# Let $K$ be a simplicial complex (that need not be finite). Prove $|K|$ is Hausdorff.

As the title makes clear, I'm trying to solve a question which asks me to show the topological realisation of a simplicial complex is Hausdorff.

The question reminds me that a subset of $$|K|$$ is open if it intersects every simplex in an open set, and hints that the standard simplex has a natural metric as a subset of $$\mathbb{R}^n$$.

This is the first problem on simplicial complexes I'm facing, and so I'm struggling to think of how I can go about constructing disjoint open sets containing two different points in $$|K|$$.

I'd really appreciate it if someone could help me get started with this.

The space $$|K|$$ can be modelled in this way: $$|K|$$ is the set of maps $$f$$ from $$V$$ (the vertex set) to $$\Bbb R_{\ge0}$$ with the properties that $$\{v\in V:f(v)>0\}$$ is the vertex set of a simplex in $$K$$ (and so finite) and $$\sum_{v\in V}f(v)=1$$.
In this model, $$|K|$$ has a metric $$d$$ defined by $$d(f,g)=\max_{v\in V}|f(v)-g(v)|$$. In general, the topology of $$|K|$$ is not the metric topology induced by $$d$$ (it is when $$K$$ is finite; it may not be if $$K$$ is infinite). But the open sets of $$|K|$$ with respect to this metric are open in the standard topology on $$|K|$$ (the standard topology is finer than the metric topology).