Q50·Section II (Mathematics)·50 of 180The circle is not simply connected; the square disk isLet S1={(x,y)∈R2/x2+y2=1}S^1=\{(x,y)\in\mathbf R^2/x^2+y^2=1\}S1={(x,y)∈R2/x2+y2=1} be the unit circle and D={(x,y)∈R2/∣x∣≤1,∣y∣≤1}D=\{(x,y)\in\mathbf R^2/|x|\le1,|y|\le1\}D={(x,y)∈R2/∣x∣≤1,∣y∣≤1}. Then: OptionsAS1S^1S1 is homeomorphic to R\mathbf RRBS1S^1S1 is not homeomorphic to DCS1S^1S1 is homeomorphic to an interval in R\mathbf RRDD is homeomorphic to an interval in R\mathbf RRPick an option to check your answer.Show answerShow solution Report a problem with this question← Q49All questionsQ51 →Keyboard: A–D answer · S solution · ←/→ previous/next