Discussion
Let \(\mathbb{R}^{3}\) be a topological space with the usual topology and \(\mathbb{Q}\) denote the set of rational numbers. Define the subspaces X, Y, Z and W of \(\mathbb{R}^{3}\) as follows:
\(X = \{(x,y,z) \in \mathbb{R}^{3} : |x|+|y|+|z| \in \mathbb{Q}\}\)
\(Y = \{(x,y,z) \in \mathbb{R}^{3} : xyz = 1\}\)
\(Z = \{(x,y,z) \in \mathbb{R}^{3} : x^{2}+y^{2}+z^{2} = 1\}\)
\(W = \{(x,y,z) \in \mathbb{R}^{3} : xyz = 0\}\)
Which of the following statements is correct?
Answer: D
The correct statement is X is NOT homeomorphic to W.
W, the union of the three coordinate planes, is connected. X is the origin together with countably many octahedral surfaces \(|x|+|y|+|z| = q\), one for each positive rational q. It is not connected: for any irrational r, the sets where \(|x|+|y|+|z| \lt r\) and \(\gt r\) split X into two disjoint non-empty open pieces. Connectedness is preserved by homeomorphisms, so X and W are not homeomorphic.
The other statements fail for similar reasons: Z, the unit sphere, is compact while W is not; Y (xyz = 1) has four connected components while W is connected; and X has infinitely many components (indeed it is not locally connected near the origin) while Y has four.
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