\b[Always true.] Simplicial <initial segments> are nested: $C(k)\subseteq C(l)$ when $k\leq l$. Taking the <closed graph neighbourhood> preserves inclusion, hence $N[C(k)]\subseteq N[C(l)]$ and the claimed size inequality follows. No extremal theorem is needed for this step.

Solution

Past exam of the mathematics course of the University of Cambridge / 2005 / iii / Paper 10 / 1 / i / Solution

 Solution (source code)