I am interested in triangulating $\mathbb{R}^n$ using the standard triangulation of $[0, 1]^n$.
By a triangulation of a subset $X \subseteq \mathbb{R}^n$, I mean a set of $n$-simplices $S$ whose union equals $X$, such that if $s_0 \in S$ has vertex set $V_0$ and $s_1\in S$ has vertex set $V_1,$ then $s_0 \cap s_1$ is a simplex with vertex set $V_0 \cap V_1$.
For each permutation $\pi\in S_n$, the simplex $T_\pi$ is defined by $0 \leq x_{\pi(1)} \leq \dots \leq x_{\pi(n)} \leq 1$. The collection of these $n!$ simplices, which we denote $T$, is the standard triangulation of $[0, 1]^n$.
For each $(k_1, \dots, k_n)\in\mathbb{Z}^n$, let $I(k_1, \dots, k_n) = [k_1, k_1+1] \times \dots \times [k_n, k_n+1]$. There is an isometry $[0, 1]^n \to I(k_1, \dots, k_n)$ defined by $(x_1, \dots, x_n) \mapsto (x_1+k_1, \dots, x_n+k_n)$. This isometry carries $T$ to a triangulation $T(k_1, \dots, k_n)$ of $I(k_1, \dots, k_n)$.
Let $T'$ be the union of all $T(k_1, \dots, k_n)$'s. I want to show that $T'$ is a triangulation of $\mathbb{R}^n$.
This comes down to the following. If $I(k_1, \dots, k_n)$ and $I(l_1, \dots, l_n)$ share the common face $F$, I want to show that the restriction of $T(k_1, \dots, k_n)$ to $F$ agrees with the restriction of $T(l_1, \dots, l_n)$ to $F$.
I have been stuck for a while without making much progress. Anything would be a great help! Alternatively, if anyone knows of a simpler way of triangulating $\mathbb{R}^n$, I would be interested to hear.