Show that a set defines a simplicial complex 
Let 
  
  
*
  
*$A_n := \left\{-n,\cdots,-1,1,\cdots,n\right\}$
  
*$\Delta_n := \left\{ B \subseteq A \; \big\vert \; \#(\{-i,i\}\cap B)\leq 1 \; \forall 1 \leq i \leq n \right\}$ 
Show that $\Delta_n$ is a simplical complex and calculate its
  $f$vector.

This exercise should be somehow related to combinatorics. Unfortunately I did only find definitions related to geometric constructs, and did not understand them entirely. 
Could you please tell me what I need to show that this set is a simplicial complex? How can that $f$ vector be calculated? 
 A: Since you said you've been able to show that $Δ_n$ is a simplicial complex, I give here the solution for other people that come by
$Δ_n:=\{B⊆A_n\mid \#(\{−i,i\}∩B)≤1 \text{ for all }1≤i≤n\}$ is the set of subsets of $A_n$ which contain at most one of $i,-i$, for each $i$ between $1$ and $n$. It is clear that a subset of such a set $B$ satisfies the same property, so $Δ_n$ is a simplicial complex. We see that each $\{i\}$ and $\{-i\}$ is an element of $Δ_n$, so the vertex set of $Δ_n$ is just $A_n$. 
This simplicial complex has a nice geometric interpretation: If we think of $i$ as the $i$-th unit basis vector of $\Bbb R^n$, and $-i$ its negation, then a $B$ in $Δ_n$ is the convex hull of some of these vectors, as long as it doesn't contain $i$ and $-i$ at the same time. That means a simplex is a standard-$k$-simplex or its reflection in the subspace spanned by a subset of $\{1,...,n\}$. The realization is actually 
$\{x\in\Bbb R^n\mid ||x||_1=1\}$ and this is homeomorphic to the $n-1$-dimensional sphere.
A $k$-simplex, that is an element of $Δ_n$ of size $k+1$, contains for $k+1$ numbers $i$ between $1$ and $n$ exactly one of $i,−i$, and the number of $k+1$-subsets of $\{1,...,n\}$ is $\binom n{k+1}$. Then for each choice of $k+1$ numbers we can choose if either $i$ or $−i$ is in that set, i.e. $2^{k+1}$ possibilities. So $f_k=2^{k+1}\binom n{k+1}$. Note that the total number of simplices is $3^n$ so we have the identity
$$3^n=\sum_{k=0}^{n-1}2^{k+1}\binom n{k+1}$$
