Q50·Section II (Mathematics)·50 of 190(−1,1)(-1,1)(−1,1) is completely metrizable even though its usual metric is not completeFor x,y∈Rx,y\in\mathbb{R}x,y∈R, let d(x,y)=∣x−y∣d(x,y)=|x-y|d(x,y)=∣x−y∣. Then: OptionsA(−1,1)(-1,1)(−1,1) is complete in dddB(−1,1)(-1,1)(−1,1) is not complete in ddd, but there is a complete metric on (−1,1)(-1,1)(−1,1) inducing the same topology on (−1,1)(-1,1)(−1,1) as of dddCevery metric on (−1,1)(-1,1)(−1,1) inducing the same topology as that of ddd is completeDno metric on (−1,1)(-1,1)(−1,1) inducing the same topology as that of ddd is completePick 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