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 January 18th, 2012, 06:17 AM #1 Member   Joined: Jan 2012 Posts: 57 Thanks: 0 metric space Let (X,d) be a metric space. I have already shown that d'=d(x,y)/1+d(x,y) is also a metric on X. my question is: Show that a sequence x_n in X, n greater than or equal to 1, is a Cauchy sequence with respect to d if and only if it is a Cauchy sequence with respect to d'
 January 18th, 2012, 04:10 PM #2 Senior Member   Joined: Nov 2011 Posts: 100 Thanks: 0 Re: metric space Go ahead and write out what the definitions give you.. it shouldn't be too much more. Maybe I'll work through one direction, and you can do the other.. If a sequence is Cauchy, this tells you: Given $\varepsilon>0$, there exists $N\in\mathbb{N}$ such that for any $n\geq m\geq N$, we have $d(x_n,x_m)<\varepsilon$. Now, we consider the same sequence under the metric $d'$: $d'(x_n,x_m)=\frac{d(x_n,x_m)}{1+d(x_n,x_m)}\le q d(x_n,x_m)<\varepsilon$. Now that wasn't too bad, was it? Can you do the other direction?
 January 18th, 2012, 05:43 PM #3 Member   Joined: Jan 2012 Posts: 57 Thanks: 0 Re: metric space Thanks I got the other direction
 January 18th, 2012, 05:49 PM #4 Senior Member   Joined: Nov 2011 Posts: 100 Thanks: 0 Re: metric space Great!
 January 18th, 2012, 06:10 PM #5 Member   Joined: Jan 2012 Posts: 57 Thanks: 0 Re: metric space Are you any good with complete metric spaces

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