# Second category but not locally residual

Let $X$ be a Baire space.

A subset $E\subset X$ is said to be of first category if it can be expressed as the union of countably many nowhere dense subsets of $X$. Then $E$ is said to be of second category if it is not of first category.

A subset $E\subset X$ is said to be residual if its complement $X\backslash E$ is of first category.
Then $E\subset X$ is said to be locally residual if there exists a nonempty open subset $U\subset X$ such that $E\cap U$ is residual in $U$.

My question is: if a subset is of second category, does it have to be locally residual?

There might be some counterexamples. I have no idea. Thank you!

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I believe that second category is a well known definition compared to residual. Could you please add that definition as well? –  Asaf Karagila Jan 29 '12 at 11:02
@AsafKaragila Thank you! I added the definition. –  Pengfei Jan 29 '12 at 13:28
I edited the revision since using LaTeX to emphasis words is just plain wrong. One * for italics and ** for bold. :-) –  Asaf Karagila Jan 29 '12 at 13:32
Are you sure this is the definition of locally residual? Locally usually implies that something happens in a neighborhood of every point... –  Asaf Karagila Jan 29 '12 at 20:09
@Asaf: The answer’s no either way. –  Brian M. Scott Jan 29 '12 at 22:51

I believe a counterexample is given by taking $X = [0,1]$ and $E$ to be a Vitali set.

I'll use $+,-$ to denote translation mod 1 (so really I'm taking $X = S^1$ if you like). Also, let $Q = \mathbb{Q} \cap [0,1]$.

First, $E$ is second category. If it were first category, then so would be $E + q$ for every $q \in Q$. Since $X = \bigcup_{q \in Q} E + q$, $X$ would be first category, contradicting the Baire category theorem.

Now suppose there is an interval $(a,b)$ such that $E \cap (a,b)$ is residual in $(a,b)$. Let $(c,d)$ be any other interval of length less than $b-a$. There is a rational $q$ such that $(c,d) + q \subset (a,b)$. Then $(E \cap (c,d)) + q \subset (a,b) \backslash E$, so $(E \cap (c,d)) + q$ is first category in $(a,b)$. Thus $E \cap (c,d)$ is first category in $(a-q, b-q)$ and hence also first category in $X$. But $E$ is a finite union of sets of this form, and $E$ is not first category. This is absurd, so $E$ is not locally residual.

Edit: However, if $E$ has the Baire property, then Proposition I.8.26 of Kechris's Classical Descriptive Set Theory gives a positive answer to your question: every second-category set with the Baire property is locally residual. In particular, this holds for Borel sets.

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Thank you! The book provides the exact propositions I want. –  Pengfei Jan 30 '12 at 2:04

It does not. Here is a counterexample in $I=[0,1]$.

$I$ has only $2^\omega$ open sets, so it has only $2^\omega$ $G_\delta$-sets and hence only $2^\omega$ pairs $\langle U,G\rangle$ such that $U$ is open and $G$ is a dense $G_\delta$-subset of $U$; let $\{\langle U_\xi,G_\xi\rangle\:\xi<2^\omega\}$ be an enumeration of these pairs.

Now construct a set $E$ by recursion as follows. Suppose that $\eta<2^\omega$, and for each $\xi<\eta$ we’ve chosen distinct $x_\xi,y_\xi\in G_\xi$. Then

$$|\{x_\xi:\xi<\eta\}\cup\{y_\xi:\xi<\eta\}|\le|\eta|<2^\omega=|G_\eta|\;,$$

so we can choose distinct $$x_\eta,y_\eta\in G_\eta\setminus\Big(\{x_\xi:\xi<\eta\}\cup\{y_\xi:\xi<\eta\}\Big)\;.$$ Clearly the recursion goes through to $2^\omega$.

Now let $E=\{x_\xi:\xi<2^\omega\}$. If $G$ is a dense $G_\delta$-subset of $I$, then $G=G_\xi$ for some $\xi<2^\omega$, and $x_\xi\in E\cap G_\xi$, so $E$ meets every dense $G_\delta$-subset of $I$ and therefore must be second category in $I$. On the other hand, if $U$ is any non-empty open set in $I$, and $G$ is any dense $G_\delta$-subset of $U$, then $\langle U,G\rangle=\langle U_\xi,G_\xi\rangle$ for some $\xi<2^\omega$, and $y_\xi\in G_\xi\setminus E$, so $E\cap U\ne G$. Thus, $E$ is not locally residual.

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Thank you for the clear construction! –  Pengfei Jan 30 '12 at 2:05