Totally Imperfect vs Dense We say a set $X$ is totally imperfect in a set $E$ if it contains no non-empty perfect subset of $E$.
Let $X$ be a dense subset of $E$ such that $E \backslash X$ is also dense in $E$. Is $X$ necessarily totally imperfect in $E$?
I am particularly interested in the special case $E = \mathbb{R}$.
Edit: Originally I asked the question if $X$ was simply a dense proper subset of $E$ with no restrictions on its complement.
 A: Others have commented that being dense with a dense complement is not enough. In fact, totally imperfect sets of reals and their complements satisfy a much stronger type of density.
THEOREM: Let $X$ be a totally imperfect subset of the reals $\mathbb{R}$. Then, for each open interval $(a,b)$, we have:
(a) The outer Lebesgue measure of $X \cap (a,b)$ is $b - a$.
(b) The outer Lebesgue measure of $(\mathbb{R}\setminus X) \cap (a,b)$ is $b - a$.
PROOF: If the inner Lebesgue measure of $X \cap (a,b)$ were positive, then this intersection would contain a Lebesgue measurable set of positive measure, hence a closed set of positive Lebesgue measure, hence a perfect set of positive Lebesgue measure (Cantor-Bendixson), which contradicts $\mathbb{R}\setminus X$ having nonempty intersection with every perfect set. Therefore, the inner Lebesgue measure of $X \cap (a,b)$ is zero. The same argument works for $\mathbb{R}\setminus X$.
REMARK: This proves that every totally imperfect set in the reals fails to be Lebesgue measurable, even nowhere locally Lebesgue measurable, and in a rather strong way since finite additivity is enough to get a contradiction.
(FYI, I haven't had a chance to read about how to post math symbols like 'intersect' yet, and copying/pasting from other posts didn't work. Others should feel free to edit my comments using the appropriate math symbols.)
A: Take $E = \mathbb R$ and $X = \mathbb R \backslash \{ 0 \}$. Clearly $X$ is a proper dense subset of $E$ and contains a perfect proper subset, so that $X$ is not totally imperfect. 
To the new question, hmm. The Cantor set is a good example of a perfect set that is nowhere dense. Choose one such set in $\mathbb R$ and call it $Y$. Now consider $F = \mathbb R \backslash Y$. The question now becomes : can you find a subset $Z$ of this $F$ such that $Z$ and its complement are dense in $F$? Afterwards to get such an $X$ in $\mathbb R$ you simply take $Z \cup Y$, which will have the desired properties, because since $Z$ is dense in $F$ and $F$ is dense in $\mathbb R$, $Z$ is dense in $\mathbb R$, but then so does $Z \cup Y$. Also, $Z \cup Y$ contains $Y$, and $(Z \cup Y)^c = Z^c \cap Y^c \supseteq Z^c$ which is dense in $\mathbb R \backslash Y$, and clearly $Y^c$ is dense in $Y^c$, so finding that $Z$ such that $(Z \cup Y)^c$ is also dense in $\mathbb R$ means we're done. 
One example of this would be to let $Y$ be the cantor set and $Z$ to be the irrationals not in the Cantor set. Note that this basically means the set I'd obtain would be the Cantor set union the irrationals.
Hope that helps,
Hm. I am not happy about my answer ; it can generate counter examples to your question but it doesn't work in its generality. Intersection of dense sets is not a dense set in general. Although the Cantor set union with the irrationals does work.
