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

Always true. Simplicial initial segments are nested: $C(k)\subseteq C(l)$ when $k\le l$. 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.