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

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 .

• $$Z$$ 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 $$q.$$