If $E \subset R$ and $m(E) \gt 0$ , can we find a surjective function that maps $E$ to $R$ ?

Question

If $m(E)=0$ , then $E$ the cardinality can be rather large . How about the inverse condition ? If $E \subset R$ and $m(E) \gt 0$ , then we can assert that $E$ is uncountable . But can we find a surjective function that maps $E$ to $R$ ? Or if $m(E)=a \gt 0$ , can we find such a surjective function $f$ maps $E$ to $(0,a)$ ?

Answer 1

If $m(E)>0,$ then there exists a binary system of pairwise disjoint compact subsets of $E$ of positive measure: We start with disjoint $K_0,K_1\subset E.$ Then we split $K_0$ into disjoint $K_{00}$ and $K_{01},$ and split $K_1$ into disjoint $K_{10}$ and $K_{11}.$ All these sets have positive measure so we can keep going, a la Cantor. See binary tree

Every binary sequence generates a nested sequence of compact sets as above. For example $0,1,1,0,1,0,\dots$ generates

$$K_0,K_{01},K_{011},K_{0110},K_{01101}, K_{011010},\dots $$

These sets are nested, compact and nonempty. Hence their intersection is nonempty. Since the set of binary sequences has the cardinality of $\mathbb R$ we end up with with $\mathbb R$ many nonempty pairwise compact subsets of $E.$ Therefore the cardinality of $E$ is at least that of $\mathbb R .$ Schroeder-Bernstein then shows the cardinality of $E$ is the cardinality of $\mathbb R.$