Is every (finite) simplicial complex the nerve of some covering?

Question

I need to prove that for every finite simplicial complex $\Delta$ exists a Hausdorff paracompact space $X$ and a good covering $\mathfrak{U}$ of $X$ such that the nerve of the complex is $\Delta$. I don't need the construction itself but to know whether it is true or not so if you give me a reference to cite I would be grateful as well. I guess it is true and I have tried to prove it this way: Suppose the complex to be $n$-th dimensional.

• For each point ($0$-simplex) $A_k$ take an open $n$-ball what we will write as $A_k$ too. (Different points have disjoint balls).
• For each 1-simplex $\{A_0,A_1\}$ take a proper open ball $B_0\subset A_0$ and another $B_1\subset A_1$ and quotient by $B_0\equiv B_1$ (if there is a different 1-simplex $\{A_0,A_1'\}$ take an open ball $B_0' \subset A_0\setminus B_0$).
• By induction, given a k-simplex $\{A_0,A_1,\ldots, A_k\}$, if we want to construct the intersection corresponding to a k+1 simplex $\{A_0,A_1,\ldots, A_k, A_{k+1}\}$, we take an open ball inside the intersection $B_1\subset A_0\cap A_1 \cap \ldots \cap A_k$ and an open ball $B_2\subset A_{k+1}$ and quotient by $B_1\equiv B_2$. This process is finite because the complex is finite dimensional. Finite intersections are contractible because they are defined to be identified balls.

Nevertheless, I think it is not Hausdorff, because if I have balls $A_1, A_2$ and I glue two sub-balls $B_1, B_2$ then there are points in $\partial (A_1\setminus B_1)$ that cannot be separated from points in $\partial(A_2\setminus B_2)$

Answer 1

This is true (and the finiteness assumption is unnecessary): you can take $X$ to be the geometric realization of $\Delta$. For each vertex $x\in X$, associate to it the union $U_x$ of the interiors of all the simplices that have $x$ as a vertex. An intersection $U_{x_1}\cap \dots \cap U_{x_n}$ of these sets is the union of the interiors of all simplices that contain each $x_i$, which is nonempty iff $\{x_1,\dots,x_n\}\in\Delta$, so the nerve of the covering formed by the $U_x$'s is $\Delta$. Moreover, when $U_{x_1}\cap \dots \cap U_{x_n}$ is nonempty, it is contractible because it deformation-retracts onto the interior of the simplex with vertices $x_1,\dots,x_n$ (just linearly move the barycentric coordinates corresponding to other vertices to $0$).