrary graph G = (V, E). We will prove that
the connectivity relation over V is refexive, symmetric, and
transitive.
To show that connectivity is refexive, consider any v ∈ V. Then
the singleton path v is a path from v to itself. Therefore, v is
connected to itself, as required.
To show that connectivity is symmetric, consider any x, y ∈ V
where x is connected to y. We need to show that y is connected
to x. Since x is connected to y, there is some path x, v₁, …, vₙ, y
from x to y. Then y, vₙ, …, v₁, x is a path from y back to x, so y is
connected to x.
Finally, to show that connectivity is transitive, let x, y, z ∈ V be
arbitrary nodes where x is connected to y and y is connected to
z. We will prove that x is connected to z. Since x is connected to
y, there is a path x, u₁, …, uₙ, y from x to y. Since y is connected
to z, there is a path y, v₁, …, vₖ, z from y to z. Then the path
x, u₁, …, uₙ, y, v₁, …, vₖ, z goes from x to z. Thus x is connected to z,as required.

Colby Messinger answered 2 years ago

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