The following is an exercise from Lee's Introduction to Topological Manifolds:
Show that an abstract simplicial complex is the vertex scheme of a Euclidean complex if and only if it is finite-dimensional, locally finite, and countable.
The author provides a hint:
"If the complex has dimension $n$, let the vertices be the points $v_k = (k, k^2, k^3,\ldots, k^{2n+1}) \in \mathbb{R}^{2n+1}$. Use the fundamental theorem of algebra to show that no $2n+2$ vertices lie in a proper affine subspace, so any set of $2n+2$ vertices is affinely independent. If two simplices $\sigma,\tau$ with vertices in this set intersect, let $\sigma_0,\tau_0$ be the smallest face of each containing an intersection point, and consider the set consisting of all the vertices of $\sigma_0$ and $\tau_0$."
I've followed this hint through to the bolded section, but the strategy in bold seems more complicated than it needs to be. Its aim is clearly to verify the intersection property of simplicial complexes; i.e.
- If $\sigma$ and $\tau$ are simplices in a simplicial complex $K$, then $\sigma \cap \tau$ is either empty or a face of both $\sigma$ and $\tau$.
Wouldn't the following simpler argument suffice? If $\sigma,\tau$ are two simplices of dimension at most $n$ with vertices in the set $\{v_k\}$, then the union of their vertex sets is at most $2n+2$ elements and so spans a simplex $\phi$, of which $\sigma$ and $\tau$ are both faces. Thus $\sigma \cap \tau$ is a face of both $\sigma$ and $\tau$.