Does a compact set with positive Lebesgue measure mean it containing an open set? Let $A \subset \mathbb{R}^n$ be a compact set with positive Lebesgue measure on $\mathbb{R}^n$. Can we find an open set $B \subset \mathbb{R}^n$ such that $B \subset A$?
PS: I know that if the compactness removed, the answer is no, since $A$ can be any compact set removing all the rational points.
 A: Check out the Smith-Volterra-Cantor set or variations thereof. Note that $B=\emptyset$ is, of course, always possible.
A: Although the other answer is really a good answer, it's worth noting that you can cook up a lot of examples in a similar manner: choose your favorite compact set $C$ (in $\mathbb R^n$ or some other nicely behaved space) with positive measure. Now, make a list of countably many open sets such that every non-empty open set contains one on your list (e.g. the set of balls of rational radius centered at rational coordinates).
Let's decide that we're okay with losing some $\varepsilon$ of area from $C$ during this construction. We are now going to nibble away at $C$ by removing open sets (preserving compactness, even in the limiting case). Choose a point in your first open set and remove from $C$ a ball with area $\varepsilon 2^{-1}$ around that point. Choose a point in your second set and remove a ball around it with area $\varepsilon 2^{-2}$. Choose a point in your third set and remove a ball around it with area $\varepsilon 2^{-3}$. Do this for your entire list. You've at most removed an area of $\varepsilon$. However, no non-empty open set is a subset of the remaining set, since no set on your list is a subset of the remaining set.
Basically, the fact that we're in a second-countable space means that there really aren't so many open sets, so we can just make a list of enough open sets and deal with them individually. This is a nice thing to keep in mind if you want to make pathological examples from scratch.
