Is this specific example of a graph having 10 vertices and 25 edges planar or non-planar?

Question

Let $N = \{1, 2, 3, 4, 5\}$
Let $G$ be an un-directed graph defined as follows:

• Let $VERTS(G)$ denote the vertex set of $G$.
• $N$ subset of $VERTS(G)$
• For all $k$ ∈ $N$, $N/\{k\}$ ∈ $VERTS(G)$
  • Specifically,
    • {2, 3, 4, 5} ∈ Verts(G)
    • {1, 3, 4, 5} ∈ Verts(G)
    • {1, 2, 4, 5} ∈ Verts(G)
    • {1, 2, 3, 5} ∈ Verts(G)
    • {1, 2, 3, 4} ∈ Verts(G)

The vertices noted above are the only elements of Verts(G) G has a total of 10 verticies.

For all k ∈ N, $\{k, N/\{k\}\}$ is an edge of $G$.
For example, $\{1, \{2, 3, 4, 5\}\}$ is an edge of $G$

Also, For all $k$ in $N$, for all $p$ ∈ N/{k} {N/{k}, p} is an edge of $G$.
For example, {{2, 3, 4, 5}, 2} is an edge of G.
G has a total of 25 edges.

Question: is $G$ a planar graph? Can it be drawn on the Cartesian plane without any two edges crossing each other?

I drew the graph in paint, and started adding in edges, but I didn't finish adding edges. It looks non-planar, I think:

Answer 1

Without having to read beyond the title, the answer must be no. A simple planar graph on $n$ vertices can have at most $3n-6$ edges.

Answer 2

First point regarding notation : in Graph theory, it is widely standard to call $V(G)$ the set of vertices of a graph $G$, with cardinality $n$. And $E(G)$ its edge set with wardinality $m$. I would recommend keeping the standard notations, rather than "self-documenting" ones.

You are only describing, in the very convoluted way, the complete bipartite graph $K_{5,5}$.

Each vertex $i \in \{1,2,3,4,5\}$ is connected to $N\setminus\{i\}$ by your first rule and all $N\setminus\{j\neq i\}$ by the second rule.

As stated, as $m\leq3n-6$ for any planar graph, then this is not planar. Even if you delete two edges, it won't be as it would still include $K_{3,3}$ as a subgraph.