Discussion
Let \(S^{1} = \{(x, y) \in \mathbf{R}^{2} \,/\, x^{2} + y^{2} = 1\}\) be the unit circle and
\[ D = \{(x, y) \in \mathbf{R}^{2} \,/\, |x| \le 1, |y| \le 1\} \]
Then :
Answer: B
The true statement is S1 is not homeomorphic to D.
D is the filled square, which is two-dimensional, while \(S^{1}\) is a curve. Removing two points from \(S^{1}\) disconnects it, but removing any two points from the filled square leaves it connected, so the two are not homeomorphic.
\(S^{1}\) is compact while R is not, so they are not homeomorphic. Removing one point leaves \(S^{1}\) connected, but removing an interior point disconnects any interval, so \(S^{1}\) is not homeomorphic to an interval. The square D is not homeomorphic to an interval either, by the same cut-point argument.
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