I now have an answer, which I am going to outline as follows. The most challenging part is showing that $\bigtriangleup(x^1,\ldots,x^k)\subseteq\bigcup_{E\in\mathscr E}E$ (in particular, $\mathscr E\neq\varnothing$), so I am going to focus on this.
In other words: I need to find for every point in $\bigtriangleup(x^1,\ldots,x^k)$ a face of some member of the subdivision $\mathscr T$ that has $k$ vertices and fully lies in $\bigtriangleup(x^1,\ldots,x^k)$.
For example, if $k=3$, we want the triangle on the base $\bigtriangleup(x^1,x^2,x^3)$ to be intersected by a tetrahedron from the subdivision of the “big” tetrahedron $\bigtriangleup(x^1,x^2,x^3,x^4)$ not just at a single point, not just at a line segment, but actually a smaller triangle!
The case $k=1$ is rather trivial, so assume from now on that $k\geq 2$. In what follows, if $T\in\mathscr T$, then denote the vertices of the simplex $T$ by $y^1(T),\ldots,y^{k+1}(T)$, so that $$T=\bigtriangleup(y^1(T),\ldots,y^{k+1}(T)).$$
Step 1$\quad$ Define, for each $\ell\in\{1,\ldots,k-1\}$, $$\mathcal I_{\ell}=\{I\,|\,I\subseteq\{1,\ldots,k+1\}\text{ and }\#I=\ell\}.$$ That is, $\mathcal I_{\ell}$ is simply the family of all the $\ell$-element subsets of the index set $\{1,\ldots,k+1\}$. For each $I\in\mathcal I_{\ell}$, define $$\mathscr T_{I}\equiv\{T\in\mathscr T\,|\,y^i(T)\in\bigtriangleup(x^1,\ldots,x^k)\text{ for each $i\in I$}\}.$$Furthermore, let $$W_T\equiv\bigtriangleup(\{y^i(T)\}_{i\in I})$$ for each $\ell\in\{1,\ldots,k-1\}$, $I\in\mathcal I_{\ell}$, and $T\in\mathscr T_I$. Finally, define $$W\equiv\bigcup_{\ell=1}^{k-1}\bigcup_{I\in\mathcal I_{\ell}}\bigcup_{T\in\mathscr T_I}W_T.$$ Intuitively, $W$ is the set of all points that can be expressed as a linear combination of at most $k-1$ vertices of some member of the subdivision $\mathscr T$. (Remark: I am pretty sure that the set $\mathscr T_I$ is never empty, but I didn’t bother checking that in detail because it is not really important for the purposes of the proof—you’ll see soon why not.)
Step 2$\quad$ Fix, for a moment, $\ell\in\{1,\ldots,k-1\}$, $I\in\mathcal I_{\ell}$, and $T\in\mathscr T_I$. It is not difficult to check that $W_T\subseteq\bigtriangleup(x^1,\ldots,x^k)$ and that $W_T$ is closed in $\bigtriangleup(x^1,\ldots,x^k)$. The next step is to show that $W_T$ has empty interior in $\bigtriangleup(x^1,\ldots,x^k)$ (with respect to the relative Euclidean topology). Indeed, if one had for some $w\in W_T$ a small ball around $w$ contained in $W_T$, then one could define, for any $\varepsilon\in(0,1]$, $$a^i\equiv(1-\varepsilon)w+\varepsilon x^i$$ for each $i\in\{1,\ldots,k\}$. (Geometrically, think about “shrinking” the face $\bigtriangleup(x^1,\ldots,x^k)$ around $w$.) It is easily seen that the affine independence of $x^1,\ldots,x^k$ implies that of $a^1,\ldots,a^k$. But for small enough $\varepsilon$, $\{a^1,\ldots,a^k\}$ would be fully contained in $W_T$, which is a simplex with $\ell<k$ vertices—this would contradict affine independence. It follows that $W_T$ has empty (relative) interior and it is nowhere dense in $\bigtriangleup(x^1,\ldots,x^k)$.
Step 3$\quad$ Since the union defining $W$ is finite, it follows that $W$ is also closed and nowhere dense in $\bigtriangleup(x^1,\ldots,x^k)$.
Step 4$\quad$ Fix any $w\in\bigtriangleup(x^1,\ldots,x^k)$. The goal is to show that $w\in E$ for some $E\in\mathscr E$. Since $W$ is nowhere dense in $\bigtriangleup(x^1,\ldots,x^k)$, one can find for any $m\in\mathbb N$ some $w^{(m)}\in\bigtriangleup(x^1,\ldots,x^k)\setminus W$ such that $$\lVert w^{(m)}-w\rVert<\frac{1}{m},$$ where $\lVert\cdot\rVert$ is the $n$-dimensional Euclidean norm. Clearly, the sequence $(w^{(m)})_{m\in\mathbb N}$ converges to $w$. Since the union of the members of $\mathscr T$ yields $\bigtriangleup(x^1,\ldots,x^k,x^{k+1})$, there exists some $T^{(m)}\in\mathscr T$ such that $w^{(m)}\in T^{(m)}$, so that $$w^{(m)}=\sum_{i=1}^{k+1}\mu_i^{(m)}y^i(T^{(m)})$$ for some weight vector $(\mu_i^{(m)})_{i=1}^{k+1}$ whose elements are non-negative and sum up to $1$. Let $$I^{(m)}\equiv\{i\in\{1,\ldots,k+1\}\,|\,\mu_i^{(m)}>0\}.$$ Now, since $w^{(m)}$ takes zero weight from $x^{k+1}$ in the “big” simplex $\bigtriangleup(x^1,\ldots,x^k,x^{k+1})$, it follows that $y^i(T^{(m)})\in\bigtriangleup(x^1,\ldots,x^k)$ for every $i\in I^{(m)}$. This implies two things:
- $\#I^{(m)}>k-1$, which follows from the fact that $w^{(m)}\notin W$; and
- $\#I^{(m)}<k+1$, because not all of the affinely independent vertices $y^1(T^{(m)}),\ldots,y^{k+1}(T^{(m)})$ can lie in the proper face $\bigtriangleup(x^1,\ldots,x^k)$.
Thus, $\#I^{(m)}=k$. Showdown time: since $\mathscr T$ is finite, there exists a subsequence $(T^{(m_r)})_{r\in\mathbb N}$ whose terms are all the same; let me denote it by $T^0$. By the foregoing, $T^0$ has precisely $k$ vertices lying in $\bigtriangleup(x^1,\ldots,x^k)$; let the simplex spanned by these $k$ vertices be denoted as $E^0$. Clearly, $E^0\in\mathscr E$. Also, $E^0$ is closed in $\bigtriangleup(x^1,\ldots,x^k)$ and the subsequence $(w^{(m_r)})_{r\in\mathbb N}$ in $E^0$ converges to $w$. Therefore, $w\in E^0$, and the proof is complete. Any comments on it are greatly appreciated.
Best Answer
The short answer is: the maps $\partial$ are defined on the free abelian groups generated by $n$-simplices, hence it makes sense to speak about sums of simplices in these groups.
I'll try to give a down to earth answer, for which we will need to detour first. Consider the set $\mathbb{Z}[X]$ of integer polynomials. An element there is an expression $$ a_0 + a_1 X + \ldots + a_nX^n $$ where each $a_i$ is an integer, and $X$ is an indeterminate. Even though we might be quite familiar with thinking of polynomials in this way, the terms 'expression' and 'indeterminate' are informal. A rigorous definition of $\mathbb{Z}[X]$ could be the set of sequences of integers that are eventually zero,
$$ \mathbb{Z}[X] = \{(a_n)_{n \geq 0} \in \mathbb{Z}^{\mathbb{N}_0} : \text{there exists $k \in \mathbb{N}_0$ such that $a_n = 0$ for all $k \geq n$}\}. \tag{1} $$
Given integers $a_1,\ldots,a_n\in \mathbb{Z}$, we then define
$$ a_0 + a_1 X + \ldots + a_nX^n := (a_0,a_1,\ldots,a_n,0,0,\ldots) \tag{2}. $$
All the operations on polynomials can be defined in terms of sequences as in $(1)$ and are compatible with the usual notation $(2)$. This is a way to formally capture the idea of having some unknown element $X$ that can be 'scaled' by integers, multiplied by itself, and such that we can make sense of the sums of expressions like this.
In the same spirit, suppose you have a set $X$ and you want to have another set $M$ such that:
A way to do this is to define the free abelian group $\mathbb{Z}^{(X)}$ with basis $X$. This concept is closely related to bases of vector spaces. The definition (technically, one definition, there are 'many' equivalent ones) is as follows: as a set $\mathbb{Z}^{(X)}$ consists of functions $f \colon X \to \mathbb{Z}$ such that $f(x) \neq 0$ for finitely many $x \in X$. In other words, the elements of $\mathbb{Z}^{(X)}$ are finitely supported functions $\mathbb{Z} \to X$. We can make sense of sum and multiplication of elements here in the same way we do to define a vector space structure on $\mathbb{R}^X$. Namely:
You can check that these operations define once again elements of $\mathbb{Z}^{(X)}$.
Now, let's go back to the analogy with polynomials; in particular, to the relation between definition $(1)$ and notation $(2)$. Even though formally the free abelian group is defined as finitely supported functions, we want to think of it as "$\mathbb{Z}$-linear combinations of $X$". To do this, we write $$ x := \chi_{\{x\}}, \quad \chi_{\{x\}}(y) = \begin{cases}1 &\text{if $x=y$}\\ 0 &\text{otherwise}\end{cases} $$ You can check that, following this notation, every element of $\mathbb{Z}^{(X)}$ can be written as a finite sum $$ a_1 x_1 + \ldots +a_n x_n $$ for some $x_i \in X$ and integers $a_i$. Moreover, two such expressions are equal if the elements of $X$ appearing are the same, and the coefficients accompanying each element coincide.
In this case, we consider $\mathbb{Z}^{(C_n)}$ with $C_n$ the set of $n$-simplices. In $\mathbb{Z}^{(C_n)}$ it makes perfect sense to speak of $3 \sigma + 22 \tau$ for some pair of simplices $\sigma,\tau$ or any expression of this sort; likewise the maps $\partial_k$ are well defined.
I hope this gives a little bit more context to the first sentence of this answer.