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Oct 2, 2022 at 10:58 vote accept an_ordinary_mathematician
Sep 30, 2022 at 10:36 comment added YCor (My previous comment justifies that $d'$ extends $d$ on $Y\times Y$.)
Sep 30, 2022 at 9:48 comment added YCor @MattF. Yes it does. If $x,x'\in Y$, then $D(x,x')\ge \delta(x,x')$ (first using $d\ge \delta$, then using the triangle inequality). Since $d\ge\delta$ it follows that $d'\ge\delta$ on $Y\times Y$. The other inequality is clear by choosing $y=x$, $y'=x'$.
Sep 28, 2022 at 8:22 history edited YCor CC BY-SA 4.0
fixed typo
Sep 27, 2022 at 14:14 history answered YCor CC BY-SA 4.0