Homeomorphic images of an "almost" basic open set in $2^{\omega}$ Let $X\subseteq 2^{\omega}$ such that $X=O\setminus M$ for some basic open set $O$ and a meager set $M$. Suppose that $Y\subseteq 2^{\omega}$ is homeomorphic to $X$. Does it follow that $Y=O'\setminus M'$ for some basic open set $O'$ and meager set $M'$?
 A: Not necessarily. 
First for any finite $0-1$ sequence $s$ let $O_s = \{x \in 2^\omega: s \subseteq x\}$. And for any $x \in 2^\omega$ and $n \in \omega$ let $x/n \in 2^\omega$ be the sequence such that for each $i\in\omega$, $x/n(i) = x(i+n)$. $x/n$ just cuts the first $n$ elements of $x$.
Now let $f$ be the constant $0$ function and $X = 2^\omega - \{f\}$ and $Y = O_{\langle 1\rangle} \cup O_{\langle 0, 0\rangle} - \{f\}$. Now let $\sigma: Y \rightarrow X$ with $\sigma(x) = x$ for $x\in O_{\langle 1\rangle}$ and $\sigma(x) = x/1$ for $x \in O_{\langle 0, 0\rangle}$. $\sigma$ is clearly bijective. To see that it is continuous, you can take a sequence of converging functions, and see that after some step, they fall in one "component" of $Y$ and then it becomes easy. To see that it's inverse is continuous, take another convergent sequence of functions and look at their first coordinates, then again it will be straightforward as above. So $\sigma$ is a homeomorphism.
Also note that $Y$ is not of the form $Y = O'-M'$ for some basic open set $O'$ and meager set $M'$. Because if it were so, then the only choice for $O'$ would be $O_\emptyset$, but then we must have $O_{\langle 0, 1\rangle} \subseteq M'$, which is a contradiction.

As you requested, I think I have an example, where the $O'$ can't even be any open set. First for $x,y \in 2^\omega$, let $x*y \in 2^\omega$ be such that $x*y(2n+1) = x(n)$ and $x*y(2n) = y(n)$. $x*y$ is just the sequence which has $x$ as it's odd-indexed subsequence and has $y$ as it's even-indexed subsequence.
Now let $f$ and $X$ be as above, and let $\sigma:X \rightarrow 2^\omega$ with $\sigma(x) = x*f$. You can again see that $\sigma$ is a homeomorphism onto it's range. The range of $\sigma$ which we denote with $Y$, is the set of all functions which are $0$ in their even-indexed subsequence minus $f$. You can see that $Y$ is itself nowhere dense. So it is meager. And so it can't have the above representation.
A: I just realized that there is an even easier counterexample, namely let $X=2^{\omega}$ and let $Y$ be any perfect meager subset of $2^{\omega}$ (such sets can be obtained, for example, by partitioning $2^{\omega}$ to a union of a Lebesgue null set and a meager set, which must contain a perfect set).
