Show simplicial complex is Hausdorff

Question

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.

Answer 1

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 $|K|$.