Let: $$D_t = \{(x,y,t) \in \mathbb{R}^3 : x^2 +y^2 \leq 1\},$$ $$S=\{(p,q,0)\in \mathbb{R}^3: p^2 +q^2<1, p\in \mathbb{Q}, q\in \mathbb{Q} \}$$ $$I(a,b) \text{ - open line segment between points } a, b$$ Now let's define set $X$: $$X=D_0 \cup \Biggl( \bigcup_{(p,q,0)\in S} I\big((p,q,0),(p,q,p)\big) \Biggr) \subset \mathbb{R}^3$$ Check if $X$ is completely metrizable.
I have an answer of this task but I don't understand one statement:
Let: $$A=X\cap \{(x,y,z): z=\frac 12\}=\{(p,q,\frac 12): p\ge \frac 12; \; p^2+q^2<1; \; p,q\in \mathbb Q\}=$$ $$=\bigcup\limits_{p,q\in \mathbb Q} \{(p,q,\frac 12): p\ge \frac 12; \; p^2+q^2<1\}$$ So $A$ is the countable sum of points i.e. closed and boundary set.
Assume that $X$ is completely metrizable. Then $A$ as closed subset $X$ is also completely metrizable. However $A$ isn't a boundary set so Baire's condition is not satisfied. So we got a contradiction - $X$ is not completely metrizable.
Can someone explain to me why $ A $ is not a boundary set?
EDIT: $X$ is a boundary set when Int$(X)=\varnothing$
I don´t know how they define boundary set. However, it is easy to check that $A$ is not a Baire space. For if you pick an enumeration $a_n$ of the points of $A$, and you pick the open sets $A-\{a_n\}$, they are all dense open sets in $A$, and their intersection is empty. Thus $A$ is not a Baire space.