Prove topological equivalence between two distances












0














Let $d(x,y)=|x-y|$ and $h(x,y)=|x-y|^{frac{1}{3}}$ be two distances on $mathbb{R}$.



I want to prove that they are topologically equivalent



I supposed that $f(x)=x$ and $g(x)=x^{frac{1}{3}}$



so we can say that



$f(x)leq g(x)$ $forall xin [0,1]$ and $f(x)geq g(x)$ $forall xgeq 1$



Will this information lead me somewhere ? If yes, how would I proceed ?






general-topology







share|cite|improve this question







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Any Bany is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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    0














    Let $d(x,y)=|x-y|$ and $h(x,y)=|x-y|^{frac{1}{3}}$ be two distances on $mathbb{R}$.



    I want to prove that they are topologically equivalent



    I supposed that $f(x)=x$ and $g(x)=x^{frac{1}{3}}$



    so we can say that



    $f(x)leq g(x)$ $forall xin [0,1]$ and $f(x)geq g(x)$ $forall xgeq 1$



    Will this information lead me somewhere ? If yes, how would I proceed ?






    general-topology







    share|cite|improve this question







    New contributor




    Any Bany is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.























      0












      0








      0


      1





      Let $d(x,y)=|x-y|$ and $h(x,y)=|x-y|^{frac{1}{3}}$ be two distances on $mathbb{R}$.



      I want to prove that they are topologically equivalent



      I supposed that $f(x)=x$ and $g(x)=x^{frac{1}{3}}$



      so we can say that



      $f(x)leq g(x)$ $forall xin [0,1]$ and $f(x)geq g(x)$ $forall xgeq 1$



      Will this information lead me somewhere ? If yes, how would I proceed ?






      general-topology







      share|cite|improve this question







      New contributor




      Any Bany is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.











      Let $d(x,y)=|x-y|$ and $h(x,y)=|x-y|^{frac{1}{3}}$ be two distances on $mathbb{R}$.



      I want to prove that they are topologically equivalent



      I supposed that $f(x)=x$ and $g(x)=x^{frac{1}{3}}$



      so we can say that



      $f(x)leq g(x)$ $forall xin [0,1]$ and $f(x)geq g(x)$ $forall xgeq 1$



      Will this information lead me somewhere ? If yes, how would I proceed ?






      general-topology




      general-topology






      share|cite|improve this question







      New contributor




      Any Bany is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.











      share|cite|improve this question







      New contributor




      Any Bany is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.









      share|cite|improve this question




      share|cite|improve this question






      New contributor




      Any Bany is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.









      asked 2 days ago









      Any BanyAny Bany

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      New contributor





      Any Bany is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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          2 Answers
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          active

          oldest

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          1














          Here, I use the term "$d$-ball" to indicate a set of the form $bigl{yinBbb R:d(x,y)<rbigr}$--specifically, I'd call that the $d$-ball centered at $x$ with radius $r$--and similarly use the term "$h$-ball" to indicate a set of the form $bigl{yinBbb R:h(x,y)<rbigr}.$



          To show topological equivalence, we want to show the following:




          1. Given any $d$-ball $B$ and any $yin B,$ there exists some $h$-ball $B'$ such that $yin B'subseteq B.$

          2. Given any $h$-ball $B$ and any $yin B,$ there exists some $d$-ball $B'$ such that $yin B'subseteq B.$


          Your first observation makes one direction fairly easy. Pick an arbitrary $d$-ball $B,$ with center $x$ and radius $r,$ and take any $yin B.$ Find a radius $r'$ such that the $d$-ball centered at $y$ with radius $r_0$ is a subset of $B.$ Call this $d$-ball $B_0.$ Now, using your first observation, show that by choosing a radius $r'$ such that $0<r'<1$ and $r'le r_0,$ then the $h$-ball centered at $y$ of radius $r'$ is then a subset of $B_0$ containing $y,$ so you've proved 1.



          To prove 2, we proceed similarly, but have more work to do. Start with an arbitrary $h$-ball $B,$ with center $x$ and radius $r,$ and take any $yin B.$ Your goal is to find some $r'>0$ such that whenever $d(y,z)<r',$ we have $h(x,z)<r.$ In other words, we need $|x-z|^{frac13}<r$ whenever $|y-z|<r'.$



          By triangle inequality, we have $$|x-z|^{frac13}le |x-y|^{frac13}+|y-z|^{frac13}$$ right away, and recalling our choice of $y,$ we have that $|x-y|^{frac13}<r,$ so that $0<r-|x-y|^{frac13}.$ Consequently, if we can ensure that $|y-z|^{frac13}<r-|x-y|^{frac13},$ we'll have $|x-z|^{frac13}<r$ as desired. Can you take it from there?






          share|cite|improve this answer



















          • 1




            If we choose $r'=(r-|x-y|)^{frac{1}{3}}$ it will be done right ?
            – Any Bany
            2 days ago










          • Exactly! Nicely done.
            – Cameron Buie
            2 days ago










          • You said 0<r'<1 , it should be r'>1
            – Any Bany
            2 days ago










          • Oh, dear! I did things backward. I'll edit accordingly.
            – Cameron Buie
            2 days ago










          • So now it would be $r'=(r-|x-y|^{frac{1}{3}})^3$ right ? Thanks btw
            – Any Bany
            2 days ago





















          0














          Observe that for any sequences $x_n$ and $y_n$,
          $$|x_n - y_n| to 0 iff |x_n - y_n|^{1/3} to 0$$
          as $ntoinfty$.






          share|cite|improve this answer





















          • Can't we prove that $exists alpha ,beta $ such that $alpha d(x,y)leq h(x,y)leq beta d(x,y)$
            – Any Bany
            2 days ago










          • For $|x| < 1$, $|x| le |x|^{1/3}$. You cannot get a constant for the reverse inequality.
            – ncmathsadist
            2 days ago











          Your Answer





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          2 Answers
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          active

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          2 Answers
          2






          active

          oldest

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          active

          oldest

          votes






          active

          oldest

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          1














          Here, I use the term "$d$-ball" to indicate a set of the form $bigl{yinBbb R:d(x,y)<rbigr}$--specifically, I'd call that the $d$-ball centered at $x$ with radius $r$--and similarly use the term "$h$-ball" to indicate a set of the form $bigl{yinBbb R:h(x,y)<rbigr}.$



          To show topological equivalence, we want to show the following:




          1. Given any $d$-ball $B$ and any $yin B,$ there exists some $h$-ball $B'$ such that $yin B'subseteq B.$

          2. Given any $h$-ball $B$ and any $yin B,$ there exists some $d$-ball $B'$ such that $yin B'subseteq B.$


          Your first observation makes one direction fairly easy. Pick an arbitrary $d$-ball $B,$ with center $x$ and radius $r,$ and take any $yin B.$ Find a radius $r'$ such that the $d$-ball centered at $y$ with radius $r_0$ is a subset of $B.$ Call this $d$-ball $B_0.$ Now, using your first observation, show that by choosing a radius $r'$ such that $0<r'<1$ and $r'le r_0,$ then the $h$-ball centered at $y$ of radius $r'$ is then a subset of $B_0$ containing $y,$ so you've proved 1.



          To prove 2, we proceed similarly, but have more work to do. Start with an arbitrary $h$-ball $B,$ with center $x$ and radius $r,$ and take any $yin B.$ Your goal is to find some $r'>0$ such that whenever $d(y,z)<r',$ we have $h(x,z)<r.$ In other words, we need $|x-z|^{frac13}<r$ whenever $|y-z|<r'.$



          By triangle inequality, we have $$|x-z|^{frac13}le |x-y|^{frac13}+|y-z|^{frac13}$$ right away, and recalling our choice of $y,$ we have that $|x-y|^{frac13}<r,$ so that $0<r-|x-y|^{frac13}.$ Consequently, if we can ensure that $|y-z|^{frac13}<r-|x-y|^{frac13},$ we'll have $|x-z|^{frac13}<r$ as desired. Can you take it from there?






          share|cite|improve this answer



















          • 1




            If we choose $r'=(r-|x-y|)^{frac{1}{3}}$ it will be done right ?
            – Any Bany
            2 days ago










          • Exactly! Nicely done.
            – Cameron Buie
            2 days ago










          • You said 0<r'<1 , it should be r'>1
            – Any Bany
            2 days ago










          • Oh, dear! I did things backward. I'll edit accordingly.
            – Cameron Buie
            2 days ago










          • So now it would be $r'=(r-|x-y|^{frac{1}{3}})^3$ right ? Thanks btw
            – Any Bany
            2 days ago


















          1














          Here, I use the term "$d$-ball" to indicate a set of the form $bigl{yinBbb R:d(x,y)<rbigr}$--specifically, I'd call that the $d$-ball centered at $x$ with radius $r$--and similarly use the term "$h$-ball" to indicate a set of the form $bigl{yinBbb R:h(x,y)<rbigr}.$



          To show topological equivalence, we want to show the following:




          1. Given any $d$-ball $B$ and any $yin B,$ there exists some $h$-ball $B'$ such that $yin B'subseteq B.$

          2. Given any $h$-ball $B$ and any $yin B,$ there exists some $d$-ball $B'$ such that $yin B'subseteq B.$


          Your first observation makes one direction fairly easy. Pick an arbitrary $d$-ball $B,$ with center $x$ and radius $r,$ and take any $yin B.$ Find a radius $r'$ such that the $d$-ball centered at $y$ with radius $r_0$ is a subset of $B.$ Call this $d$-ball $B_0.$ Now, using your first observation, show that by choosing a radius $r'$ such that $0<r'<1$ and $r'le r_0,$ then the $h$-ball centered at $y$ of radius $r'$ is then a subset of $B_0$ containing $y,$ so you've proved 1.



          To prove 2, we proceed similarly, but have more work to do. Start with an arbitrary $h$-ball $B,$ with center $x$ and radius $r,$ and take any $yin B.$ Your goal is to find some $r'>0$ such that whenever $d(y,z)<r',$ we have $h(x,z)<r.$ In other words, we need $|x-z|^{frac13}<r$ whenever $|y-z|<r'.$



          By triangle inequality, we have $$|x-z|^{frac13}le |x-y|^{frac13}+|y-z|^{frac13}$$ right away, and recalling our choice of $y,$ we have that $|x-y|^{frac13}<r,$ so that $0<r-|x-y|^{frac13}.$ Consequently, if we can ensure that $|y-z|^{frac13}<r-|x-y|^{frac13},$ we'll have $|x-z|^{frac13}<r$ as desired. Can you take it from there?






          share|cite|improve this answer



















          • 1




            If we choose $r'=(r-|x-y|)^{frac{1}{3}}$ it will be done right ?
            – Any Bany
            2 days ago










          • Exactly! Nicely done.
            – Cameron Buie
            2 days ago










          • You said 0<r'<1 , it should be r'>1
            – Any Bany
            2 days ago










          • Oh, dear! I did things backward. I'll edit accordingly.
            – Cameron Buie
            2 days ago










          • So now it would be $r'=(r-|x-y|^{frac{1}{3}})^3$ right ? Thanks btw
            – Any Bany
            2 days ago
















          1












          1








          1






          Here, I use the term "$d$-ball" to indicate a set of the form $bigl{yinBbb R:d(x,y)<rbigr}$--specifically, I'd call that the $d$-ball centered at $x$ with radius $r$--and similarly use the term "$h$-ball" to indicate a set of the form $bigl{yinBbb R:h(x,y)<rbigr}.$



          To show topological equivalence, we want to show the following:




          1. Given any $d$-ball $B$ and any $yin B,$ there exists some $h$-ball $B'$ such that $yin B'subseteq B.$

          2. Given any $h$-ball $B$ and any $yin B,$ there exists some $d$-ball $B'$ such that $yin B'subseteq B.$


          Your first observation makes one direction fairly easy. Pick an arbitrary $d$-ball $B,$ with center $x$ and radius $r,$ and take any $yin B.$ Find a radius $r'$ such that the $d$-ball centered at $y$ with radius $r_0$ is a subset of $B.$ Call this $d$-ball $B_0.$ Now, using your first observation, show that by choosing a radius $r'$ such that $0<r'<1$ and $r'le r_0,$ then the $h$-ball centered at $y$ of radius $r'$ is then a subset of $B_0$ containing $y,$ so you've proved 1.



          To prove 2, we proceed similarly, but have more work to do. Start with an arbitrary $h$-ball $B,$ with center $x$ and radius $r,$ and take any $yin B.$ Your goal is to find some $r'>0$ such that whenever $d(y,z)<r',$ we have $h(x,z)<r.$ In other words, we need $|x-z|^{frac13}<r$ whenever $|y-z|<r'.$



          By triangle inequality, we have $$|x-z|^{frac13}le |x-y|^{frac13}+|y-z|^{frac13}$$ right away, and recalling our choice of $y,$ we have that $|x-y|^{frac13}<r,$ so that $0<r-|x-y|^{frac13}.$ Consequently, if we can ensure that $|y-z|^{frac13}<r-|x-y|^{frac13},$ we'll have $|x-z|^{frac13}<r$ as desired. Can you take it from there?






          share|cite|improve this answer














          Here, I use the term "$d$-ball" to indicate a set of the form $bigl{yinBbb R:d(x,y)<rbigr}$--specifically, I'd call that the $d$-ball centered at $x$ with radius $r$--and similarly use the term "$h$-ball" to indicate a set of the form $bigl{yinBbb R:h(x,y)<rbigr}.$



          To show topological equivalence, we want to show the following:




          1. Given any $d$-ball $B$ and any $yin B,$ there exists some $h$-ball $B'$ such that $yin B'subseteq B.$

          2. Given any $h$-ball $B$ and any $yin B,$ there exists some $d$-ball $B'$ such that $yin B'subseteq B.$


          Your first observation makes one direction fairly easy. Pick an arbitrary $d$-ball $B,$ with center $x$ and radius $r,$ and take any $yin B.$ Find a radius $r'$ such that the $d$-ball centered at $y$ with radius $r_0$ is a subset of $B.$ Call this $d$-ball $B_0.$ Now, using your first observation, show that by choosing a radius $r'$ such that $0<r'<1$ and $r'le r_0,$ then the $h$-ball centered at $y$ of radius $r'$ is then a subset of $B_0$ containing $y,$ so you've proved 1.



          To prove 2, we proceed similarly, but have more work to do. Start with an arbitrary $h$-ball $B,$ with center $x$ and radius $r,$ and take any $yin B.$ Your goal is to find some $r'>0$ such that whenever $d(y,z)<r',$ we have $h(x,z)<r.$ In other words, we need $|x-z|^{frac13}<r$ whenever $|y-z|<r'.$



          By triangle inequality, we have $$|x-z|^{frac13}le |x-y|^{frac13}+|y-z|^{frac13}$$ right away, and recalling our choice of $y,$ we have that $|x-y|^{frac13}<r,$ so that $0<r-|x-y|^{frac13}.$ Consequently, if we can ensure that $|y-z|^{frac13}<r-|x-y|^{frac13},$ we'll have $|x-z|^{frac13}<r$ as desired. Can you take it from there?







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited 2 days ago

























          answered 2 days ago









          Cameron BuieCameron Buie

          85.1k771155




          85.1k771155








          • 1




            If we choose $r'=(r-|x-y|)^{frac{1}{3}}$ it will be done right ?
            – Any Bany
            2 days ago










          • Exactly! Nicely done.
            – Cameron Buie
            2 days ago










          • You said 0<r'<1 , it should be r'>1
            – Any Bany
            2 days ago










          • Oh, dear! I did things backward. I'll edit accordingly.
            – Cameron Buie
            2 days ago










          • So now it would be $r'=(r-|x-y|^{frac{1}{3}})^3$ right ? Thanks btw
            – Any Bany
            2 days ago
















          • 1




            If we choose $r'=(r-|x-y|)^{frac{1}{3}}$ it will be done right ?
            – Any Bany
            2 days ago










          • Exactly! Nicely done.
            – Cameron Buie
            2 days ago










          • You said 0<r'<1 , it should be r'>1
            – Any Bany
            2 days ago










          • Oh, dear! I did things backward. I'll edit accordingly.
            – Cameron Buie
            2 days ago










          • So now it would be $r'=(r-|x-y|^{frac{1}{3}})^3$ right ? Thanks btw
            – Any Bany
            2 days ago










          1




          1




          If we choose $r'=(r-|x-y|)^{frac{1}{3}}$ it will be done right ?
          – Any Bany
          2 days ago




          If we choose $r'=(r-|x-y|)^{frac{1}{3}}$ it will be done right ?
          – Any Bany
          2 days ago












          Exactly! Nicely done.
          – Cameron Buie
          2 days ago




          Exactly! Nicely done.
          – Cameron Buie
          2 days ago












          You said 0<r'<1 , it should be r'>1
          – Any Bany
          2 days ago




          You said 0<r'<1 , it should be r'>1
          – Any Bany
          2 days ago












          Oh, dear! I did things backward. I'll edit accordingly.
          – Cameron Buie
          2 days ago




          Oh, dear! I did things backward. I'll edit accordingly.
          – Cameron Buie
          2 days ago












          So now it would be $r'=(r-|x-y|^{frac{1}{3}})^3$ right ? Thanks btw
          – Any Bany
          2 days ago






          So now it would be $r'=(r-|x-y|^{frac{1}{3}})^3$ right ? Thanks btw
          – Any Bany
          2 days ago













          0














          Observe that for any sequences $x_n$ and $y_n$,
          $$|x_n - y_n| to 0 iff |x_n - y_n|^{1/3} to 0$$
          as $ntoinfty$.






          share|cite|improve this answer





















          • Can't we prove that $exists alpha ,beta $ such that $alpha d(x,y)leq h(x,y)leq beta d(x,y)$
            – Any Bany
            2 days ago










          • For $|x| < 1$, $|x| le |x|^{1/3}$. You cannot get a constant for the reverse inequality.
            – ncmathsadist
            2 days ago
















          0














          Observe that for any sequences $x_n$ and $y_n$,
          $$|x_n - y_n| to 0 iff |x_n - y_n|^{1/3} to 0$$
          as $ntoinfty$.






          share|cite|improve this answer





















          • Can't we prove that $exists alpha ,beta $ such that $alpha d(x,y)leq h(x,y)leq beta d(x,y)$
            – Any Bany
            2 days ago










          • For $|x| < 1$, $|x| le |x|^{1/3}$. You cannot get a constant for the reverse inequality.
            – ncmathsadist
            2 days ago














          0












          0








          0






          Observe that for any sequences $x_n$ and $y_n$,
          $$|x_n - y_n| to 0 iff |x_n - y_n|^{1/3} to 0$$
          as $ntoinfty$.






          share|cite|improve this answer












          Observe that for any sequences $x_n$ and $y_n$,
          $$|x_n - y_n| to 0 iff |x_n - y_n|^{1/3} to 0$$
          as $ntoinfty$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 2 days ago









          ncmathsadistncmathsadist

          42.4k259102




          42.4k259102












          • Can't we prove that $exists alpha ,beta $ such that $alpha d(x,y)leq h(x,y)leq beta d(x,y)$
            – Any Bany
            2 days ago










          • For $|x| < 1$, $|x| le |x|^{1/3}$. You cannot get a constant for the reverse inequality.
            – ncmathsadist
            2 days ago


















          • Can't we prove that $exists alpha ,beta $ such that $alpha d(x,y)leq h(x,y)leq beta d(x,y)$
            – Any Bany
            2 days ago










          • For $|x| < 1$, $|x| le |x|^{1/3}$. You cannot get a constant for the reverse inequality.
            – ncmathsadist
            2 days ago
















          Can't we prove that $exists alpha ,beta $ such that $alpha d(x,y)leq h(x,y)leq beta d(x,y)$
          – Any Bany
          2 days ago




          Can't we prove that $exists alpha ,beta $ such that $alpha d(x,y)leq h(x,y)leq beta d(x,y)$
          – Any Bany
          2 days ago












          For $|x| < 1$, $|x| le |x|^{1/3}$. You cannot get a constant for the reverse inequality.
          – ncmathsadist
          2 days ago




          For $|x| < 1$, $|x| le |x|^{1/3}$. You cannot get a constant for the reverse inequality.
          – ncmathsadist
          2 days ago










          Any Bany is a new contributor. Be nice, and check out our Code of Conduct.










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          Mario Kart Wii

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          Please consider supporting The Cutting Room Floor on Patreon. Thanks for all your support! Mario Kart Wii From The Cutting Room Floor Jump to: navigation, search Mario Kart Wii Developer : Nintendo Publisher : Nintendo Platform : Wii Released in JP : April 10, 2008 Released in US : April 27, 2008 Released in EU : April 11, 2008 Released in AU : April 24, 2008 Released in KR : April 30, 2009 This game has unused animations. This game has unused areas. This game has unused code. This game has unused objects. This game has unused modes / minigames. This game has unused graphics. This game has unused models. This game has unused items. This game has unused music. This game has regional differences. This game has a notes page This game has a prerelease article Mario Kart Wii is the sixth game in the series (not counting Arcade editions), which intro
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          What does “Dominus providebit” mean?

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          5 Does "Dominus providebit" mean "The Lord will provide" or "The Lord provides"? I once had a bit of an argument with a guy who studied Latin over that. It's an inscription that appears on the rim of some old Swiss coins that I have. meaning latin-to-english-translation grammar-identification share | improve this question edited Jan 18 at 15:27 V2Blast 105 4 asked Jan 17 at 22:49 Donald K. Donald K. 26 1
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          Antonio Litta Visconti Arese

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          Antonio Litta Visconti Arese I duca Litta, VII marchese di Gambolò In carica 1797 – 1820 Predecessore Pompeo Litta Visconti Arese Successore Alfonso Litta Visconti Arese Trattamento Sua grazia Altri titoli Duca Litta Conte del Regno napoleonico d'Italia Patrizio milanese Don Nascita Milano, 2 maggio 1748 Morte Milano, 14 agosto 1820 Dinastia Litta Visconti Arese Padre Pompeo Litta Visconti Arese Madre Maria Elisabetta Visconti Borromeo Consorte Barbara Barbiano di Belgioioso Religione cattolicesimo Antonio Litta Visconti Arese , I duca Litta, VII marchese di Gambolò (Milano, 2 maggio 1748 – Milano, 14 agosto 1820), è stato un nobile e politico italiano. Composizione "Il pastore svizzero" di Pietro Morlacchi, con dedica Indice 1 Biografia 2 Onorificenze 3 Araldica 4 Albero genealogico 5 Note 6 Altri progetti Biografia | Antonio Litta Visco
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