Discussion
Let \(X = \{(x, y) \in \mathbf{R}^{2} : x^{2} + y^{2} \lt 1\}\) \(\cup \{(x, y) \in \mathbf{R}^{2} : (x - 2)^{2} + y^{2} \lt 1\}\) and \(Y = \{(x, y) \in \mathbf{R}^{2} : x^{2} + y^{2} \lt 1\}\) U \(\{(x, y) \in \mathbf{R}^{2} : (x - 3)^{2} + y^{2} \lt 1\}\) and \(Z = \{(x, y) \in \mathbf{R}^{2} : x^{2} + y^{2} \lt 1, x \ne 0\}\) be subspaces of the Euclidean space \(\mathbf{R}^{2}\). Then :
Answer: A
The answer is X and Y are homeomorphic to Z.
Each set is a union of two disjoint open pieces, each homeomorphic to an open disc. X is two open unit discs centred at (0, 0) and (2, 0); they touch only at (1, 0), which belongs to neither, so they are disjoint. Y is two disjoint open discs centred at (0, 0) and (3, 0). Z is the open unit disc with the segment \(x = 0\) removed, leaving two open half-discs, and an open half-disc is homeomorphic to an open disc.
A homeomorphism can be built piece by piece, so all three spaces are homeomorphic. The distance between the pieces, or whether their closures touch, does not affect the topology of the subspaces themselves.
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