Continuous collections of arcs

Question

Let $X$ be a separable metric space. Suppose there is a mapping $f:X\to C$ of $X$ onto the Cantor set $C$, whose point preimages are arcs (homeomorphic to $[0,1]$), and such that if $c_n\to c$ in $C$ then $f^{-1}\{c_n\}\to f^{-1}\{c\}$ in the Hausdorff distance.

It is not necessarily the case that $X$ is homeomorphic to the topological product $C\times [0,1]$, as $X$ could contain bent arcs limiting toward a straight arc.

Question: Does $X$ contain a copy of $C\times [0,1]$?

For my application we can assume $X\subset \mathbb R^2$ (if that helps).

Answer 1

Let $F(c)=f^{-1}(c)$, $F: C\to K(X)$, where $K(X)$ is the hyperspace of $X$. Then $F$ is continuous by your condition and thus $F(C)$ is compact. Consequently $\bigcup F(C)$ is compact as well and $\bigcup F(C)=X$. Hence $X$ is complete.

By a result of van Douwen (Uncountably many pairwise disjoint copies of one metrizable compactum in another, Topology Appl.51(1993), 87–91) the answer is positive:

Theorem (van Douwen) If a separable completely metrizable space $X$ contains uncountably many pairwise disjoint copies of a compact space $K$, then $X$ contains a copy of $K\times C$.