From Rotman's Algebraic Topology
Prove that $\text{st} (p)$ is an open subset of $|K|$, where $|K|$ is the underlying space of a simplicial complex and $\text{st} (p)$ is defined as $\bigcup_{s \in K, p \in \text{Vert}(s)} s^{\circ}$.
I can see that $\text{st}(p)$ is open iff $|K| - \text{st}(p)$ is closed in $|K|$, which is true iff $s' \cap (|K| - \text{st}(p))$ is closed for each $s' \in K$.
If $s'$ doesn't contain the vertex $p$, then then $s' \cap (|K| - \text{st}(p)) = s'$, which is closed in $s'$.
But if $s'$ contains the vertex $p$, then $s' \cap (|K| - \text{st}(p)) = s' - s' \cap \text{st}(p)$ is closed is what needs to be shown.
Either that or $s' \cap \text{st}(p)$ must be shown to be open. I can see this is true in visualizable space, but I'm not sure how to show this for any arbitrary simplex of higher dimensions.
Anyone have any ideas?
Let $p\in s'$. Then $p$ has vertices $p_0=p,p_1,\ldots,p_d$. Then $s'\cap(|K|-\mathrm{st}(p))$ is then the $(d-1)$-simplex spanned by $p_1,\ldots,p_d$. It's a facet of $s'$ and so a closed subset of $s'$.