Edgeless Graph of Order n has n Components

Theorem

Let $N_n$ denote the edgeless graph with $n$ vertices.

Then $N_n$ has $n$ components.

Proof

Because there are $n$ vertices in $N_n$, it cannot have more than $n$ components.

Aiming for a contradiction, suppose $N_n$ has fewer than $n$ components.

That would mean that at least $2$ vertices of $N_n$ are connected.

But to be connected, they need to be joined by at least one edge.

But that contradicts the fact that $N_n$ has no edges.

Hence by Proof by Contradiction it is not possible for $N_n$ to have fewer than $n$ components.

Thus it must be the case that $N_n$ has exactly $n$ components.

$\blacksquare$