Does there exist a nonempty measurable set \(E \subset \mathbb{R}\) satisfying the following two properties:
given distinct \(x,y\in E\), there exists \(z\notin E\) strictly between \(x\) and \(y\);
\(E\) has no isolated points?
The distinctness requirement is necessary. If \(x=y\in E\), there is no point outside \(E\) between them, whether “between” is interpreted strictly or inclusively. Without that qualification no nonempty set can satisfy (a).
Solution
1. Take the middle-thirds Cantor set.
Start with \(C_0=[0,1]\) and obtain \(C_{n+1}\) by removing the open middle third of each component interval of \(C_n\). Then \(C_n\) is a union of \(2^n\) disjoint closed intervals, each of length \(3^{-n}\), by induction. Set \(E=C=\bigcap_{n\geq0}C_n\). This is closed, hence Lebesgue measurable [@Fol13], and nonempty since \(0\in C_n\) for every \(n\).
2. Verify property (a).
Let \(x<y\) belong to \(C\) and choose \(n\) with \(3^{-n}<y-x\). If \((x,y)\subset C\), it is contained in \(C_n\). An interval contained in the disjoint union defining \(C_n\) must lie in a single component interval: meeting two components would force it to contain the gap between them. But a component of length \(3^{-n}\) cannot contain \((x,y)\). This contradiction shows that some \(z\in(x,y)\) lies outside \(C\), proving (a).
3. Verify property (b).
Every endpoint of a component interval of \(C_n\) survives all later stages, because only open middle thirds are removed; it therefore belongs to \(C\). For \(x\in C\) and \(\varepsilon>0\), choose \(n\) with \(3^{-n}<\varepsilon\) and let \(J\) be the component interval of \(C_n\) containing \(x\). At least one of the two endpoints of \(J\) is different from \(x\), lies in \(C\), and is at distance at most \(3^{-n}<\varepsilon\) from \(x\). Thus \(x\) is not isolated.
Therefore the answer is yes; the standard Cantor set satisfies both requested properties.