“Every planar graph has a vertex of degree at most 5” So what's wrong in this case?

As far as I know, a planar graph is simply defined as a graph that can be drawn in the plane with none of its edges crossing. This I understand.

However, I came across a problem that says: "Every planar graph has a vertex of degree at most five."

In the image above there are nine vertices, one of which has degree 8. None of the edges cross, so why is this not a planar graph?

• The problem you cite says only that there is at least one vertex of degree at most five. That is different than saying all vertices have degree at most $5$. – Casteels Apr 15 '15 at 12:17
• This is planar graph. "Every planar graph has a vertex of degree at most five." in this graph there is also a vertex with a degree less then 5 as you can see. It's easy to prove that if all vertices have a degree higher then 5, so 6 or higher that this graph cannot be planar. You can prove this using Eulers formula for planar graphs. – Nescrio Apr 15 '15 at 12:17

That theorem is meant to be read as follows: "For all planar graphs G, there exists a vertex $v$ of $G$ with degree $5$ or less." You are reading it as "For all planar graphs G and for all vertices $v$ of $G$, $v$ has degree $5$ or less"