Set $\mathfrak{c} = 2^{\aleph_0}$. Say that a set $X \subseteq \mathbb{R}$ has the property $P(X)$ if for any open interval $(a, b)$ we have $|(a, b) \cap X| = \mathfrak{c}$ and $|(a, b) \setminus X| = \mathfrak{c}$. Let $A, B \subseteq \mathbb{R}$ with $P(A)$ and $P(B)$. Does it follow that $A$ and $B$ are homeomorphic?
This question is inspired by https://math.stackexchange.com/a/2805635/491450.
Let $X=C+\mathbb Q$ where $C$ is the Cantor set. Although $|(a,b)\cap X|=|(a,b)\setminus X|=\mathbb c$ for every interval $(a,b)$, the set $X$ is not homeomorphic to its complement.
Proof. $X$ is the union of countably many totally disconnected compact sets (translates of $C$). If $\mathbb R\setminus X$ were also the union of countably many totally disconnected compact sets, then $\mathbb R$ would be the union of countably many totally disconnected compact sets. Since totally disconnected compact subsets of $\mathbb R$ are closed and nowhere dense, this would contradict the Baire category theorem.
Why is $|(a,b)\cap X|=\mathfrak c$? It is a property of the Cantor set $C$ that every neighborhood of $0$ contains a subset of $C$ which is similar to $C$ and therefore has cardinality $\mathfrak c$. Therefore, if we choose $t\in(a,b)\cap\mathbb Q$, then every neighborhood of $t$ contains a subset of $C+t$ which is similar to $C$ and has cardinality $\mathfrak c$.
Why is $|(a,b)\setminus X|=\mathfrak c$? The set $(a,b)\setminus X$ is a $G_\delta$ set which is dense in $(a,b)$ and therefore uncountable. By a classical theorem, an uncountable $G_\delta$ subset (or even an uncountable Borel subset) of $\mathbb R$ contains a nonempty perfect set which has cardinality $\mathfrak c$. Alternatively, since $X$ has Lebesgue measure zero, the set $(a,b)\setminus X$ has positive Lebesgue measure, whence it contains a closed subset of positive measure, which has cardinality $\mathfrak c$.
