Embedding nowhere dense subsets of $[0,1]$ in the Cantor set Suppose $A$ is a closed, nowhere dense subset of the unit inteveral $[0,1]$ such that $\{0,1\}\subseteq A$. Let $C\subseteq [0,1]$ be the usual middle third Cantor set. Is there an order-preserving homeomorphism $f:[0,1]\to[0,1]$ such that $f(A)\subseteq C$?
Can we also choose $f$ so that it takes each component of $[0,1]\backslash A$ to some component of $[0,1]\backslash C$?
 A: The answer to the second question is no. It could be, for example, that $A=\{0,1\}$. In that case we'd need to send the interval $(0,1)$ to a component of the complement of $C$, and hence $f$ would not be surjective.
I'm pretty sure that the answer to the first question is yes. Doubtless someone will either post a link or inform me I'm all wet. It's a fairly simple construction as these things go. 
I'm not going to attempt to set up precise notation for all this; I'm just going to indicate how the first few steps in the construction go.
Given $I=(a,b)$, let's say that $c$ is "near the center of $I$" if $c$ is in the middle third of $I$. A crucial point is that if $c$ is near the center of $I$ then $(a,c)$ and $(c,b)$ both have lenght no larger than two thirds of the length of $I$.
We begin. Since $A$ is nowhere dense there exists $p\in[0,1]\setminus A$ such that $p$ is near the center of $[0,1]$. And now there exists an open interval $I$ with $p\in I\subset[0,1]\setminus A$. Say $I=(a,b)$. We define $f(x)$ for $x\in[a,b]$ by saying that $f(a)=1/3$, $f(b)=2/3$, and requiring that $f$ be linear (or rather affine) between $a$ and $b$.
Now there is a point $q\in[0,a]\setminus A$ such that $q$ is near the center of $[0,a]$. And there is an open interval $J$ with $q\in J\subset[0,a]\setminus A$. Define $f$ on the closure of $J$ to be an increasing affine function taking $J$ to $(1/9,2/9)$. Do something similar on $[b,1]$.
And now do something similar on the two components of $[0,a]\setminus J$, and something similar on the two components of $[b,1]$ minus the interval we removed in the "something similar" in the previous paragraph.
And at the next stage we do something similar on each of eight intervals.
Repeat forever. We've defined $f$ on a dense open set $V\subset [0,1]$ with $V\cap A=\emptyset$. It's easy to see that if $x$ is a point of $[0,1]\setminus V$ then the left and right limits of $f$ at $x$ are equal. So $f$ extends to a strictly increasing continuous map $f:[0,1]\to[0,1]$. And $f(V)$ is the complement of the Cantor set, so $f(A)$ must be contained in $C$.
I have no idea whether this is clear or not. It would be very easy to convince you standing at a blackboard...
