Do the metrics $d$ and $frac{d}{1+d}$ induce the same uniformity?












2












$begingroup$


Let $d_1$ and $d_2$ be two metrics on the same set $M$. Then $d_1$ and $d_2$ are called uniformly equivalent if the identity maps $i:(M,d_1)rightarrow(M,d_2)$ and $i^{-1}:(M,d_2)rightarrow(M,d_1)$ are uniformly continuous. Now this textbook gives the following exercise:




Given any metric space $(M,d)$, show that the metric $rho=frac{d}{1+d}$ is always uniformly equivalent to $d$[.]




My question is, is the result of the exercise correct? Because two metrics are uniformly equivalent if and only if they induce the same uniformity, and if two metrics induce the same uniformity then they have the same bounded sets. Yet all sets are bounded with respect to $frac{d}{1+d}$, whereas all sets need not be bounded with respect to $d$.



Where am I going wrong?






general-topology metric-spaces examples-counterexamples uniform-continuity uniform-spaces







share|cite|improve this question











$endgroup$












  • $begingroup$
    Presumably you need $i^{-1}$ to be uniformly continuous as well?
    $endgroup$
    – copper.hat
    Dec 23 '18 at 2:42
















2












$begingroup$


Let $d_1$ and $d_2$ be two metrics on the same set $M$. Then $d_1$ and $d_2$ are called uniformly equivalent if the identity maps $i:(M,d_1)rightarrow(M,d_2)$ and $i^{-1}:(M,d_2)rightarrow(M,d_1)$ are uniformly continuous. Now this textbook gives the following exercise:




Given any metric space $(M,d)$, show that the metric $rho=frac{d}{1+d}$ is always uniformly equivalent to $d$[.]




My question is, is the result of the exercise correct? Because two metrics are uniformly equivalent if and only if they induce the same uniformity, and if two metrics induce the same uniformity then they have the same bounded sets. Yet all sets are bounded with respect to $frac{d}{1+d}$, whereas all sets need not be bounded with respect to $d$.



Where am I going wrong?






general-topology metric-spaces examples-counterexamples uniform-continuity uniform-spaces







share|cite|improve this question











$endgroup$












  • $begingroup$
    Presumably you need $i^{-1}$ to be uniformly continuous as well?
    $endgroup$
    – copper.hat
    Dec 23 '18 at 2:42














2












2








2





$begingroup$


Let $d_1$ and $d_2$ be two metrics on the same set $M$. Then $d_1$ and $d_2$ are called uniformly equivalent if the identity maps $i:(M,d_1)rightarrow(M,d_2)$ and $i^{-1}:(M,d_2)rightarrow(M,d_1)$ are uniformly continuous. Now this textbook gives the following exercise:




Given any metric space $(M,d)$, show that the metric $rho=frac{d}{1+d}$ is always uniformly equivalent to $d$[.]




My question is, is the result of the exercise correct? Because two metrics are uniformly equivalent if and only if they induce the same uniformity, and if two metrics induce the same uniformity then they have the same bounded sets. Yet all sets are bounded with respect to $frac{d}{1+d}$, whereas all sets need not be bounded with respect to $d$.



Where am I going wrong?






general-topology metric-spaces examples-counterexamples uniform-continuity uniform-spaces







share|cite|improve this question











$endgroup$




Let $d_1$ and $d_2$ be two metrics on the same set $M$. Then $d_1$ and $d_2$ are called uniformly equivalent if the identity maps $i:(M,d_1)rightarrow(M,d_2)$ and $i^{-1}:(M,d_2)rightarrow(M,d_1)$ are uniformly continuous. Now this textbook gives the following exercise:




Given any metric space $(M,d)$, show that the metric $rho=frac{d}{1+d}$ is always uniformly equivalent to $d$[.]




My question is, is the result of the exercise correct? Because two metrics are uniformly equivalent if and only if they induce the same uniformity, and if two metrics induce the same uniformity then they have the same bounded sets. Yet all sets are bounded with respect to $frac{d}{1+d}$, whereas all sets need not be bounded with respect to $d$.



Where am I going wrong?






general-topology metric-spaces examples-counterexamples uniform-continuity uniform-spaces




general-topology metric-spaces examples-counterexamples uniform-continuity uniform-spaces






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 23 '18 at 4:27







Keshav Srinivasan

















asked Dec 23 '18 at 2:25









Keshav SrinivasanKeshav Srinivasan

2,33621444




2,33621444












  • $begingroup$
    Presumably you need $i^{-1}$ to be uniformly continuous as well?
    $endgroup$
    – copper.hat
    Dec 23 '18 at 2:42


















  • $begingroup$
    Presumably you need $i^{-1}$ to be uniformly continuous as well?
    $endgroup$
    – copper.hat
    Dec 23 '18 at 2:42
















$begingroup$
Presumably you need $i^{-1}$ to be uniformly continuous as well?
$endgroup$
– copper.hat
Dec 23 '18 at 2:42




$begingroup$
Presumably you need $i^{-1}$ to be uniformly continuous as well?
$endgroup$
– copper.hat
Dec 23 '18 at 2:42










2 Answers
2






active

oldest

votes


















3












$begingroup$

The textbook you linked distinguishes between $3$ types of equivalence:



•strong equivalence



•uniform equivalence &



•equivalence.



Evidently the condition that the identity map and its inverse be uniformly continuous corresponds to uniform equivalence. But strong equivalence is stronger, and is the one for which bounded sets are preserved.



As pointed out in the text, $rhole Csigma$ may fail under uniform equivalence. In particular, it may fail when $sigma =frac{rho}{1+rho}$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    But don't uniformly equivalent metrics induce the same uniformity? And don't metrics which induce the same uniformity have the same bounded sets?
    $endgroup$
    – Keshav Srinivasan
    Dec 23 '18 at 3:57










  • $begingroup$
    Apparently not.
    $endgroup$
    – Chris Custer
    Dec 23 '18 at 3:58










  • $begingroup$
    So which part is wrong? Is it not true that two metrics are uniformly equivalent if and only if they induce the same uniformity? Or is it not true that metrics that induce the same uniformity have the same bounded sets?
    $endgroup$
    – Keshav Srinivasan
    Dec 23 '18 at 3:59










  • $begingroup$
    I didn't see the "uniformity" terminology anywhere, though admittedly I only looked at a small portion.
    $endgroup$
    – Chris Custer
    Dec 23 '18 at 4:02










  • $begingroup$
    Uniformities aren't discussed in that book at all. But see here: en.wikipedia.org/wiki/Uniform_space
    $endgroup$
    – Keshav Srinivasan
    Dec 23 '18 at 4:08



















1












$begingroup$

Yes, the exercise is correct, and $mathcal{U}(d) = mathcal{U}(d')$ where $d'=frac{d}{1+d}$. In the latter uniformity all subsets of $X$ are $d'$-bounded but the $mathcal{U}(d')$-bounded sets are not always the same as the $d'$-bounded sets. $d$ and $d'$ induce the same uniform structure (and thus the same uniform-bounded and totally-bounded bornologies) but not the same metric bornologies.






share|cite|improve this answer











$endgroup$













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    2 Answers
    2






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    oldest

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    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    3












    $begingroup$

    The textbook you linked distinguishes between $3$ types of equivalence:



    •strong equivalence



    •uniform equivalence &



    •equivalence.



    Evidently the condition that the identity map and its inverse be uniformly continuous corresponds to uniform equivalence. But strong equivalence is stronger, and is the one for which bounded sets are preserved.



    As pointed out in the text, $rhole Csigma$ may fail under uniform equivalence. In particular, it may fail when $sigma =frac{rho}{1+rho}$.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      But don't uniformly equivalent metrics induce the same uniformity? And don't metrics which induce the same uniformity have the same bounded sets?
      $endgroup$
      – Keshav Srinivasan
      Dec 23 '18 at 3:57










    • $begingroup$
      Apparently not.
      $endgroup$
      – Chris Custer
      Dec 23 '18 at 3:58










    • $begingroup$
      So which part is wrong? Is it not true that two metrics are uniformly equivalent if and only if they induce the same uniformity? Or is it not true that metrics that induce the same uniformity have the same bounded sets?
      $endgroup$
      – Keshav Srinivasan
      Dec 23 '18 at 3:59










    • $begingroup$
      I didn't see the "uniformity" terminology anywhere, though admittedly I only looked at a small portion.
      $endgroup$
      – Chris Custer
      Dec 23 '18 at 4:02










    • $begingroup$
      Uniformities aren't discussed in that book at all. But see here: en.wikipedia.org/wiki/Uniform_space
      $endgroup$
      – Keshav Srinivasan
      Dec 23 '18 at 4:08
















    3












    $begingroup$

    The textbook you linked distinguishes between $3$ types of equivalence:



    •strong equivalence



    •uniform equivalence &



    •equivalence.



    Evidently the condition that the identity map and its inverse be uniformly continuous corresponds to uniform equivalence. But strong equivalence is stronger, and is the one for which bounded sets are preserved.



    As pointed out in the text, $rhole Csigma$ may fail under uniform equivalence. In particular, it may fail when $sigma =frac{rho}{1+rho}$.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      But don't uniformly equivalent metrics induce the same uniformity? And don't metrics which induce the same uniformity have the same bounded sets?
      $endgroup$
      – Keshav Srinivasan
      Dec 23 '18 at 3:57










    • $begingroup$
      Apparently not.
      $endgroup$
      – Chris Custer
      Dec 23 '18 at 3:58










    • $begingroup$
      So which part is wrong? Is it not true that two metrics are uniformly equivalent if and only if they induce the same uniformity? Or is it not true that metrics that induce the same uniformity have the same bounded sets?
      $endgroup$
      – Keshav Srinivasan
      Dec 23 '18 at 3:59










    • $begingroup$
      I didn't see the "uniformity" terminology anywhere, though admittedly I only looked at a small portion.
      $endgroup$
      – Chris Custer
      Dec 23 '18 at 4:02










    • $begingroup$
      Uniformities aren't discussed in that book at all. But see here: en.wikipedia.org/wiki/Uniform_space
      $endgroup$
      – Keshav Srinivasan
      Dec 23 '18 at 4:08














    3












    3








    3





    $begingroup$

    The textbook you linked distinguishes between $3$ types of equivalence:



    •strong equivalence



    •uniform equivalence &



    •equivalence.



    Evidently the condition that the identity map and its inverse be uniformly continuous corresponds to uniform equivalence. But strong equivalence is stronger, and is the one for which bounded sets are preserved.



    As pointed out in the text, $rhole Csigma$ may fail under uniform equivalence. In particular, it may fail when $sigma =frac{rho}{1+rho}$.






    share|cite|improve this answer











    $endgroup$



    The textbook you linked distinguishes between $3$ types of equivalence:



    •strong equivalence



    •uniform equivalence &



    •equivalence.



    Evidently the condition that the identity map and its inverse be uniformly continuous corresponds to uniform equivalence. But strong equivalence is stronger, and is the one for which bounded sets are preserved.



    As pointed out in the text, $rhole Csigma$ may fail under uniform equivalence. In particular, it may fail when $sigma =frac{rho}{1+rho}$.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Dec 23 '18 at 3:18

























    answered Dec 23 '18 at 3:03









    Chris CusterChris Custer

    13.6k3827




    13.6k3827












    • $begingroup$
      But don't uniformly equivalent metrics induce the same uniformity? And don't metrics which induce the same uniformity have the same bounded sets?
      $endgroup$
      – Keshav Srinivasan
      Dec 23 '18 at 3:57










    • $begingroup$
      Apparently not.
      $endgroup$
      – Chris Custer
      Dec 23 '18 at 3:58










    • $begingroup$
      So which part is wrong? Is it not true that two metrics are uniformly equivalent if and only if they induce the same uniformity? Or is it not true that metrics that induce the same uniformity have the same bounded sets?
      $endgroup$
      – Keshav Srinivasan
      Dec 23 '18 at 3:59










    • $begingroup$
      I didn't see the "uniformity" terminology anywhere, though admittedly I only looked at a small portion.
      $endgroup$
      – Chris Custer
      Dec 23 '18 at 4:02










    • $begingroup$
      Uniformities aren't discussed in that book at all. But see here: en.wikipedia.org/wiki/Uniform_space
      $endgroup$
      – Keshav Srinivasan
      Dec 23 '18 at 4:08


















    • $begingroup$
      But don't uniformly equivalent metrics induce the same uniformity? And don't metrics which induce the same uniformity have the same bounded sets?
      $endgroup$
      – Keshav Srinivasan
      Dec 23 '18 at 3:57










    • $begingroup$
      Apparently not.
      $endgroup$
      – Chris Custer
      Dec 23 '18 at 3:58










    • $begingroup$
      So which part is wrong? Is it not true that two metrics are uniformly equivalent if and only if they induce the same uniformity? Or is it not true that metrics that induce the same uniformity have the same bounded sets?
      $endgroup$
      – Keshav Srinivasan
      Dec 23 '18 at 3:59










    • $begingroup$
      I didn't see the "uniformity" terminology anywhere, though admittedly I only looked at a small portion.
      $endgroup$
      – Chris Custer
      Dec 23 '18 at 4:02










    • $begingroup$
      Uniformities aren't discussed in that book at all. But see here: en.wikipedia.org/wiki/Uniform_space
      $endgroup$
      – Keshav Srinivasan
      Dec 23 '18 at 4:08
















    $begingroup$
    But don't uniformly equivalent metrics induce the same uniformity? And don't metrics which induce the same uniformity have the same bounded sets?
    $endgroup$
    – Keshav Srinivasan
    Dec 23 '18 at 3:57




    $begingroup$
    But don't uniformly equivalent metrics induce the same uniformity? And don't metrics which induce the same uniformity have the same bounded sets?
    $endgroup$
    – Keshav Srinivasan
    Dec 23 '18 at 3:57












    $begingroup$
    Apparently not.
    $endgroup$
    – Chris Custer
    Dec 23 '18 at 3:58




    $begingroup$
    Apparently not.
    $endgroup$
    – Chris Custer
    Dec 23 '18 at 3:58












    $begingroup$
    So which part is wrong? Is it not true that two metrics are uniformly equivalent if and only if they induce the same uniformity? Or is it not true that metrics that induce the same uniformity have the same bounded sets?
    $endgroup$
    – Keshav Srinivasan
    Dec 23 '18 at 3:59




    $begingroup$
    So which part is wrong? Is it not true that two metrics are uniformly equivalent if and only if they induce the same uniformity? Or is it not true that metrics that induce the same uniformity have the same bounded sets?
    $endgroup$
    – Keshav Srinivasan
    Dec 23 '18 at 3:59












    $begingroup$
    I didn't see the "uniformity" terminology anywhere, though admittedly I only looked at a small portion.
    $endgroup$
    – Chris Custer
    Dec 23 '18 at 4:02




    $begingroup$
    I didn't see the "uniformity" terminology anywhere, though admittedly I only looked at a small portion.
    $endgroup$
    – Chris Custer
    Dec 23 '18 at 4:02












    $begingroup$
    Uniformities aren't discussed in that book at all. But see here: en.wikipedia.org/wiki/Uniform_space
    $endgroup$
    – Keshav Srinivasan
    Dec 23 '18 at 4:08




    $begingroup$
    Uniformities aren't discussed in that book at all. But see here: en.wikipedia.org/wiki/Uniform_space
    $endgroup$
    – Keshav Srinivasan
    Dec 23 '18 at 4:08











    1












    $begingroup$

    Yes, the exercise is correct, and $mathcal{U}(d) = mathcal{U}(d')$ where $d'=frac{d}{1+d}$. In the latter uniformity all subsets of $X$ are $d'$-bounded but the $mathcal{U}(d')$-bounded sets are not always the same as the $d'$-bounded sets. $d$ and $d'$ induce the same uniform structure (and thus the same uniform-bounded and totally-bounded bornologies) but not the same metric bornologies.






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      Yes, the exercise is correct, and $mathcal{U}(d) = mathcal{U}(d')$ where $d'=frac{d}{1+d}$. In the latter uniformity all subsets of $X$ are $d'$-bounded but the $mathcal{U}(d')$-bounded sets are not always the same as the $d'$-bounded sets. $d$ and $d'$ induce the same uniform structure (and thus the same uniform-bounded and totally-bounded bornologies) but not the same metric bornologies.






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        Yes, the exercise is correct, and $mathcal{U}(d) = mathcal{U}(d')$ where $d'=frac{d}{1+d}$. In the latter uniformity all subsets of $X$ are $d'$-bounded but the $mathcal{U}(d')$-bounded sets are not always the same as the $d'$-bounded sets. $d$ and $d'$ induce the same uniform structure (and thus the same uniform-bounded and totally-bounded bornologies) but not the same metric bornologies.






        share|cite|improve this answer











        $endgroup$



        Yes, the exercise is correct, and $mathcal{U}(d) = mathcal{U}(d')$ where $d'=frac{d}{1+d}$. In the latter uniformity all subsets of $X$ are $d'$-bounded but the $mathcal{U}(d')$-bounded sets are not always the same as the $d'$-bounded sets. $d$ and $d'$ induce the same uniform structure (and thus the same uniform-bounded and totally-bounded bornologies) but not the same metric bornologies.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 23 '18 at 6:24

























        answered Dec 23 '18 at 6:14









        Henno BrandsmaHenno Brandsma

        110k347117




        110k347117






























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