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.


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.

  • $\begingroup$ But why this set has empty interior? $\endgroup$ – M. Rahmat Mar 27 '16 at 4:17
  • 1
    $\begingroup$ If it contained any open set, projection in the first coordinate would give an open subset of $\mathbb R$ contained in $K$, which is a contradiction. $\endgroup$ – Alex S Mar 27 '16 at 4:24

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 $x \to (...,x_i,...,y_j,...) \to y$.

  • $\begingroup$ You might like to explain why $A$ is connected. Although the trick is pretty well known, it may not be obvious to someone who hasn't seen it. $\endgroup$ – Nate Eldredge Mar 27 '16 at 3:47
  • $\begingroup$ You presumably mean "for some $k$" rather than "for some i"? $\endgroup$ – ForgotALot Mar 27 '16 at 3:49
  • $\begingroup$ @ForgotALot: Thanks. $\endgroup$ – copper.hat Mar 27 '16 at 3:51
  • $\begingroup$ @NateEldredge: Good suggestion, I'm not sure my explanation is entirely clear, but it should give the idea. $\endgroup$ – copper.hat Mar 27 '16 at 4:06
  • $\begingroup$ Another way of writing $A$ is $A=\Bbb R^m \setminus \Bbb Q^m$. $\endgroup$ – Greg Martin Mar 27 '16 at 4:10

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