Prove that no disconnected region exists 
Ten points in space, no three of which are collinear, are connected, each oneto all the others, by a total of $45$ line segments. The resulting framework $F$ will be “disconnected” into two disjoint nonempty parts by the removal of one point from the interior of each of the 9 segments emanating from any one vertex off. Prove that $F$ cannot be similarly disconnected by the removal of only $8$ points from the interiors of the 45 segments.

I thought it was simple to prove this since you must disconnect at least $9$ segments in order to disconnect a vertex.
 A: Let there be $n$ points, to generalize on the $n=10$ case of the post. Suppose we remove a number of edges, so as to create a result having $k>1$ connected components remaining. If the sizes of the components are $a_1,a_2,\cdots a_k$ then how many edges do we need to remove to do this? For each ordered pair $(s,t)$ with $1 \le s<t\le k,$ we must remove each of the $a_sa_t$ edges which connect a vertex in the component going with size $a_s$ to another vertex in the component going with size $a_t.$ So in all we must remove at least
$$\sum_{s<t} a_sa_t$$
edges. We can get a lower bound for this by only using the pairs which have $a_1$ in them, that is, the above sum is at least
$$a_1a_2+a_1a_3+\cdots +a_1a_k=a_1(n-a_1),$$
the last because $a_1+a_2+ \cdots +a_k=n.$
Now we need to note that the minimal value of $a_1(n-a_1)$ occurs when $a_1=1$ or $a_1=n-1$ and is $n-1.$ The upshot is that, no matter how we delete edges so as to create a result having more than one connected component, we must in all cases remove at least $n-1$ edges to do it. Applied to $n=10$ case it shows we need to delete at least nine edges.
A: Ok, so we want to show that to cut a complete Graph on n (here 10) vertices into disjoint pieces we need to remove at least n-1 edges.

Say we have removed m edges and cut the graph into disjoint pieces. Then there are 2 vertices $a_1, a_n$ so that the graph without those edges (let's call it G) doesn't contain a path between the two. 
Now we look at the n-1 paths of the form 
$a_1 a_n, a_1 a_2 a_n, a_1 a_3 a_n, ... , a_1 a_{n-1} a_n$
These paths are all contained in our original graph (it was complete) and no two of them share any edges. for $a_1$ and $a_n$ to be disconnected in G we must have removed at least one edge in each of the paths, so m >= n-1
