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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          The Binding of Isaac: Rebirth/Afterbirth

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          Please consider supporting The Cutting Room Floor on Patreon. Thanks for all your support! The Binding of Isaac: Rebirth/Afterbirth From The Cutting Room Floor < The Binding of Isaac: Rebirth Jump to: navigation, search This is a sub-page of The Binding of Isaac: Rebirth . Afterbirth is the first expansion pack for Rebirth and features a variety of improvements. Contents 1 Unused Graphics 1.1 Old Man Keeper 1.2 Skinless Hush 1.3 Miscellaneous Unused Hush Sprites 1.4 Generosity achievement 1.5 Afterbirth+ Room Editor 1.6 Bearded Ultra Greed 1.7 Early Ultra Greed 1.8 Unused Brownie Expression 1.9 Frail's Spinning Head 1.10 Mom's Dead Hand Alt 2 Unused Audio 2.1 Unused Dead Womb Layer 2.2 Unused Hush Battle Outro 2.3 The Stain Grunt Unused Graphics Old Man Keeper Image file: gfxuimain me
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