Problem about the existence of a continuous surjective map Let $F$ be a closed set in $\Bbb R^2$, $F\neq \varnothing,\Bbb R^2$, and $F^\circ\neq \varnothing$,
does there exist a continuous surjective map from $\Bbb R\times \Bbb Z$ to $F^{\circ-}-F^\circ$？
 A: Not always.
One way to see this is to find $F$ such that $\overline{F^\circ} - F^\circ$ has uncountably many connected components. Since a continuous image of $\mathbb{R} \times \mathbb{Z}$ has only countably many connected components, this $F$ will be a counterexample.
To get $F$, the basic idea is to take a bunch of open boxes whose closure is a bunch of closed boxes plus a Cantor set, and to do this in such a way that each point in the Cantor set will constitute a connected component of $\overline{F^\circ} - F^\circ$. (There may be a simpler way than the one I'm about to describe, but this is what came to mind.)
Let
$$I_0 = (0,1)$$
$$I_1 = \left(0,\frac{1}{3}\right) \cup \left(\frac{2}{3},1\right)$$
$$I_2 = \left(0,\frac{1}{9}\right) \cup \left(\frac{2}{9},\frac{1}{3}\right) \cup \left(\frac{2}{3},\frac{7}{9}\right) \cup \left(\frac{8}{9},1\right)$$
$$\dots$$
Then, for each $n$, let
$$B_n = I_n \times \left(\frac{1}{2^{2n+1}},\frac{1}{2^{2n}}\right).$$
Finally, let $U = \bigcup_{n \in \mathbb{N}}B_n$ and $F = \overline{U}$. 
It's not too hard to see that $F^\circ = U$, so $\overline{F^\circ} = F$ and $\overline{F^\circ} - F^\circ = F - U$.
Notice that each $B_n$ contains finitely many little open rectangles, say $B_n^1, \dots, B_n^{2^n}$. Then each $\overline{B_n^i}$ is a clopen subset of $F$, and $\overline{B_n^i} - B_n^i$ is a clopen subset of $F-U$. Let $D_n^i = \overline{B_n^i} - B_n^i$.
We want to show that $F-U$ has uncountably many connected components. To see this, let $C = \bigcap_{n \in \mathbb{N}}\overline{I_n}$ (the typical middle-thirds Cantor set). It's not too hard to see that $F-U$ is just $C \times \{0\}$ together with all the $D_n^i$. Each point of $C \times \{0\}$ is a connected component in $F-U$ because each neighborhood of a point in $C \times \{0\}$ will contain some $D_n^i$.
