Let $\mathbb{S}^1=\{x\in\mathbb{R}^2:\ \|x\|_2=1\}$.
My question is: Is it possible to define a incomplete metric on $\mathbb{S}^1$, i.e. a metric such that $\mathbb{S}^1$ is not complete.
Thank you.
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1Any metric, or should it induce the standard topology? – Daniel Fischer Jun 28 '13 at 12:47
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Any metric @DanielFischer – Tomás Jun 28 '13 at 12:51
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1Then take any bijection with an incomplete metric space of cardinality $\mathfrak{c}$. For example $[0,,2\pi)$. – Daniel Fischer Jun 28 '13 at 12:53
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Let $f \colon (0,1) \to \mathbb S^1$ be a bijection. Define $d(x,y) = |f^{-1}(x) - f^{-1}(y)|$, then $f$ is an isometry, hence $(\mathbb S^1, d)$ is not complete.
martini
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If you want it to induce the standard topology, no, you can't. It is compact, and therefore, any metric inducing it will be complete (and totally bounded).
Cameron Buie
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