Discussion
Consider the following subsets of the Euclidean space \(\mathbb{R}^{4}\):
\[ \begin{aligned} S &= \{(x_{1}, x_{2}, x_{3}, x_{4}) \in \mathbb{R}^{4} : \\ &\qquad x_{1}^{2} + x_{2}^{2} + x_{3}^{2} - x_{4}^{2} = 0\}, \\ T &= \{(x_{1}, x_{2}, x_{3}, x_{4}) \in \mathbb{R}^{4} : \\ &\qquad x_{1}^{2} + x_{2}^{2} + x_{3}^{2} - x_{4}^{2} = 1\}, \text{ and} \\ U &= \{(x_{1}, x_{2}, x_{3}, x_{4}) \in \mathbb{R}^{4} : \\ &\qquad x_{1}^{2} + x_{2}^{2} + x_{3}^{2} - x_{4}^{2} = -1\}. \end{aligned} \]
Then, which one of the following is TRUE?
Answer: D
S and T are connected, but U is not connected.
Write \(r^{2} = x_{1}^{2} + x_{2}^{2} + x_{3}^{2}\).
- S: \(x_{4} = \pm r\), a double cone whose two halves meet at the origin, so it is connected (every point joins the origin along a straight segment in S).
- T: \(r^{2} = 1 + x_{4}^{2}\). For each value of \(x_{4}\) the slice is a 2-sphere of radius \(\sqrt{1 + x_{4}^{2}}\), so T is homeomorphic to \(S^{2} \times \mathbb{R}\), which is connected.
- U: \(x_{4}^{2} = 1 + r^{2} \ge 1\), so \(x_{4} \ge 1\) or \(x_{4} \le -1\). The two sheets are separated by the slab \(|x_{4}| \lt 1\), so U is not connected.
Every other option misjudges one of these three.
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