A noncompact algebraic line needs a deleted boundary triangulation

The increasing open disks $\{z:\mathrm{\mid }z\mathrm{\mid }<n\}$ for positive integers $n$ cover $\mathbb{C}$, and no finite subfamily covers: a finite union of such disks is contained in the largest of them. Hence $\mathbb{C}$ is not compact, and by [F1] it is not homeomorphic to the realization of any finite simplicial complex.

Let $K$ be the boundary complex of the octahedron with vertices $\pm e_1,\pm e_2,\pm e_3$ and eight triangular faces, one for each choice of signs. Its realization is $\{x\in {\mathbb{R}}^3:\mathrm{\mid }{x}_1\mathrm{\mid }+\mathrm{\mid }{x}_2\mathrm{\mid }+\mathrm{\mid }{x}_3\mathrm{\mid }=1\}$: every listed triangular face satisfies the equation and every point of the ${\mathrm{\ell }}^1$-sphere lies in such a face. The radial normalization $x\mapsto x\mathrm{/}\mathrm{\parallel }x{\mathrm{\parallel }}_2$ maps this realization bijectively onto the unit sphere $S^2$, with inverse $u\mapsto u\mathrm{/}(\mathrm{\mid }{u}_1\mathrm{\mid }+\mathrm{\mid }{u}_2\mathrm{\mid }+\mathrm{\mid }{u}_3\mathrm{\mid })$; both maps are continuous, so by [F2] it is a homeomorphism. Deleting the subcomplex consisting of the single vertex $e_3$ identifies $\mathrm{\mid }K\mathrm{\mid }\setminus \{e_3\}$ with $S^2\setminus \{e_3\}$.

For $(a,b)\in {\mathbb{R}}^2$ put $s=a^2+b^2$ and map $(a,b)$ to $(2a,2b,s-1)\mathrm{/}(s+1)$. Its squared Euclidean norm is $(4s+(s-1{)}^2)\mathrm{/}(s+1{)}^2=(s+1{)}^2\mathrm{/}(s+1{)}^2=1$, so the image lies on $S^2$, and its third coordinate $(s-1)\mathrm{/}(s+1)$ is strictly less than $1$. The map $u\mapsto (u_1,u_2)\mathrm{/}(1-u_3)$ is defined on $S^2\setminus \{e_3\}$ and is inverse to the previous formula by direct substitution; both are continuous. Composing with the identification $\mathbb{C}\cong {\mathbb{R}}^2$ and with step 1.2 gives a homeomorphism $\mathbb{C}\cong S^2\setminus \{e_3\}\cong \mathrm{\mid }K\mathrm{\mid }\setminus \{e_3\}$, a finite triangulation with one boundary vertex deleted; step 1.1 shows that deleting a boundary simplex is necessary for a noncompact space.