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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.

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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).

Metric topologies are Hausdorff; topologies finer than Hausdorff topologies are also Hausdorff.

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