Show simplicial complex is Hausdorff I have a simplicial complex $K$ and I need to show that its topological realisation $|K|$ is Hausdorff. And $K$ need not be finite.
I have very little idea on how to get started on this. Only that if $x,y \in |K|$, I need to find disjoint open sets containing $x$ and $y$. I also know $|K|$ is a quotient space formed by "glueing" simplices together along their faces, so I have the quotient collapsing map, $p: K \to |K|$ and so my open sets $U, V$ are open iff their pre-images under this map are open.
Any help on how to get started would be much appreciated.
 A: I read the following solution in $\textit{Elements of Algebraic Topology}$ by Munkres.
If $x$ is a point of polyhedron $|K|$ (the topological space of simplicial complex), then $x$ is interior to exactly one simplex of $K$, whose vertices are (say) $a_0, \cdots, a_n$. Then 
$$
x= \sum_{i=0}^{n}t_ia_i
$$
where, $t_i>0 \; \forall i$ and $\sum_{i=1}^{n}t_i =1$. If $v$ is an arbitrary vertex of $K$, we define the $\textbf{barycentric coordinate} \; t_v(x)$ of $x$ w.r.t. $v$ as:
$$
t_{v}(x)={\displaystyle \left\{{\begin{array}{lr}0 \quad \text{ if } v \text{ is none of } \{a_0,\cdots,a_n \}  \\ t_i \quad \text{ if } v \text{ is one of } \{a_0,\cdots,a_n \} \text{ (say) } a_i \end{array}}  \right.}
$$
For a fixed $v$, the function $t_v(x)$ is continuous when restricted to a fixed simplex $\sigma $ of $K$, since either it is equal to $0$ on $\sigma$ or equals the barycentric coordinates $t_i(x)$ of $x$ w.r.t $\{a_0,\cdots,a_n \}$ ( which are continuous function of $x$). We use another result that a map $f : |K| \rightarrow X$ is continuous $\iff \; f|_{\sigma}$ is continous for each simplex $\sigma \in K$ (use the result $f^{-1}(C) \cap \sigma = (f|_{\sigma}) ^{-1}(C) $ for any closed set $C$ in $X$).
Hence, the function $t_v(x)$ is continuous on $K$.
$\textbf{|K| is Hausdorff:} $
Let $ x_0,x_1 (x_0 \neq x_1)$ be two points in $|K|$. There exist atleast one vertex $v$ in $K$ s.t. $t_v(x_0) \neq t_v(x_1)$. Now, we choose any real no. $r$ between these two. The set $\{x:t_{v}(x)<r \}$ and $\{x:t_{v}(x)> r \}$ are two open sets. (Why are these open? Since, range of the continuous function $t_v(x)$ is $[0,1)$, and $[0,r)$ and $(r,1)$ is open in $[0,1)$)
A: Let $q:D\to|K|$ denote the quotient map, where $D$ is the disjoint union $\coprod_σ|\Delta_\sigma^n|$ of (the realizations of) all simplices in $K$. The quotient space consists of equivalence classes, and each class contains a unique point in the interior of some simplex, this is the point in the simplex of lowest dimension (We take the interior of $Δ^0$ to be $Δ^0$ itself). For the two classes, let $x,y$ denote these two points. In order to find disjoint open neighbourhood $N(x)$ and $N(y)$, let us construct disjoint saturated open sets in $D$.
For each simplex $\sigma$, we will find an open subset $U_σ$ of $\Delta_σ^n$ containing the set $[x]\capΔ_σ^n$ as well as the points identified with $U_\tau$ for each face $\tau$ of $σ$. Denote the union of all $U_σ$ for the $σ$ of dimension $\le n$ by $N^n(x)$
Start at dimension $0$: If $\{x\}$ is one $0$-simplex $σ$, let $N^0(x)=\sigma$, otherwise $N^o(x)=\emptyset$.
Now assume by induction that we have constructed the disjoint open saturated sets $N^n(x)$ and $N^n(y)$.


*

*Let $V_σ(x)$ and $V_σ(y)$ denote the sets of points in the faces of $Δ_σ^{n+1}$ which are identified with $N^n(x),N^n(y)$ respectively. Note that if $x\in\mathring{Δ_σ}$, then $N^n(x)$ is empty.  

*If one of the $V_σ$ is non-empty, we can thicken it a bit by stretching it towards the barycenter of the simplex. These thickenings will remain disjoint.

*If $V_σ(x)$ is non-empty and $y\in\mathring{Δ_σ}$, then we can keep the thickening of $V_σ(x)$ small enough, so that it is disjoint to an open ball around $y$ within 
$\mathring{Δ_σ}$

*If both $x$ and $y$ are in $\mathring{Δ_σ}$, then they have disjoint open balls within the interior of $σ$.


If we do this for every simplex of dimension $n+1$, we obtain disjoint open saturated sets in the disjoint union of all simplices of dimension $\le n+1$. This completes the induction.
In the end, $N(x):=\bigcup_n N^n(x)$ and $N(y):=\bigcup_n N^n(y)$ have disjoint open images in $|K|$.
A: Yet another approach...
I will call $|K|_d$ the metric space associated to $|K|$. Consider the function $\eta : |K| \rightarrow |K|_d$ defined to be $\eta(\alpha)=\alpha$ for $\alpha  \in |K|$. 
By proving $\eta$ continuous, we show that every open set in $|K|_d$ is open in $|K|$, equivalently, the metric topology is coarser than the topology of $|K|$ (that is the final topology on $|K|$ with respect to the family of inclusions $inc_\sigma \colon |\sigma| \rightarrow |K|$). But $|K|_d$ is Hausdorff since it is a metric space, so $|K|$ is Hausdorff too. 
So let's show that $\eta$ is continuous. For every $\sigma \in S_K$, $\eta \circ inc_\sigma \colon |\sigma| \rightarrow |K|_d$ is an isometry, therefore continuous. The (so called) universal property of the final topology imply $\eta$ is continuous. 
