$\operatorname{diam}(K) \ge \varepsilon$. But $\bigcap_{n\in {\mathbb N}} {\varphi}_n^{-1}(b_n)= \set{u} $, a contradiction.

\(2) By (1) and the fact that function $p$ is uniformly continuous.

\[limitconnected\] If $B_{n} \subseteq A_{n}$, for $n \in {\mathbb N}$, are $R$-connected subsets and $({\varphi}_{n}^{-1}(B_{n}))_{n \in {\mathbb N}}$ converges in $\operatorname{\mathcal K}( \mathbb A )$ to $K$, then $K$ is $R$-connected.

Let $U, U'$ be clopen, nonempty subsets of $ \mathbb A $, with some positive distance $\delta$, such that $U\cap K, U'\cap K $ partition $K$. Consider the open neighborhood $O = \setnew{ C \in \operatorname{\mathcal K}( \mathbb A )}{C \subseteq U \cup U', C \cap U \neq \emptyset, C \cap U' \neq \emptyset}$ of $K$ in $\operatorname{\mathcal K}(\mathbb A)$. Let $n \in {\mathbb N}$ be such that $ {\varphi}_{n}^{-1}(B_{n}) \in O$, and $\operatorname{diam}( {\varphi}_n^{-1}(a))<\delta$ for each $a \in A_{n}$: such a $n$ exists by . Then each $ {\varphi}_{n}^{-1}(a)$ for $a \in B_{n}$ is either contained in $U$ or in $U'$, as the distance between the two clopen sets is greater than $ \operatorname{diam}( {\varphi}_n^{-1}(a))$, and $U, U'$ each contain at least one such set, since $ {\varphi}_n^{-1}(B_{n})$ has nonempty intersection with both $U$ and $U'$. It follows that ${\varphi}_n[U]\cap B_n, {\varphi}_n[U']\cap B_n$ partition 