Let $K$ be a simplicial complex. If $L$ is a subcomplex of $K$ , then $\left|L\right|_{d}$ is a closed subset of $\left|K\right|_{d} $ .
Some definitions:
A simplicial complex is defined to be a set $K$ consisting of finite sets called simplices such that every subset of a simplex is also a simplex. A vertex of $K$ is defined to be a simplex containing only 1 point. We denote the set of all vertices of $K$ by $V(K)$. A subcomplex $L$ of a simplicial complex $K$ is a subset of $K$ that is a simplicial complex itself. We define $\left|K\right|_{d} $ to be the set of all functions $\alpha:V\left(K\right)\rightarrow [0,1]$ such that
$\left\{ v\in V\left(K\right):\alpha\left(v\right)\neq0\right\} \in K$ , and
$\sum_{v\in V\left(K\right)}\alpha\left(v\right)=1$.
We can define a metric $d$ on $\left|K\right|_d$ by $d\left(\alpha,\beta\right)=\sqrt{\sum_{v\in V\left(K\right)}\left[\alpha\left(v\right)-\beta\left(v\right)\right]^{2}}$ for all $\alpha,\beta\in\left|K\right|_d$. So $\left|K\right|_{d} $ is now a topological space. If $L$ is a subcomplex of $K$, then we can can consider $\left|L\right|_{d} $ as a subspace of $\left|K\right|_{d} $ in the following way. If $\alpha\in\left|L\right|$ , then $\left\{ v\in V\left(L\right):\alpha\left(v\right)\neq0\right\} \in L$ , and $\sum_{v\in V\left(L\right)}\alpha\left(v\right)=1$ . Now define a function $\alpha':V\left(K\right)\rightarrow [0,1]$ by $\alpha'(v)=\alpha\left(v\right)$ for all $v\in V\left(L\right)$ and $\alpha\left(v\right)=0$ for all $v\in V\left(K\right)-V\left(L\right)$ . Then $\alpha'\in\left|K\right|_d$ .
These definitions are all introduced in Spanier's book. He claims that if $L$ is a subcomplex of $K$ , then $\left|L\right|_{d}$ is a closed subset of $\left|K\right|_{d} $. I tried proving it, but could not do it.