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Timeline for Quotient of metric spaces

Current License: CC BY-SA 3.0

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Sep 4, 2020 at 10:00 comment added Robert Frost @burtonpeterj for any metric $d$, $\lambda d$ is a metric too, right, for any arbitrary positive real $\lambda$?
Feb 5, 2015 at 20:58 comment added burtonpeterj We can find such a $\lambda$ only if the quotient map is Lipschitz. So why can we assume the quotient map is Lipschitz?
Feb 5, 2015 at 11:33 comment added user2173168 For example, consider a compact space $X$, $X$ has a metric $d$, $d$ therefore has an upper bound. $X/\sim$ is metrizable, it can be endowed with a metric $d'$. It is trivial to prove, there is a $\lambda>0$, such that $\lambda d'$ is smaller than $d$.
Feb 5, 2015 at 11:27 comment added user2173168 because when the metric $d$ has an upper bound, you can take $\lambda d$ to make the metric arbitrarily small.
Feb 4, 2015 at 17:16 comment added burtonpeterj Why is this trivially true when the metric has an upper bound?
Feb 4, 2015 at 7:57 history edited user2173168 CC BY-SA 3.0
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Feb 4, 2015 at 7:42 history answered user2173168 CC BY-SA 3.0