Could you please help me to solve this problem? Especially a)
Let $$D_t = \{(x,y,t) \in \mathbb{R}^3 : x^2 +y^2 \leq 1\}$$ and
$$S=\{(p,q,0)\in \mathbb{R}^3: p^2 +q^2<1, p\in \mathbb{Q}, q\in \mathbb{Q} \}$$
Now let's define set $X$:
$$X=D_0 \cup \Biggl( \bigcup_{(p,q,0)\in S} I\big((p,q,0),(p,q,p)\big) \Biggr) \subset \mathbb{R}^3 $$
$I(a,b)$ denotes a closed line segment between $a$ and $b$.
a) Check if X is completely metrizable.
b) Prove that $Y= X \cup D_1$ is connected.
c) Prove that $Z= \overline X \cup D_1$ is contractible. Here $\overline X$ means closure of $X$ in euclidean space $\mathbb{R}^3$.
My attempt:
ad. a)
I suppose that X is not completely metrizable, because if it were, we would have that $A=\bigcup_{(p,q,0)\in S} I\big((p,q,0),(p,q,p)\big)$ has empty interior by Baire's theorem but I can't see the actual contradiction.
ad. c)
Intuitively, we can contract each line segment (except the one that connects both discs) to the lower disc, then contract the upper disc to the point where it connects with the line segment that is left, then contract the segment to the disc and at the end contract the lower disc. Is this correct? I do not need the explicit formula but just sort of a description of how to contract the space.
a) To finish your argument, remark that $A$ contains a dense open subset $X\setminus D_0$ of $X$.
b) Suppose to the contrary that $Y$ is not connected. Then there exist two nonempty disjoint open subsets $U$ and $V$ of $Y$, such that $U\cap V=Y$. Since $Y=X\cup D_1$ and each of the sets $X$ and $D_1$ is (linearly) connected, the only possibility is that one of the sets $U$ and $V$ is $X$, and the other is $D_1$. This implies $D_1$ is an open subset of $Y$. Therefore there exists an open subset $W$ of $\Bbb R^3$ such that $W\cap Y=D_1$. For each point $a\in D_1$ choose a number $\varepsilon_a>0$ such that an open cubic neighborhood $Q_a=a+(-\varepsilon_a, \varepsilon_a)^3$ is contained in $W$. Since $\{Q_a:a\in D_1\}$ is an open cover of a compact set $D_1$, there exists a finite subset $F$ of $D_1$ such that $$Q=\bigcup \{Q_a:a\in F\}\supset D_1.$$ Put $\varepsilon=\min\{\varepsilon_a:a\in F\}$ and pick arbitrary rational numbers $1-\varepsilon<p<1$ and $0<q<\sqrt{1-p^2}$. Then
$$(p,q,p)\in Q\cap X\subset W\cap X=W\cap Y\cap X=D_1\cap X=\varnothing,$$
a contradiction.
c) Your argument looks OK, but I’m going to check it.