A connected set with no interior but positive measure? Hi everyone: Is there a set $E$ in $\mathbb{R}^{m}$ for $m\geq2$ and with respect to the (regular) Euclidean topology such that:
(1) The interior of $E$ is empty
(2) The Lebesgue measure of $E$ is positive
(3) $E$ is connected? Thanks for your help.
 A: Let $K\subset\mathbb R$ be a fat cantor set (which has positive measure, but empty interior). Consider $(K\times I)\cup(I\times\{0\})\subset\mathbb R^2$, where $I$ is the unit interval. This is connected (because of $I\times\{0\}$), and has the same Lebesgue measure as $K$. It also has empty interior because $K$ has empty interior. To get that the result holds in $\mathbb R^n$ for all $n>1$, simply cross this set with the unit interval $n-2$ more times. 
A: For $m=1$ the answer is no, the connected sets are intervals and if the Lebesgue measure is non zero then the interval must contain an open set.
For $m>1$, let $A= \{ x | x_k \in \mathbb{Q}^c \text{ for some } k \}$. Then
$A$ is connected, has empty interior and the complement has measure zero.
Following Nate's suggestion:
To see why $A$ is connected, in fact path connected, pick two points $x,y \in A$.
Suppose $x_i$ is irrational. Suppose $y_i$ is irrational and let $j$ be another
index and let $\alpha$ be irrational. Consider the path, where only one component is changed at a time, $x \to (...,x_i,...,\alpha,...) \to (...,y_i,...,\alpha,...) \to y$. The $\alpha$ is in the $j$th position.
Now suppose $y_i$ is rational and $y_j$ is irrational. As above, consider
the path 
$x \to (...,x_i,...,y_j,...) \to y$.
