For simple planar graph

Question

Let $G$ be a simple planar graph (without assuming "conneceted").

Then, without assuming connectedness, is it still true that $e\le3v-6$ when $v\ge3$?

Certain problem states just "simple planar".

But, its solution uses the inequality $e\le3v-6$ when $v\ge3$.

Give some advice. Thank you.

Answer 1

Let there be $k$ components, $k\ge1$, and $e,v$ edges and vertices in totality respectively. Let the $i^{th}$ component have $e_i$ edges and $v_i$ vertices. Since the graph is simple and planar, each component is simple and planar and the relation $e_i\le3v_i-6$ holds for each component. Thus,$$\sum_{i=1}^ke_i\le\sum_{i=1}^k(3v_i-6)\\\implies e\le3n-6k\le3n-6$$

Answer 2

If you have a simple planar graph that's not connected, you can add some edges to connect its components and get a simple, planar, connected graph. The new graph has $e \le 3v-6$, so the old graph does as well.

We can make some stronger statements if we want to. If the graph has $c$ connected components, then $e \le 3v-6 - (c-1)$, and Euler's formula $f - e + v = 2$ can be adjusted to $f - e + v - c = 1$.