Prove topological equivalence between two distances












0












$begingroup$


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









$endgroup$

















    0












    $begingroup$


    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









    $endgroup$















      0












      0








      0


      1



      $begingroup$


      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









      $endgroup$




      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













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 5 at 16:30









      Pedro AlvarèsPedro Alvarès

      946




      946






















          2 Answers
          2






          active

          oldest

          votes


















          1












          $begingroup$

          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











          $endgroup$









          • 1




            $begingroup$
            If we choose $r'=(r-|x-y|)^{frac{1}{3}}$ it will be done right ?
            $endgroup$
            – Pedro Alvarès
            Jan 5 at 18:07










          • $begingroup$
            Exactly! Nicely done.
            $endgroup$
            – Cameron Buie
            Jan 5 at 18:12










          • $begingroup$
            You said 0<r'<1 , it should be r'>1
            $endgroup$
            – Pedro Alvarès
            Jan 5 at 20:19










          • $begingroup$
            Oh, dear! I did things backward. I'll edit accordingly.
            $endgroup$
            – Cameron Buie
            Jan 5 at 20:40










          • $begingroup$
            So now it would be $r'=(r-|x-y|^{frac{1}{3}})^3$ right ? Thanks btw
            $endgroup$
            – Pedro Alvarès
            Jan 5 at 21:34





















          0












          $begingroup$

          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









          $endgroup$













          • $begingroup$
            Can't we prove that $exists alpha ,beta $ such that $alpha d(x,y)leq h(x,y)leq beta d(x,y)$
            $endgroup$
            – Pedro Alvarès
            Jan 5 at 16:48










          • $begingroup$
            For $|x| < 1$, $|x| le |x|^{1/3}$. You cannot get a constant for the reverse inequality.
            $endgroup$
            – ncmathsadist
            Jan 5 at 18:21












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          2 Answers
          2






          active

          oldest

          votes








          2 Answers
          2






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          1












          $begingroup$

          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











          $endgroup$









          • 1




            $begingroup$
            If we choose $r'=(r-|x-y|)^{frac{1}{3}}$ it will be done right ?
            $endgroup$
            – Pedro Alvarès
            Jan 5 at 18:07










          • $begingroup$
            Exactly! Nicely done.
            $endgroup$
            – Cameron Buie
            Jan 5 at 18:12










          • $begingroup$
            You said 0<r'<1 , it should be r'>1
            $endgroup$
            – Pedro Alvarès
            Jan 5 at 20:19










          • $begingroup$
            Oh, dear! I did things backward. I'll edit accordingly.
            $endgroup$
            – Cameron Buie
            Jan 5 at 20:40










          • $begingroup$
            So now it would be $r'=(r-|x-y|^{frac{1}{3}})^3$ right ? Thanks btw
            $endgroup$
            – Pedro Alvarès
            Jan 5 at 21:34


















          1












          $begingroup$

          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











          $endgroup$









          • 1




            $begingroup$
            If we choose $r'=(r-|x-y|)^{frac{1}{3}}$ it will be done right ?
            $endgroup$
            – Pedro Alvarès
            Jan 5 at 18:07










          • $begingroup$
            Exactly! Nicely done.
            $endgroup$
            – Cameron Buie
            Jan 5 at 18:12










          • $begingroup$
            You said 0<r'<1 , it should be r'>1
            $endgroup$
            – Pedro Alvarès
            Jan 5 at 20:19










          • $begingroup$
            Oh, dear! I did things backward. I'll edit accordingly.
            $endgroup$
            – Cameron Buie
            Jan 5 at 20:40










          • $begingroup$
            So now it would be $r'=(r-|x-y|^{frac{1}{3}})^3$ right ? Thanks btw
            $endgroup$
            – Pedro Alvarès
            Jan 5 at 21:34
















          1












          1








          1





          $begingroup$

          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











          $endgroup$



          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 Jan 5 at 20:42

























          answered Jan 5 at 17:04









          Cameron BuieCameron Buie

          86.7k773161




          86.7k773161








          • 1




            $begingroup$
            If we choose $r'=(r-|x-y|)^{frac{1}{3}}$ it will be done right ?
            $endgroup$
            – Pedro Alvarès
            Jan 5 at 18:07










          • $begingroup$
            Exactly! Nicely done.
            $endgroup$
            – Cameron Buie
            Jan 5 at 18:12










          • $begingroup$
            You said 0<r'<1 , it should be r'>1
            $endgroup$
            – Pedro Alvarès
            Jan 5 at 20:19










          • $begingroup$
            Oh, dear! I did things backward. I'll edit accordingly.
            $endgroup$
            – Cameron Buie
            Jan 5 at 20:40










          • $begingroup$
            So now it would be $r'=(r-|x-y|^{frac{1}{3}})^3$ right ? Thanks btw
            $endgroup$
            – Pedro Alvarès
            Jan 5 at 21:34
















          • 1




            $begingroup$
            If we choose $r'=(r-|x-y|)^{frac{1}{3}}$ it will be done right ?
            $endgroup$
            – Pedro Alvarès
            Jan 5 at 18:07










          • $begingroup$
            Exactly! Nicely done.
            $endgroup$
            – Cameron Buie
            Jan 5 at 18:12










          • $begingroup$
            You said 0<r'<1 , it should be r'>1
            $endgroup$
            – Pedro Alvarès
            Jan 5 at 20:19










          • $begingroup$
            Oh, dear! I did things backward. I'll edit accordingly.
            $endgroup$
            – Cameron Buie
            Jan 5 at 20:40










          • $begingroup$
            So now it would be $r'=(r-|x-y|^{frac{1}{3}})^3$ right ? Thanks btw
            $endgroup$
            – Pedro Alvarès
            Jan 5 at 21:34










          1




          1




          $begingroup$
          If we choose $r'=(r-|x-y|)^{frac{1}{3}}$ it will be done right ?
          $endgroup$
          – Pedro Alvarès
          Jan 5 at 18:07




          $begingroup$
          If we choose $r'=(r-|x-y|)^{frac{1}{3}}$ it will be done right ?
          $endgroup$
          – Pedro Alvarès
          Jan 5 at 18:07












          $begingroup$
          Exactly! Nicely done.
          $endgroup$
          – Cameron Buie
          Jan 5 at 18:12




          $begingroup$
          Exactly! Nicely done.
          $endgroup$
          – Cameron Buie
          Jan 5 at 18:12












          $begingroup$
          You said 0<r'<1 , it should be r'>1
          $endgroup$
          – Pedro Alvarès
          Jan 5 at 20:19




          $begingroup$
          You said 0<r'<1 , it should be r'>1
          $endgroup$
          – Pedro Alvarès
          Jan 5 at 20:19












          $begingroup$
          Oh, dear! I did things backward. I'll edit accordingly.
          $endgroup$
          – Cameron Buie
          Jan 5 at 20:40




          $begingroup$
          Oh, dear! I did things backward. I'll edit accordingly.
          $endgroup$
          – Cameron Buie
          Jan 5 at 20:40












          $begingroup$
          So now it would be $r'=(r-|x-y|^{frac{1}{3}})^3$ right ? Thanks btw
          $endgroup$
          – Pedro Alvarès
          Jan 5 at 21:34






          $begingroup$
          So now it would be $r'=(r-|x-y|^{frac{1}{3}})^3$ right ? Thanks btw
          $endgroup$
          – Pedro Alvarès
          Jan 5 at 21:34













          0












          $begingroup$

          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









          $endgroup$













          • $begingroup$
            Can't we prove that $exists alpha ,beta $ such that $alpha d(x,y)leq h(x,y)leq beta d(x,y)$
            $endgroup$
            – Pedro Alvarès
            Jan 5 at 16:48










          • $begingroup$
            For $|x| < 1$, $|x| le |x|^{1/3}$. You cannot get a constant for the reverse inequality.
            $endgroup$
            – ncmathsadist
            Jan 5 at 18:21
















          0












          $begingroup$

          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









          $endgroup$













          • $begingroup$
            Can't we prove that $exists alpha ,beta $ such that $alpha d(x,y)leq h(x,y)leq beta d(x,y)$
            $endgroup$
            – Pedro Alvarès
            Jan 5 at 16:48










          • $begingroup$
            For $|x| < 1$, $|x| le |x|^{1/3}$. You cannot get a constant for the reverse inequality.
            $endgroup$
            – ncmathsadist
            Jan 5 at 18:21














          0












          0








          0





          $begingroup$

          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









          $endgroup$



          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 Jan 5 at 16:35









          ncmathsadistncmathsadist

          43.1k261103




          43.1k261103












          • $begingroup$
            Can't we prove that $exists alpha ,beta $ such that $alpha d(x,y)leq h(x,y)leq beta d(x,y)$
            $endgroup$
            – Pedro Alvarès
            Jan 5 at 16:48










          • $begingroup$
            For $|x| < 1$, $|x| le |x|^{1/3}$. You cannot get a constant for the reverse inequality.
            $endgroup$
            – ncmathsadist
            Jan 5 at 18:21


















          • $begingroup$
            Can't we prove that $exists alpha ,beta $ such that $alpha d(x,y)leq h(x,y)leq beta d(x,y)$
            $endgroup$
            – Pedro Alvarès
            Jan 5 at 16:48










          • $begingroup$
            For $|x| < 1$, $|x| le |x|^{1/3}$. You cannot get a constant for the reverse inequality.
            $endgroup$
            – ncmathsadist
            Jan 5 at 18:21
















          $begingroup$
          Can't we prove that $exists alpha ,beta $ such that $alpha d(x,y)leq h(x,y)leq beta d(x,y)$
          $endgroup$
          – Pedro Alvarès
          Jan 5 at 16:48




          $begingroup$
          Can't we prove that $exists alpha ,beta $ such that $alpha d(x,y)leq h(x,y)leq beta d(x,y)$
          $endgroup$
          – Pedro Alvarès
          Jan 5 at 16:48












          $begingroup$
          For $|x| < 1$, $|x| le |x|^{1/3}$. You cannot get a constant for the reverse inequality.
          $endgroup$
          – ncmathsadist
          Jan 5 at 18:21




          $begingroup$
          For $|x| < 1$, $|x| le |x|^{1/3}$. You cannot get a constant for the reverse inequality.
          $endgroup$
          – ncmathsadist
          Jan 5 at 18:21


















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