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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.

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No. Take $X$ to be an interval union the rationals (for example). –  t.b. Jul 11 '11 at 15:25
    
Re: updated question. Again, no: The irrationals contain a homeomorphic copy of the Cantor set, see e.g. my answer here. –  t.b. Jul 11 '11 at 16:05
    
Thanks, I was worried about the effect of my edit on Patrick's reputation too, but I didn't think it good to ask a whole new question... I hope my new edit clarifies things... –  Condor Jul 11 '11 at 16:05
    
I edited my answer and there is comments. People who would try to downvote my answer will probably read the comments before and understand. Thanks for worrying though. –  Patrick Da Silva Jul 11 '11 at 18:11
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I guess somebody should point out that if $X$ is a Borel (or analytic) subset of $\mathbb{R}$ which is totally imperfect, then $X$ is countable. –  user83827 Jul 11 '11 at 18:33

2 Answers 2

up vote 2 down vote accepted

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.

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Yes that is correct thanks... I hope you don't mind if I change my question slightly to remove such trivial cases... –  Condor Jul 11 '11 at 15:52
    
As ccc pointed out in a comment to the question, any uncountable Borel or analytic set has a perfect subset, see Wikipedia and Kechris (13.6), p.82 and Exercise (14.13), p.88. The proofs are not trivial at all. –  t.b. Jul 11 '11 at 19:46
    
So it seems. Thanks –  Patrick Da Silva Jul 11 '11 at 19:48

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.)

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I found your post perfectly readable. Nevertheless I typeset it in order to show you that you can type (La)TeX in pretty much the same way as usual. If you click on the edit link on the bottom left of your post, you can look at the source again and see in detail what I did. Nice argument, by the way! –  t.b. Jul 11 '11 at 21:17
    
Are you considering sets $X$ such that both $X$ and $\mathbb{R} \backslash X$ are totally imperfect? These sets are usually called Bernstein sets and also fail to have the property of Baire in a strong way (since any nonmeager BP set contains a perfect set). –  user83827 Jul 11 '11 at 21:34
    
ccc -- you're corrrect. I inadvertently wound up talking about Bernstein sets, which form a subcollection (although a very important subcollection) of the collection of totally imperfect sets. For more about totally imperfect sets, see the survey article "Classical theory of totally imperfect spaces" by Brown and Cox [Real Analysis Exchange 7 (1981-82), 185-232] and the remarks I posted at <groups.google.com/group/sci.math/msg/36cdfd0c767b1ab3>; (see a follow-up post in that thread for a correction). –  Dave L. Renfro Jul 11 '11 at 21:41
    
@Dave: Thanks for the link and the article! –  user83827 Jul 11 '11 at 21:51

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