How to prove that these subspaces of $\Bbb R^3$ are pairwise non-homeomorphic?

Question

Let $\mathbb{R}^3$ be usual topological space and $\mathbb{Q}$ the set of rational numbers. Define $X,Y,Z,$ and $W$ as follows

$$\begin{align}X&=\{(x,y,z)\in\Bbb R^3\mid |x|+|y|+|z|\in\mathbb Q\}\\ Y&=\{(x,y,z)\in\Bbb R^3\mid xyz=1\}\\ Z&=\{(x,y,z)\in\Bbb R^3\mid x^2+y^2+z^2=1\}\\ W&=\{(x,y,z)\in\Bbb R^3\mid xyz=0\} \end{align}$$

Which of the following statements is correct?

$a.$ $X$ is homeomorphic to $Y.$

$b.$ $Z$ is homeomorphic to $W.$

$c.$ $Y$ is homeomorphic to $W.$

$d.$ $X$ is not homeomorphic to $W.$

According to me $X$ is NOT connected but $W$ is connected so answer is option $d?$ Am I right? Thank you .

Answer 1

Yes, d is true and a, b, c are false. More precisely, these four spaces are pairwise non-homeomorphic, "because":

• $Z$ is compact but $X,Y,W$ are not even bounded ($X$ is not closed either).
• $W$ is connected but (see below) $X,Y$ are not.
• $Y$ has four connected components, whereas $X$ has more (in fact: infinitely many). Let us detail this last point.
  • The four connected components of $Y$ are:
    • $Y_{+,+}:=\{(x,y,z)\in Y\mid x>0,y>0\},$
    • $Y_{+,-}:=\{(x,y,z)\in Y\mid x>0,y<0\},$
    • $Y_{-,+}:=\{(x,y,z)\in Y\mid x<0,y>0\},$
    • $Y_{-,-}:=\{(x,y,z)\in Y\mid x<0,y<0\}.$
  • The connected components of $X$ are $X_q:=\{(x,y,z)\in X\mid |x|+|y|+|z|=q\}$ for each non-negative rational $q.$