# Show equivalence statements simplicial complexes

Let $K$ be a simplicial complex in $\mathbb{R}^n$ and $|K|$ its polyhedron (union of all simplices equipped with subspace topology). Show that the following statements are equivalent:

(i) $|K|$ is path connected.

(ii) $|K|$ is connected.

(iii) For any $\sigma,\tau \in K$ there exists a family $\sigma = \eta_1,\eta_2,\dots,\eta_p = \tau$ of simplices in $K$ with $\eta_k \cap \eta_{k + 1} \neq \emptyset$ for any $k = 1,\dots,p-1$.

So (i) implies (ii) trivially since every path connected space is connected. Further assuming (iii) one can construct paths from any point $p$ to $q$ in $|K|$. Hence (iii) implies (i). Now left to show is (ii) implies (iii). However I do not quite see this at the moment. I thought maybe I could work with locally path connectedness or something like that. Has anyone a hint for me or would it make more sence instead of showing (i) implies (ii) implies (iii) implies (i) to show that (i) and (ii) are equivalent and (iii) and (i) are equivalent?

Hint: The relationship between $\sigma$ and $\tau$ described in (iii) is an equivalence relation on $K$. For each equivalence class, take the union of all its elements (as subsets of $|K|$) and show these sets are disjoint and closed in $|K|$.