# A set which is neither meagre nor comeagre in any interval.

As the title suggests, I'm interested in a subset of the real line which is neither meagre nor comeagre in any interval. Does anyone have an example?

Added. See the comments for some discussion about Bernstein sets, which may provide a solution. I don't know anything about these sets, so a detailed answer or reference in that connection would certainly be welcome.

Here are some thoughts of mine and of one of my classmates regarding the problem. My original idea was as follows: Let $\mathcal V$ be a Vitali set (i.e., $\mathcal V$ contains exactly one element from each additive coset of $\mathbb Q$), choose dense subset $S$ of $\mathbb Q$ whose complement $\mathbb Q \setminus S$ in $\mathbb Q$ is also dense, and let $E = \mathcal{V} + S$. Then $E^c = \mathcal{V} + \mathbb{Q}\setminus S$, and it seems like $E$ should have the required property, but I have been unable to prove that this is so.

My classmate had another idea. He worked in $[-1,1]\setminus\{0\}$, seeking a set $E$ with the following properties.

1. The complement of $E$ is obtained from $E$ via reflection through the origin, that is, $E^c = -E$.
2. If $E \cap I$ is the intersection of $E$ with some dyadic subinterval $I \subset [-1,1]$ of the $k$th generation, then $E$ is similar to this intersection, in the sense that $E = 2^k(E\cap I) + a$ for some real number $a$; that is, the intersection of $E$ with any dyadic subinterval is a scaled translate of $E$.

If I'm not mistaken, it is (relatively) straightforward to verify that any set $E$ satisfying 1. and 2. provides an example. As far as the existence of such a set is concerned, my classmate made an argument, but it was fairly complicated, and I cannot recall it well enough off the top of my head to verify its correctness.

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Have you tried a Bernstein set kind of trick? – Asaf Karagila Mar 11 '12 at 21:36
@Asaf From what the Wikipedia entry says, that looks promising. However, that entry doesn't have very much information, and I'm not otherwise familiar with Bernstein sets. Maybe you could elaborate? – Nick Strehlke Mar 11 '12 at 21:39
You formulate an argument to show that there are only $2^{\aleph_0}$ many perfect sets, therefore only that many co-perfect, well-order them and inductively pick elements which are in the $\alpha$-th perfect set and $\alpha$-th co-perfect set. The result is a set which is not contained nor containing a perfect set. If you have an argument why there are only $2^{\aleph_0}$ many meagre sets (locally meagre actually, but perhaps generally meagre would work too.) the argument should hold. – Asaf Karagila Mar 11 '12 at 22:00
I think Asaf is onto the exact correct idea, though slightly off in the details. To construct a Bernstein set start with an enumeration $\{ P_\alpha : \alpha < \mathfrak{c} \}$ of the perfect subsets of $\mathbb{R}$, and recursively pick for each $\alpha < \mathfrak{c}$ distinct $x_\alpha , y_\alpha \in P_\alpha \setminus \{ x_\xi , y_\xi : \xi < \alpha \}$. As each $P_\alpha$ has cardinality $\mathfrak{c}$, you never run out. Then your Bernstein set can be taken to be $B = \{ x_\alpha : \alpha < \mathfrak{c} \}$. (...cont) – arjafi Mar 11 '12 at 22:41
(...cont) It is relatively easy to show that Bernstein sets do not have the Baire Property. This comes does to the fact that every co-meagre set must include an uncountable $G_\delta$-set, which itself must include a copy of the Cantor set; being a perfect set, we know that neither $B$ nor its complement can include it. A modification of this should show that Bernstein sets are neither meagre nor co-meagre on any interval. – arjafi Mar 11 '12 at 22:41

I'll try to fill in some details about the Bernstein set construction, and its connection to this problem.

Recall that $B \subseteq \mathbb{R}$ is called a Bernstein set if it neither includes nor is disjoint from any uncountable closed set; i.e., $F \setminus B$ and $F \cap B$ are both nonempty for every uncountable closed $F \subseteq \mathbb{R}$. (We may equivalently define a Bernstein set as a subset $B \subseteq \mathbb{R}$ which neither includes nor is disjoint from any perfect set.)

We can construct a Bernstein set by noticing that there are $2^{\aleph_0} = \mathfrak{c}$ many uncountable closed subsets of $\mathbb{R}$. (You have hopefully seen that there are $\mathfrak{c}$ many open subsets of $\mathbb{R}$, and therefore there are $\mathfrak{c}$ many closed subsets of $\mathbb{R}$. Moreover, for each $x \in \mathbb{R}$ the set $\mathbb{R} \setminus ( x-1 , x+1 )$ is easily seen to be an uncountable closed set, from which it follows that there are $\mathfrak{c}$ many of these.)

A secondary fact that is needed is that uncountable closed sets of $\mathbb{R}$ all have cardinality $\mathfrak{c}$.

To "construct" a Bernstein set we "enumerate" the uncountable closed subsets of $\mathbb{R}$ as $\{ F_\alpha : \alpha < \mathfrak{c} \}$. Then for each $\alpha < \mathfrak{c}$ we inductively pick distinct reals $x_\alpha , y_\alpha$ such that $$x_\alpha , y_\alpha \in F_\alpha \setminus ( \{ x_\xi : \xi < \alpha \} \cup \{ y_\xi : \xi < \alpha \} ).$$ Since $| F_\alpha | = \mathfrak{c}$ and $| \{ x_\xi : \xi < \alpha \} \cup \{ y_\xi : \xi < \alpha \} | < \mathfrak{c}$, we can always make these choices.

After $\mathfrak{c}$-many steps, we define $B = \{ x_\alpha : \alpha < \mathfrak{c} \}$. It is a Bernstein set because, for each $\alpha < \mathfrak{c}$ we have that $x_\alpha \in B \cap F_\alpha$ and $y_\alpha \in F_\alpha \setminus B$.

We now show that if $B$ is a Bernstein set, then $B$ is neither meagre nor co-meagre in any open interval $I$. The argument will be symmetric (since the complement of a Bernstein set is also a Bernstein set), and so we will show that $B$ is not co-meagre in any open interval $I$.

Suppose that $B$ is co-meager in some open interval $I$. It follows that there is a dense $G_\delta$-subset $G$ of $I$ with $G \subseteq B$. (Note that $G$ is also a $G_\delta$ subset of $\mathbb{R}$.) It follows that $G$ itself is co-meagre in $I$, and is thus uncountable. Since Borel sets all have the Perfect Set Property, there is a nonempty perfect $P \subseteq G$. We thus have $P \subseteq G \subseteq B$, contradicting the fact that $B$ does not include any uncountable closed set!

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Thanks for taking the time to explain. This seems like a very natural solution (assuming that one is familiar with Bernstein sets). – Nick Strehlke Mar 12 '12 at 6:51

I think your idea of a Vitali set does the trick.

Fix an interval $(a,b)$. Note that, by the Baire category theorem, $\mathcal{V}$ is not meager. (If it were, then so would be all its translates, of which the countable union is $\mathbb{R}$.) Hence there is some interval of length less than $b-a$ in which $\mathcal{V}$ is not meager. (If it were meager in all such intervals, it would be a countable union of meager sets, hence meager). Let $(c,d)$ be such an interval, and choose a rational $q \in S$ such that $(c+q, d+q) \subset (a,b)$. Now $\mathcal{V} + q$ is nonmeager in $(c+q, d+q)$, hence nonmeager in $(a,b)$, and so $E = \mathcal{V}+S$ is nonmeager in $(a,b)$.

By the same argument, $E^c$ is also nonmeager in every interval, so $E$ is not comeager in any interval either.

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I was hoping that this question would turn out to have a simple and elegant answer—thank you for providing one. – Nick Strehlke Mar 12 '12 at 2:02
For some comments about Bernstein sets I posted a few years ago in sci.math, see groups.google.com/group/sci.math/msg/36cdfd0c767b1ab3 Also, see groups.google.com/group/sci.math/msg/64a01ce3d199a641 for a slight correction to those comments. – Dave L. Renfro Mar 12 '12 at 18:00