Confusion of proof on nowhere dense sets I am reading the book `Measure and Category' by Oxtoby. In the first chapter, they focus solely on the real line. I am confused by the following proof:

I understand that, since $ A_1, A_2 $ are nowhere dense, given an interval $ I $, both sets are not dense on $ I $.
Therefore, there are subintervals $ I_1 ,I_2 $ that lie on the complement of $ A_1, A_2$.
The proof then says `Hence $ I_2 \subset I -(A_1 \cup A_2) $. I am struggling to see how that follows from the previous sentence. Surely, $  I -(A_1 \cup A_2) $ need not contain $  I -A_1 $?
 A: Let $x \in I_2$. Then $x \in I$ because $I_2 \subset I_1\subset I$. $x \notin A_2$ is clear. Also $x\in I_1$ so $x \notin A_1$. It follows that $x \notin A_1\cup A_2$.
A: Since $I_1\subset I\setminus A_1$ and $I_2\subset I_1\setminus A_2$ you have that
$$ I_2\subset(I\setminus A_1)\setminus A_2=I\cap (A_1)^c\cap(A_2)^c=I\cap(A_1\cup A_2)^c=I\setminus(A_1\cup A_2)$$
by deMorgan's laws.
A: To complement the answers here, I'd like to add an alternative proof to the statement in question following one of the comments. We do not require the full generality of Baire's theorem, but only the fact that $\mathbb{R}$ is separable.  One equivalence to saying that a  $A$ is nowhere dense is that its closure has empty interior: $\operatorname{int}(\overline{A})=\emptyset$. Suppose $A$ and $B$ are nowhere dense and $U$ is any open set $U\subset \overline{A\cup B}=\overline{A}\cup\overline{B}$. Then
$\Big(\overline{A}\Big)^c \cap \Big(\overline{B}\Big)^c\subset U^c$. The sets $\Big(\overline{A}\Big)^c$ and $\Big(\overline{B}\Big)^c$ are both open and dense in $\mathbb{R}$. It follows that $\Big(\overline{A}\Big)^c \cap \Big(\overline{B}\Big)^c$ is open and dense in $\mathbb{R}$. Since $U^c$ is closed and contains a dense set, it follows that $U^c=\mathbb{R}$. Consequently $U=\emptyset$. This shows that $\overline{A\cup B}$ has empty interior, that is $A\cup B$ is also nowhere dense.
