Showing that $ U_x=left{a^p:pinmathbb{Q},pxright} $ are contiguous classes for $ a>1 $












0












$begingroup$


I'm trying to show that $ (U_x,V_x) $ is a pair of contiguous classes.



Proof. Let $ U_x=left{a^p:text{$pinmathbb{Q}$ and $p<x$}right} $ and $ V_x=left{a^q:text{$pinmathbb{Q}$ and $q>x$}right} $; obviously is $ a^pleqq a^q $ (because $ amapsto a^rho $ is increasing for rational $ rho $). Let $ xi<eta $ be two separators for $ U_x $ and $ V_x $: thus we have for all $ a^p $ and $ a^q $ the chain of inequalities $ a^qleqqxi $ and $ etaleqq a^q $. My textbook says that from there we can derive $ a^q/a^pgeqqeta/xi $ and therefore $ a^{q-p}>eta/xi>1 $. [...]



I'm okay with the $ >1 $ part, but I don't get from where the author derived the strict inequality between the first two members.



Secondly, assuming what claimed (that $ a^{1-p}>xi/eta>1 $) is true, the author states something like




Every positive rational number $ rho $ can be expressed as $ rho=q-p $, where $ p<x<q $, for every real number $ x $ (because of $ mathbb{Q} $ is dense in $ mathbb{R} $). We can now note that $ infleft{a^rho:text{$ rhoinmathbb{Q} $ and $ rho>0 $}right} $ equals $ 1 $, and derive a contradiction.




Could someone explain me this apparently tedious passage? How are the density of $ mathbb{Q} $ and the $ inf{E}=infleft{a^rhodotsright}=1 $ statements used?






real-analysis exponential-function proof-explanation







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    0












    $begingroup$


    I'm trying to show that $ (U_x,V_x) $ is a pair of contiguous classes.



    Proof. Let $ U_x=left{a^p:text{$pinmathbb{Q}$ and $p<x$}right} $ and $ V_x=left{a^q:text{$pinmathbb{Q}$ and $q>x$}right} $; obviously is $ a^pleqq a^q $ (because $ amapsto a^rho $ is increasing for rational $ rho $). Let $ xi<eta $ be two separators for $ U_x $ and $ V_x $: thus we have for all $ a^p $ and $ a^q $ the chain of inequalities $ a^qleqqxi $ and $ etaleqq a^q $. My textbook says that from there we can derive $ a^q/a^pgeqqeta/xi $ and therefore $ a^{q-p}>eta/xi>1 $. [...]



    I'm okay with the $ >1 $ part, but I don't get from where the author derived the strict inequality between the first two members.



    Secondly, assuming what claimed (that $ a^{1-p}>xi/eta>1 $) is true, the author states something like




    Every positive rational number $ rho $ can be expressed as $ rho=q-p $, where $ p<x<q $, for every real number $ x $ (because of $ mathbb{Q} $ is dense in $ mathbb{R} $). We can now note that $ infleft{a^rho:text{$ rhoinmathbb{Q} $ and $ rho>0 $}right} $ equals $ 1 $, and derive a contradiction.




    Could someone explain me this apparently tedious passage? How are the density of $ mathbb{Q} $ and the $ inf{E}=infleft{a^rhodotsright}=1 $ statements used?






    real-analysis exponential-function proof-explanation







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      I'm trying to show that $ (U_x,V_x) $ is a pair of contiguous classes.



      Proof. Let $ U_x=left{a^p:text{$pinmathbb{Q}$ and $p<x$}right} $ and $ V_x=left{a^q:text{$pinmathbb{Q}$ and $q>x$}right} $; obviously is $ a^pleqq a^q $ (because $ amapsto a^rho $ is increasing for rational $ rho $). Let $ xi<eta $ be two separators for $ U_x $ and $ V_x $: thus we have for all $ a^p $ and $ a^q $ the chain of inequalities $ a^qleqqxi $ and $ etaleqq a^q $. My textbook says that from there we can derive $ a^q/a^pgeqqeta/xi $ and therefore $ a^{q-p}>eta/xi>1 $. [...]



      I'm okay with the $ >1 $ part, but I don't get from where the author derived the strict inequality between the first two members.



      Secondly, assuming what claimed (that $ a^{1-p}>xi/eta>1 $) is true, the author states something like




      Every positive rational number $ rho $ can be expressed as $ rho=q-p $, where $ p<x<q $, for every real number $ x $ (because of $ mathbb{Q} $ is dense in $ mathbb{R} $). We can now note that $ infleft{a^rho:text{$ rhoinmathbb{Q} $ and $ rho>0 $}right} $ equals $ 1 $, and derive a contradiction.




      Could someone explain me this apparently tedious passage? How are the density of $ mathbb{Q} $ and the $ inf{E}=infleft{a^rhodotsright}=1 $ statements used?






      real-analysis exponential-function proof-explanation







      share|cite|improve this question











      $endgroup$




      I'm trying to show that $ (U_x,V_x) $ is a pair of contiguous classes.



      Proof. Let $ U_x=left{a^p:text{$pinmathbb{Q}$ and $p<x$}right} $ and $ V_x=left{a^q:text{$pinmathbb{Q}$ and $q>x$}right} $; obviously is $ a^pleqq a^q $ (because $ amapsto a^rho $ is increasing for rational $ rho $). Let $ xi<eta $ be two separators for $ U_x $ and $ V_x $: thus we have for all $ a^p $ and $ a^q $ the chain of inequalities $ a^qleqqxi $ and $ etaleqq a^q $. My textbook says that from there we can derive $ a^q/a^pgeqqeta/xi $ and therefore $ a^{q-p}>eta/xi>1 $. [...]



      I'm okay with the $ >1 $ part, but I don't get from where the author derived the strict inequality between the first two members.



      Secondly, assuming what claimed (that $ a^{1-p}>xi/eta>1 $) is true, the author states something like




      Every positive rational number $ rho $ can be expressed as $ rho=q-p $, where $ p<x<q $, for every real number $ x $ (because of $ mathbb{Q} $ is dense in $ mathbb{R} $). We can now note that $ infleft{a^rho:text{$ rhoinmathbb{Q} $ and $ rho>0 $}right} $ equals $ 1 $, and derive a contradiction.




      Could someone explain me this apparently tedious passage? How are the density of $ mathbb{Q} $ and the $ inf{E}=infleft{a^rhodotsright}=1 $ statements used?






      real-analysis exponential-function proof-explanation




      real-analysis exponential-function proof-explanation






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      edited Dec 30 '18 at 17:21







      marco21

















      asked Dec 30 '18 at 14:56









      marco21marco21

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          $begingroup$

          1) Given that $xi<eta$ are two separators for $U_x$ and $V_x$, you can always find two new sepatators $hat{xi}, hat{eta}$ such that $xi < hat{xi} < hat{eta} < eta$. Now, $ a^p leqq xi < hat{xi} < hat{eta} < eta leqq a^q$. Then, $a^{q-p}>hat{eta}/ hat{xi}$ follows.



          2) Given a positive rational number $rho$ and $x in mathbb{R}$, consider the open interval $] x, x+ rho [$. Then, you can find a rational number $q in ] x, x+ rho [$ because $mathbb{Q}$ is dense in $mathbb{R}$. Set $p:= q -rho$, then $rho = q-p$.






          share|cite|improve this answer











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          • $begingroup$
            Thank you for the answer, the point 2) is now clear. I'm assuming that there are two separators for $ U_x $ and $ V_x $ to derive a contradiction: then $ a^pleqqxi<etaleqq a^q $, with "non necessarily strict" inequality. From this follows that $ a^{q-p}geqqeta/xi $. I'm looking for how to derive that this inequality is in fact strict.
            $endgroup$
            – marco21
            Dec 30 '18 at 18:21










          • $begingroup$
            Sorry, misunderstood the question at first, I assumed the fact you were trying to prove. Hope my new edit helps.
            $endgroup$
            – JAskgaard
            Dec 30 '18 at 18:33












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          $begingroup$

          1) Given that $xi<eta$ are two separators for $U_x$ and $V_x$, you can always find two new sepatators $hat{xi}, hat{eta}$ such that $xi < hat{xi} < hat{eta} < eta$. Now, $ a^p leqq xi < hat{xi} < hat{eta} < eta leqq a^q$. Then, $a^{q-p}>hat{eta}/ hat{xi}$ follows.



          2) Given a positive rational number $rho$ and $x in mathbb{R}$, consider the open interval $] x, x+ rho [$. Then, you can find a rational number $q in ] x, x+ rho [$ because $mathbb{Q}$ is dense in $mathbb{R}$. Set $p:= q -rho$, then $rho = q-p$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thank you for the answer, the point 2) is now clear. I'm assuming that there are two separators for $ U_x $ and $ V_x $ to derive a contradiction: then $ a^pleqqxi<etaleqq a^q $, with "non necessarily strict" inequality. From this follows that $ a^{q-p}geqqeta/xi $. I'm looking for how to derive that this inequality is in fact strict.
            $endgroup$
            – marco21
            Dec 30 '18 at 18:21










          • $begingroup$
            Sorry, misunderstood the question at first, I assumed the fact you were trying to prove. Hope my new edit helps.
            $endgroup$
            – JAskgaard
            Dec 30 '18 at 18:33
















          2












          $begingroup$

          1) Given that $xi<eta$ are two separators for $U_x$ and $V_x$, you can always find two new sepatators $hat{xi}, hat{eta}$ such that $xi < hat{xi} < hat{eta} < eta$. Now, $ a^p leqq xi < hat{xi} < hat{eta} < eta leqq a^q$. Then, $a^{q-p}>hat{eta}/ hat{xi}$ follows.



          2) Given a positive rational number $rho$ and $x in mathbb{R}$, consider the open interval $] x, x+ rho [$. Then, you can find a rational number $q in ] x, x+ rho [$ because $mathbb{Q}$ is dense in $mathbb{R}$. Set $p:= q -rho$, then $rho = q-p$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thank you for the answer, the point 2) is now clear. I'm assuming that there are two separators for $ U_x $ and $ V_x $ to derive a contradiction: then $ a^pleqqxi<etaleqq a^q $, with "non necessarily strict" inequality. From this follows that $ a^{q-p}geqqeta/xi $. I'm looking for how to derive that this inequality is in fact strict.
            $endgroup$
            – marco21
            Dec 30 '18 at 18:21










          • $begingroup$
            Sorry, misunderstood the question at first, I assumed the fact you were trying to prove. Hope my new edit helps.
            $endgroup$
            – JAskgaard
            Dec 30 '18 at 18:33














          2












          2








          2





          $begingroup$

          1) Given that $xi<eta$ are two separators for $U_x$ and $V_x$, you can always find two new sepatators $hat{xi}, hat{eta}$ such that $xi < hat{xi} < hat{eta} < eta$. Now, $ a^p leqq xi < hat{xi} < hat{eta} < eta leqq a^q$. Then, $a^{q-p}>hat{eta}/ hat{xi}$ follows.



          2) Given a positive rational number $rho$ and $x in mathbb{R}$, consider the open interval $] x, x+ rho [$. Then, you can find a rational number $q in ] x, x+ rho [$ because $mathbb{Q}$ is dense in $mathbb{R}$. Set $p:= q -rho$, then $rho = q-p$.






          share|cite|improve this answer











          $endgroup$



          1) Given that $xi<eta$ are two separators for $U_x$ and $V_x$, you can always find two new sepatators $hat{xi}, hat{eta}$ such that $xi < hat{xi} < hat{eta} < eta$. Now, $ a^p leqq xi < hat{xi} < hat{eta} < eta leqq a^q$. Then, $a^{q-p}>hat{eta}/ hat{xi}$ follows.



          2) Given a positive rational number $rho$ and $x in mathbb{R}$, consider the open interval $] x, x+ rho [$. Then, you can find a rational number $q in ] x, x+ rho [$ because $mathbb{Q}$ is dense in $mathbb{R}$. Set $p:= q -rho$, then $rho = q-p$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 30 '18 at 18:30

























          answered Dec 30 '18 at 18:06









          JAskgaardJAskgaard

          1467




          1467












          • $begingroup$
            Thank you for the answer, the point 2) is now clear. I'm assuming that there are two separators for $ U_x $ and $ V_x $ to derive a contradiction: then $ a^pleqqxi<etaleqq a^q $, with "non necessarily strict" inequality. From this follows that $ a^{q-p}geqqeta/xi $. I'm looking for how to derive that this inequality is in fact strict.
            $endgroup$
            – marco21
            Dec 30 '18 at 18:21










          • $begingroup$
            Sorry, misunderstood the question at first, I assumed the fact you were trying to prove. Hope my new edit helps.
            $endgroup$
            – JAskgaard
            Dec 30 '18 at 18:33


















          • $begingroup$
            Thank you for the answer, the point 2) is now clear. I'm assuming that there are two separators for $ U_x $ and $ V_x $ to derive a contradiction: then $ a^pleqqxi<etaleqq a^q $, with "non necessarily strict" inequality. From this follows that $ a^{q-p}geqqeta/xi $. I'm looking for how to derive that this inequality is in fact strict.
            $endgroup$
            – marco21
            Dec 30 '18 at 18:21










          • $begingroup$
            Sorry, misunderstood the question at first, I assumed the fact you were trying to prove. Hope my new edit helps.
            $endgroup$
            – JAskgaard
            Dec 30 '18 at 18:33
















          $begingroup$
          Thank you for the answer, the point 2) is now clear. I'm assuming that there are two separators for $ U_x $ and $ V_x $ to derive a contradiction: then $ a^pleqqxi<etaleqq a^q $, with "non necessarily strict" inequality. From this follows that $ a^{q-p}geqqeta/xi $. I'm looking for how to derive that this inequality is in fact strict.
          $endgroup$
          – marco21
          Dec 30 '18 at 18:21




          $begingroup$
          Thank you for the answer, the point 2) is now clear. I'm assuming that there are two separators for $ U_x $ and $ V_x $ to derive a contradiction: then $ a^pleqqxi<etaleqq a^q $, with "non necessarily strict" inequality. From this follows that $ a^{q-p}geqqeta/xi $. I'm looking for how to derive that this inequality is in fact strict.
          $endgroup$
          – marco21
          Dec 30 '18 at 18:21












          $begingroup$
          Sorry, misunderstood the question at first, I assumed the fact you were trying to prove. Hope my new edit helps.
          $endgroup$
          – JAskgaard
          Dec 30 '18 at 18:33




          $begingroup$
          Sorry, misunderstood the question at first, I assumed the fact you were trying to prove. Hope my new edit helps.
          $endgroup$
          – JAskgaard
          Dec 30 '18 at 18:33


















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