Discussion
Let
\(X = \{(x, y) \in \mathbf{R}^{2} : 0 \le |x| = |y| \le 1\}\)
\(Y = \{(x, y) \in \mathbf{R}^{2} : |x| + |y| = 1\}\) and
\(Z = \{(x, y) \in \mathbf{R}^{2} : x^{2} + y^{2} = 1\}\)
be subspaces of the Euclidean space \(\mathbf{R}^{2}\). Then :
Answer: C
Y is homeomorphic to Z, but not to X.
Y, the square \(|x| + |y| = 1\), is a simple closed curve; radial projection \(p \mapsto p/|p|\) is a homeomorphism from it onto the circle Z.
X is the set where \(|x| = |y| \le 1\): the two diagonals of the square \([-1, 1]^{2}\), a cross meeting at the origin. Removing the origin splits X into four pieces, while removing any one point from a circle leaves it connected. A homeomorphism preserves such cut-point behaviour, so X is homeomorphic to neither Y nor Z.
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