element $(K,x)_{{\mathbb N}}$ is given by ${\underrightarrow{\mathrm{lim}}}_{r}\Psi(B_{(K,x)}(r))$.\
First remark that by the definition of the order, the set of vertices must be the interval $[0,1,...,N_r]\subset{{\mathbb N}}$, where $N_r:=|V(B_{(K,x)}(r))|$, for both of the subcomplexes. For $\Psi(B_{(K,x)}(r))$ this is obvious. For $B_{\Psi(B_{(K,x)}(r+1))}(r)$ we will show it by induction on the second $r$. So suppose that the vertices of $B_{\Psi(B_{(K,x)}(r+1))}(r-1)$ are the interval $[0,...,N_{r-1}]$ but on the other hand there is an $k\in [N_{r-1}+1,...,N_r]$ which is not contained in $B_{\Psi(B_{(K,x)}(r+1))}(r)$, hence $k$ must be at distance $r+1$ to the root in $\Psi(B_{(K,x)}(r+1))$. Therefore, $B_{\Psi(B_{(K,x)}(r+1))}(r)$ must contain another vertex $l\in[N_r+1,..., N_{r+1}]$. But then all simplices of the form $s\cup k  \in {\Delta(\mathbb{N}_0)}$, with $s\subset\{0,...,N_{r-1}\}$, are not contained in $\Psi(B_{(K,x)}(r+1))$. This is a contradiction to the minimality of $\Psi(B_{(K,x)}(r+1))$, since by interchanging the roles of $k$ and $l$ we would get a smaller complex for $B_{(K,x)}(r+1)$. After this consideration and the definition of the order it is clear that $\Psi(B_{(K,x)}(r))=B_{\Psi(B_{(K,x)}(r+1))}(r)$.\
**Claim:** $\Psi$ is continuous.\
Let $[k]\subset\{0,1\}^{{{\mathbb N}}_0}$ be the cylinder set with a $1$ at the $k$-th position and let $\Upsilon(k)$ the simplex $\{k_0,...,k_n\}\in{\Delta(\mathbb{N}_0)})(n)$ with $k_0<k_1...<k_n$. The distance of $k_0$ to $0$ is at most $k_0$ for every minimal subcomplex of ${\Delta(\mathbb{N}_0)}$ which contains $\Upsilon(k)$. 