Suppose $(a_n) rightarrow a.$ If $a_n geq 0$ for all $n$ then $a geq 0$. Prove this result.












0












$begingroup$


I have decided to assume that $a<0$ and let $epsilon=-a>0$. Then I want to come to a contradiction.



$|a_n-a|<-a$.



Since we have $a_n>a$ from our assumption,



$a_n-a<-a$



so $a_n<0$ which is a contradiction.






sequences-and-series proof-writing







share|cite|improve this question











$endgroup$












  • $begingroup$
    What if $a_n<a$?
    $endgroup$
    – José Carlos Santos
    Oct 21 '18 at 16:11










  • $begingroup$
    Then we have $a_n>2a$
    $endgroup$
    – m.bazza
    Oct 21 '18 at 16:13










  • $begingroup$
    But this isn't a contradiction as this gives us $a_ngeq 0$
    $endgroup$
    – m.bazza
    Oct 21 '18 at 16:14










  • $begingroup$
    But we know that $a_n-ageq 0$ from our assumption, why bother with cases
    $endgroup$
    – Jakobian
    Oct 21 '18 at 16:15










  • $begingroup$
    Oh of course haha didn't realise
    $endgroup$
    – m.bazza
    Oct 21 '18 at 16:16
















0












$begingroup$


I have decided to assume that $a<0$ and let $epsilon=-a>0$. Then I want to come to a contradiction.



$|a_n-a|<-a$.



Since we have $a_n>a$ from our assumption,



$a_n-a<-a$



so $a_n<0$ which is a contradiction.






sequences-and-series proof-writing







share|cite|improve this question











$endgroup$












  • $begingroup$
    What if $a_n<a$?
    $endgroup$
    – José Carlos Santos
    Oct 21 '18 at 16:11










  • $begingroup$
    Then we have $a_n>2a$
    $endgroup$
    – m.bazza
    Oct 21 '18 at 16:13










  • $begingroup$
    But this isn't a contradiction as this gives us $a_ngeq 0$
    $endgroup$
    – m.bazza
    Oct 21 '18 at 16:14










  • $begingroup$
    But we know that $a_n-ageq 0$ from our assumption, why bother with cases
    $endgroup$
    – Jakobian
    Oct 21 '18 at 16:15










  • $begingroup$
    Oh of course haha didn't realise
    $endgroup$
    – m.bazza
    Oct 21 '18 at 16:16














0












0








0





$begingroup$


I have decided to assume that $a<0$ and let $epsilon=-a>0$. Then I want to come to a contradiction.



$|a_n-a|<-a$.



Since we have $a_n>a$ from our assumption,



$a_n-a<-a$



so $a_n<0$ which is a contradiction.






sequences-and-series proof-writing







share|cite|improve this question











$endgroup$




I have decided to assume that $a<0$ and let $epsilon=-a>0$. Then I want to come to a contradiction.



$|a_n-a|<-a$.



Since we have $a_n>a$ from our assumption,



$a_n-a<-a$



so $a_n<0$ which is a contradiction.






sequences-and-series proof-writing




sequences-and-series proof-writing






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Oct 21 '18 at 16:17







m.bazza

















asked Oct 21 '18 at 16:07









m.bazzam.bazza

827




827












  • $begingroup$
    What if $a_n<a$?
    $endgroup$
    – José Carlos Santos
    Oct 21 '18 at 16:11










  • $begingroup$
    Then we have $a_n>2a$
    $endgroup$
    – m.bazza
    Oct 21 '18 at 16:13










  • $begingroup$
    But this isn't a contradiction as this gives us $a_ngeq 0$
    $endgroup$
    – m.bazza
    Oct 21 '18 at 16:14










  • $begingroup$
    But we know that $a_n-ageq 0$ from our assumption, why bother with cases
    $endgroup$
    – Jakobian
    Oct 21 '18 at 16:15










  • $begingroup$
    Oh of course haha didn't realise
    $endgroup$
    – m.bazza
    Oct 21 '18 at 16:16


















  • $begingroup$
    What if $a_n<a$?
    $endgroup$
    – José Carlos Santos
    Oct 21 '18 at 16:11










  • $begingroup$
    Then we have $a_n>2a$
    $endgroup$
    – m.bazza
    Oct 21 '18 at 16:13










  • $begingroup$
    But this isn't a contradiction as this gives us $a_ngeq 0$
    $endgroup$
    – m.bazza
    Oct 21 '18 at 16:14










  • $begingroup$
    But we know that $a_n-ageq 0$ from our assumption, why bother with cases
    $endgroup$
    – Jakobian
    Oct 21 '18 at 16:15










  • $begingroup$
    Oh of course haha didn't realise
    $endgroup$
    – m.bazza
    Oct 21 '18 at 16:16
















$begingroup$
What if $a_n<a$?
$endgroup$
– José Carlos Santos
Oct 21 '18 at 16:11




$begingroup$
What if $a_n<a$?
$endgroup$
– José Carlos Santos
Oct 21 '18 at 16:11












$begingroup$
Then we have $a_n>2a$
$endgroup$
– m.bazza
Oct 21 '18 at 16:13




$begingroup$
Then we have $a_n>2a$
$endgroup$
– m.bazza
Oct 21 '18 at 16:13












$begingroup$
But this isn't a contradiction as this gives us $a_ngeq 0$
$endgroup$
– m.bazza
Oct 21 '18 at 16:14




$begingroup$
But this isn't a contradiction as this gives us $a_ngeq 0$
$endgroup$
– m.bazza
Oct 21 '18 at 16:14












$begingroup$
But we know that $a_n-ageq 0$ from our assumption, why bother with cases
$endgroup$
– Jakobian
Oct 21 '18 at 16:15




$begingroup$
But we know that $a_n-ageq 0$ from our assumption, why bother with cases
$endgroup$
– Jakobian
Oct 21 '18 at 16:15












$begingroup$
Oh of course haha didn't realise
$endgroup$
– m.bazza
Oct 21 '18 at 16:16




$begingroup$
Oh of course haha didn't realise
$endgroup$
– m.bazza
Oct 21 '18 at 16:16










2 Answers
2






active

oldest

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0












$begingroup$

If $a<0$ then since $a_nto a$, there is an integer $N$ such that whenever $n>N$, we must have $a_nin (a,0)$, i.e. $a_n<0.$ And this is a contradiction.



In fact, the same reasoning shows that if $a_n>0$ for all $n$, then $any$ limit point $a$ of $(a_n)$ must satisfy $age 0.$






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    This proof has a flaw since there is no restriction that $a_n>a$ (consider $a_n=(-{1over 2})^n+1$ and $a=1$) but it can be simply fixed. Just conclude from $|a_n-a|<-a$ that $$a<a_n-a<-a$$ or equivalently $$2a<a_n<0$$ which is now a contradiction. Except that, your proof is fine.






    share|cite|improve this answer









    $endgroup$













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      2 Answers
      2






      active

      oldest

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      2 Answers
      2






      active

      oldest

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      active

      oldest

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      0












      $begingroup$

      If $a<0$ then since $a_nto a$, there is an integer $N$ such that whenever $n>N$, we must have $a_nin (a,0)$, i.e. $a_n<0.$ And this is a contradiction.



      In fact, the same reasoning shows that if $a_n>0$ for all $n$, then $any$ limit point $a$ of $(a_n)$ must satisfy $age 0.$






      share|cite|improve this answer









      $endgroup$


















        0












        $begingroup$

        If $a<0$ then since $a_nto a$, there is an integer $N$ such that whenever $n>N$, we must have $a_nin (a,0)$, i.e. $a_n<0.$ And this is a contradiction.



        In fact, the same reasoning shows that if $a_n>0$ for all $n$, then $any$ limit point $a$ of $(a_n)$ must satisfy $age 0.$






        share|cite|improve this answer









        $endgroup$
















          0












          0








          0





          $begingroup$

          If $a<0$ then since $a_nto a$, there is an integer $N$ such that whenever $n>N$, we must have $a_nin (a,0)$, i.e. $a_n<0.$ And this is a contradiction.



          In fact, the same reasoning shows that if $a_n>0$ for all $n$, then $any$ limit point $a$ of $(a_n)$ must satisfy $age 0.$






          share|cite|improve this answer









          $endgroup$



          If $a<0$ then since $a_nto a$, there is an integer $N$ such that whenever $n>N$, we must have $a_nin (a,0)$, i.e. $a_n<0.$ And this is a contradiction.



          In fact, the same reasoning shows that if $a_n>0$ for all $n$, then $any$ limit point $a$ of $(a_n)$ must satisfy $age 0.$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Oct 21 '18 at 16:27









          MatematletaMatematleta

          10.2k2918




          10.2k2918























              0












              $begingroup$

              This proof has a flaw since there is no restriction that $a_n>a$ (consider $a_n=(-{1over 2})^n+1$ and $a=1$) but it can be simply fixed. Just conclude from $|a_n-a|<-a$ that $$a<a_n-a<-a$$ or equivalently $$2a<a_n<0$$ which is now a contradiction. Except that, your proof is fine.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                This proof has a flaw since there is no restriction that $a_n>a$ (consider $a_n=(-{1over 2})^n+1$ and $a=1$) but it can be simply fixed. Just conclude from $|a_n-a|<-a$ that $$a<a_n-a<-a$$ or equivalently $$2a<a_n<0$$ which is now a contradiction. Except that, your proof is fine.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  This proof has a flaw since there is no restriction that $a_n>a$ (consider $a_n=(-{1over 2})^n+1$ and $a=1$) but it can be simply fixed. Just conclude from $|a_n-a|<-a$ that $$a<a_n-a<-a$$ or equivalently $$2a<a_n<0$$ which is now a contradiction. Except that, your proof is fine.






                  share|cite|improve this answer









                  $endgroup$



                  This proof has a flaw since there is no restriction that $a_n>a$ (consider $a_n=(-{1over 2})^n+1$ and $a=1$) but it can be simply fixed. Just conclude from $|a_n-a|<-a$ that $$a<a_n-a<-a$$ or equivalently $$2a<a_n<0$$ which is now a contradiction. Except that, your proof is fine.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 30 '18 at 18:21









                  Mostafa AyazMostafa Ayaz

                  14.8k3938




                  14.8k3938






























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