Meagre and dense sets We consider subsets of $\mathbb R$. We know the following: * If $A$ is meagre then $\mathbb R \setminus A$ is not meagre. (Converse not true.) * If $A$ is meagre then $\mathbb R \setminus A$ is dense. (Converse not true.) My question: Is is true that if $\mathbb R \setminus A$ is not meagre and $\mathbb R \setminus A$ is dense, then $A$ is meagre ? My intuition says not.

The answer to your question is **NO**. Here is an example.

Let $A$ be the rationals in $[0,1]$ union all the irrationals that do not belong to $[0,1].$ That is,

$$A \;\; = \;\; \left([0,1] \cap {\mathbb Q}\right) \;\; \cup \;\; [({\mathbb R} \setminus [0,1]) \; \cap \; {\mathbb I}], $$

where ${\mathbb I} = {\mathbb R} \setminus {\mathbb Q}.$

Then

1. ${\mathbb R} \setminus A$ is not meager (contains all the irrationals in $[0,1])$

2. ${\mathbb R} \setminus A$ is dense in ${\mathbb R}$ (contains all irrationals in $[0,1]$ and all rationals outside of $[0,1])$

3. $A$ is not meager (contains all the irrationals in $[2,3])$