Approve that if there is a lower bound for each $A subseteq R $ then there is $infA$












0












$begingroup$


I must approve that if there is a lower bound for each $ Asubseteq R $ then there is $infA$. ($A neq emptyset$)



I'm sure that there is a related post for this but I would like to validate my solution to better understand the topic of $sup$ and $inf$.



My solution:



Let us define a set A which is not empty set and there is a lower bound. Let us also define a set B such that



$B = { x in R : x$ is lower bound of A $}$



$B$ is not empty because there is a $m leqslant x $ for all $xin A$.



Is $infA =m $?



Two cases:





  1. Let us consider that $ m neq infA$ then $exists e > 0$ such that
    $forall x in A$, $m+e leqslant x$.



    So, $m+e in B$ and $m+e>m$.



    We wonder if there is $supB$. If there is $supB$ then $supB = infA$



    Let us assume that there isn't $supB$. Then there is $e > 0$ for
    each $x>0$ so that



    $x leqslant M - e$ where $ x leqslant M $ for each $x in B$



    Let us define $M = m + e $, then $x leqslant m$, which is
    impossible because



    $m+e >m$. Consequently, $ x > M -e $.



    As a result, $supB=M=infA$



  2. If $infA=m$ then there is $infA$



From (1) and (2), there is always $infA$



Could you validate my solution ?



EDIT:




  1. Can we assume that $supB = infA$ ?

  2. Can we select the $M = m + e$ ?

  3. Introducing the above two cases, can we assume that there is always $infA$ ?






calculus







share|cite|improve this question











$endgroup$












  • $begingroup$
    Actually, $B$ might be empty. You should use $x in mathbb{R}$ instead.
    $endgroup$
    – lzralbu
    Jan 27 at 14:45










  • $begingroup$
    Thanks, you are right. I will edit it.
    $endgroup$
    – Dimitris Dimitriadis
    Jan 27 at 15:41
















0












$begingroup$


I must approve that if there is a lower bound for each $ Asubseteq R $ then there is $infA$. ($A neq emptyset$)



I'm sure that there is a related post for this but I would like to validate my solution to better understand the topic of $sup$ and $inf$.



My solution:



Let us define a set A which is not empty set and there is a lower bound. Let us also define a set B such that



$B = { x in R : x$ is lower bound of A $}$



$B$ is not empty because there is a $m leqslant x $ for all $xin A$.



Is $infA =m $?



Two cases:





  1. Let us consider that $ m neq infA$ then $exists e > 0$ such that
    $forall x in A$, $m+e leqslant x$.



    So, $m+e in B$ and $m+e>m$.



    We wonder if there is $supB$. If there is $supB$ then $supB = infA$



    Let us assume that there isn't $supB$. Then there is $e > 0$ for
    each $x>0$ so that



    $x leqslant M - e$ where $ x leqslant M $ for each $x in B$



    Let us define $M = m + e $, then $x leqslant m$, which is
    impossible because



    $m+e >m$. Consequently, $ x > M -e $.



    As a result, $supB=M=infA$



  2. If $infA=m$ then there is $infA$



From (1) and (2), there is always $infA$



Could you validate my solution ?



EDIT:




  1. Can we assume that $supB = infA$ ?

  2. Can we select the $M = m + e$ ?

  3. Introducing the above two cases, can we assume that there is always $infA$ ?






calculus







share|cite|improve this question











$endgroup$












  • $begingroup$
    Actually, $B$ might be empty. You should use $x in mathbb{R}$ instead.
    $endgroup$
    – lzralbu
    Jan 27 at 14:45










  • $begingroup$
    Thanks, you are right. I will edit it.
    $endgroup$
    – Dimitris Dimitriadis
    Jan 27 at 15:41














0












0








0





$begingroup$


I must approve that if there is a lower bound for each $ Asubseteq R $ then there is $infA$. ($A neq emptyset$)



I'm sure that there is a related post for this but I would like to validate my solution to better understand the topic of $sup$ and $inf$.



My solution:



Let us define a set A which is not empty set and there is a lower bound. Let us also define a set B such that



$B = { x in R : x$ is lower bound of A $}$



$B$ is not empty because there is a $m leqslant x $ for all $xin A$.



Is $infA =m $?



Two cases:





  1. Let us consider that $ m neq infA$ then $exists e > 0$ such that
    $forall x in A$, $m+e leqslant x$.



    So, $m+e in B$ and $m+e>m$.



    We wonder if there is $supB$. If there is $supB$ then $supB = infA$



    Let us assume that there isn't $supB$. Then there is $e > 0$ for
    each $x>0$ so that



    $x leqslant M - e$ where $ x leqslant M $ for each $x in B$



    Let us define $M = m + e $, then $x leqslant m$, which is
    impossible because



    $m+e >m$. Consequently, $ x > M -e $.



    As a result, $supB=M=infA$



  2. If $infA=m$ then there is $infA$



From (1) and (2), there is always $infA$



Could you validate my solution ?



EDIT:




  1. Can we assume that $supB = infA$ ?

  2. Can we select the $M = m + e$ ?

  3. Introducing the above two cases, can we assume that there is always $infA$ ?






calculus







share|cite|improve this question











$endgroup$




I must approve that if there is a lower bound for each $ Asubseteq R $ then there is $infA$. ($A neq emptyset$)



I'm sure that there is a related post for this but I would like to validate my solution to better understand the topic of $sup$ and $inf$.



My solution:



Let us define a set A which is not empty set and there is a lower bound. Let us also define a set B such that



$B = { x in R : x$ is lower bound of A $}$



$B$ is not empty because there is a $m leqslant x $ for all $xin A$.



Is $infA =m $?



Two cases:





  1. Let us consider that $ m neq infA$ then $exists e > 0$ such that
    $forall x in A$, $m+e leqslant x$.



    So, $m+e in B$ and $m+e>m$.



    We wonder if there is $supB$. If there is $supB$ then $supB = infA$



    Let us assume that there isn't $supB$. Then there is $e > 0$ for
    each $x>0$ so that



    $x leqslant M - e$ where $ x leqslant M $ for each $x in B$



    Let us define $M = m + e $, then $x leqslant m$, which is
    impossible because



    $m+e >m$. Consequently, $ x > M -e $.



    As a result, $supB=M=infA$



  2. If $infA=m$ then there is $infA$



From (1) and (2), there is always $infA$



Could you validate my solution ?



EDIT:




  1. Can we assume that $supB = infA$ ?

  2. Can we select the $M = m + e$ ?

  3. Introducing the above two cases, can we assume that there is always $infA$ ?






calculus




calculus






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 27 at 15:42







Dimitris Dimitriadis

















asked Jan 27 at 10:28









Dimitris DimitriadisDimitris Dimitriadis

548




548












  • $begingroup$
    Actually, $B$ might be empty. You should use $x in mathbb{R}$ instead.
    $endgroup$
    – lzralbu
    Jan 27 at 14:45










  • $begingroup$
    Thanks, you are right. I will edit it.
    $endgroup$
    – Dimitris Dimitriadis
    Jan 27 at 15:41


















  • $begingroup$
    Actually, $B$ might be empty. You should use $x in mathbb{R}$ instead.
    $endgroup$
    – lzralbu
    Jan 27 at 14:45










  • $begingroup$
    Thanks, you are right. I will edit it.
    $endgroup$
    – Dimitris Dimitriadis
    Jan 27 at 15:41
















$begingroup$
Actually, $B$ might be empty. You should use $x in mathbb{R}$ instead.
$endgroup$
– lzralbu
Jan 27 at 14:45




$begingroup$
Actually, $B$ might be empty. You should use $x in mathbb{R}$ instead.
$endgroup$
– lzralbu
Jan 27 at 14:45












$begingroup$
Thanks, you are right. I will edit it.
$endgroup$
– Dimitris Dimitriadis
Jan 27 at 15:41




$begingroup$
Thanks, you are right. I will edit it.
$endgroup$
– Dimitris Dimitriadis
Jan 27 at 15:41










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