A convergent sequence has precisely one accumulation point












4












$begingroup$


As a step in a proof I've been trying to show that convergence of a sequence implies it must have precisely one accumulation point. This is the definition we use for an accumulation point



Let $S$ be a set of real numbers. A real number is an accumulation point $s_0$ of $S$ if and only if for any $epsilon > 0$, there exists at least one point $t$ of $S$ such that $0 < |t-s_0| < epsilon$.



I wish to prove:




Lemma: A convergent sequence has precisely one accumulation point.




My thoughts: Since the limit of the sequence exists we know that the set cannot have 2 accumulation points or more, we will show by contradiction.



If we would have multiple accumulation points the limit does not exist because we can never get arbitrarily close to a single point (within ϵ), because there exist certain subsequences that each get arbitrarily close to at least two accumulation points, which determine the minimum distance a sequence can be from an accumulation point (so we get a lower bound and therefore we do not get convergence). There must be precisely one accumulation point.



I do not know how to make this more precise of a statement, there are a lot of words and it's not really structured. I'm looking for some help making a more rigourous argument.






real-analysis sequences-and-series limits







share|cite|improve this question









$endgroup$

















    4












    $begingroup$


    As a step in a proof I've been trying to show that convergence of a sequence implies it must have precisely one accumulation point. This is the definition we use for an accumulation point



    Let $S$ be a set of real numbers. A real number is an accumulation point $s_0$ of $S$ if and only if for any $epsilon > 0$, there exists at least one point $t$ of $S$ such that $0 < |t-s_0| < epsilon$.



    I wish to prove:




    Lemma: A convergent sequence has precisely one accumulation point.




    My thoughts: Since the limit of the sequence exists we know that the set cannot have 2 accumulation points or more, we will show by contradiction.



    If we would have multiple accumulation points the limit does not exist because we can never get arbitrarily close to a single point (within ϵ), because there exist certain subsequences that each get arbitrarily close to at least two accumulation points, which determine the minimum distance a sequence can be from an accumulation point (so we get a lower bound and therefore we do not get convergence). There must be precisely one accumulation point.



    I do not know how to make this more precise of a statement, there are a lot of words and it's not really structured. I'm looking for some help making a more rigourous argument.






    real-analysis sequences-and-series limits







    share|cite|improve this question









    $endgroup$















      4












      4








      4





      $begingroup$


      As a step in a proof I've been trying to show that convergence of a sequence implies it must have precisely one accumulation point. This is the definition we use for an accumulation point



      Let $S$ be a set of real numbers. A real number is an accumulation point $s_0$ of $S$ if and only if for any $epsilon > 0$, there exists at least one point $t$ of $S$ such that $0 < |t-s_0| < epsilon$.



      I wish to prove:




      Lemma: A convergent sequence has precisely one accumulation point.




      My thoughts: Since the limit of the sequence exists we know that the set cannot have 2 accumulation points or more, we will show by contradiction.



      If we would have multiple accumulation points the limit does not exist because we can never get arbitrarily close to a single point (within ϵ), because there exist certain subsequences that each get arbitrarily close to at least two accumulation points, which determine the minimum distance a sequence can be from an accumulation point (so we get a lower bound and therefore we do not get convergence). There must be precisely one accumulation point.



      I do not know how to make this more precise of a statement, there are a lot of words and it's not really structured. I'm looking for some help making a more rigourous argument.






      real-analysis sequences-and-series limits







      share|cite|improve this question









      $endgroup$




      As a step in a proof I've been trying to show that convergence of a sequence implies it must have precisely one accumulation point. This is the definition we use for an accumulation point



      Let $S$ be a set of real numbers. A real number is an accumulation point $s_0$ of $S$ if and only if for any $epsilon > 0$, there exists at least one point $t$ of $S$ such that $0 < |t-s_0| < epsilon$.



      I wish to prove:




      Lemma: A convergent sequence has precisely one accumulation point.




      My thoughts: Since the limit of the sequence exists we know that the set cannot have 2 accumulation points or more, we will show by contradiction.



      If we would have multiple accumulation points the limit does not exist because we can never get arbitrarily close to a single point (within ϵ), because there exist certain subsequences that each get arbitrarily close to at least two accumulation points, which determine the minimum distance a sequence can be from an accumulation point (so we get a lower bound and therefore we do not get convergence). There must be precisely one accumulation point.



      I do not know how to make this more precise of a statement, there are a lot of words and it's not really structured. I'm looking for some help making a more rigourous argument.






      real-analysis sequences-and-series limits




      real-analysis sequences-and-series limits






      share|cite|improve this question













      share|cite|improve this question











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      asked Jan 8 at 14:05









      Wesley StrikWesley Strik

      2,194424




      2,194424






















          2 Answers
          2






          active

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          5












          $begingroup$

          Suppose the sequence converges to $x$. Clearly $x$ is an accumulation point.



          Suppose on the contrary that we have a second accumulation point, $y$. Let $r = frac{|x-y|}{2}$.



          Notice that there exists $N>0$, such that for all $n>N$, then $|x_n -x|<r$.



          That is when $n>N$, we have $$|x_n - y|=|x_n-x+x-y| ge ||x_n-x|-|x-y||=||x_n-x|-2r|=2r-|x_n-x|>r$$



          Hence we cannot have infinitely many points that get arbitrarily close to $y$, hence $y$ can't be an accumulation point.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Beautiful. You always come up with such nice and pretty arguments, this is precisely the technical argument I was thinking of but I was not sure how to formulate it so impeccably, thank you.
            $endgroup$
            – Wesley Strik
            Jan 8 at 14:51





















          5












          $begingroup$

          An accumulation point of a sequence $(a_n)_n$ is not the same as the accumulation point of its set of values ${a_n : n in mathbb{N}}$.



          Indeed, consider the constant sequence $a_n = a, forall n in mathbb{N}$. Clearly $(a_n)_n$ converges to $a$, but the set ${a_n : n in mathbb{N}} = {a}$ has no accumulation points by your definition.



          The proper definition is:




          $x in mathbb{R}$ is an accumulation point of a sequence $(a_n)_n$ if for every $varepsilon > 0$ the interval $langle x-varepsilon, x+varepsilonrangle$ contains infinitely many terms of the sequence $(a_n)_n$, i.e. for every $n in mathbb{N}$ there exists $m in mathbb{N}, m > n$ such that $|x-a_m| < varepsilon$.




          Try to show your lemma now.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Ah, very useful answer indeed. That's a very good point you make. I din't know that actually, thank you.
            $endgroup$
            – Wesley Strik
            Jan 8 at 19:59














          Your Answer





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          2 Answers
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          2 Answers
          2






          active

          oldest

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          active

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          active

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          5












          $begingroup$

          Suppose the sequence converges to $x$. Clearly $x$ is an accumulation point.



          Suppose on the contrary that we have a second accumulation point, $y$. Let $r = frac{|x-y|}{2}$.



          Notice that there exists $N>0$, such that for all $n>N$, then $|x_n -x|<r$.



          That is when $n>N$, we have $$|x_n - y|=|x_n-x+x-y| ge ||x_n-x|-|x-y||=||x_n-x|-2r|=2r-|x_n-x|>r$$



          Hence we cannot have infinitely many points that get arbitrarily close to $y$, hence $y$ can't be an accumulation point.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Beautiful. You always come up with such nice and pretty arguments, this is precisely the technical argument I was thinking of but I was not sure how to formulate it so impeccably, thank you.
            $endgroup$
            – Wesley Strik
            Jan 8 at 14:51


















          5












          $begingroup$

          Suppose the sequence converges to $x$. Clearly $x$ is an accumulation point.



          Suppose on the contrary that we have a second accumulation point, $y$. Let $r = frac{|x-y|}{2}$.



          Notice that there exists $N>0$, such that for all $n>N$, then $|x_n -x|<r$.



          That is when $n>N$, we have $$|x_n - y|=|x_n-x+x-y| ge ||x_n-x|-|x-y||=||x_n-x|-2r|=2r-|x_n-x|>r$$



          Hence we cannot have infinitely many points that get arbitrarily close to $y$, hence $y$ can't be an accumulation point.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Beautiful. You always come up with such nice and pretty arguments, this is precisely the technical argument I was thinking of but I was not sure how to formulate it so impeccably, thank you.
            $endgroup$
            – Wesley Strik
            Jan 8 at 14:51
















          5












          5








          5





          $begingroup$

          Suppose the sequence converges to $x$. Clearly $x$ is an accumulation point.



          Suppose on the contrary that we have a second accumulation point, $y$. Let $r = frac{|x-y|}{2}$.



          Notice that there exists $N>0$, such that for all $n>N$, then $|x_n -x|<r$.



          That is when $n>N$, we have $$|x_n - y|=|x_n-x+x-y| ge ||x_n-x|-|x-y||=||x_n-x|-2r|=2r-|x_n-x|>r$$



          Hence we cannot have infinitely many points that get arbitrarily close to $y$, hence $y$ can't be an accumulation point.






          share|cite|improve this answer









          $endgroup$



          Suppose the sequence converges to $x$. Clearly $x$ is an accumulation point.



          Suppose on the contrary that we have a second accumulation point, $y$. Let $r = frac{|x-y|}{2}$.



          Notice that there exists $N>0$, such that for all $n>N$, then $|x_n -x|<r$.



          That is when $n>N$, we have $$|x_n - y|=|x_n-x+x-y| ge ||x_n-x|-|x-y||=||x_n-x|-2r|=2r-|x_n-x|>r$$



          Hence we cannot have infinitely many points that get arbitrarily close to $y$, hence $y$ can't be an accumulation point.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 8 at 14:22









          Siong Thye GohSiong Thye Goh

          103k1468119




          103k1468119












          • $begingroup$
            Beautiful. You always come up with such nice and pretty arguments, this is precisely the technical argument I was thinking of but I was not sure how to formulate it so impeccably, thank you.
            $endgroup$
            – Wesley Strik
            Jan 8 at 14:51




















          • $begingroup$
            Beautiful. You always come up with such nice and pretty arguments, this is precisely the technical argument I was thinking of but I was not sure how to formulate it so impeccably, thank you.
            $endgroup$
            – Wesley Strik
            Jan 8 at 14:51


















          $begingroup$
          Beautiful. You always come up with such nice and pretty arguments, this is precisely the technical argument I was thinking of but I was not sure how to formulate it so impeccably, thank you.
          $endgroup$
          – Wesley Strik
          Jan 8 at 14:51






          $begingroup$
          Beautiful. You always come up with such nice and pretty arguments, this is precisely the technical argument I was thinking of but I was not sure how to formulate it so impeccably, thank you.
          $endgroup$
          – Wesley Strik
          Jan 8 at 14:51













          5












          $begingroup$

          An accumulation point of a sequence $(a_n)_n$ is not the same as the accumulation point of its set of values ${a_n : n in mathbb{N}}$.



          Indeed, consider the constant sequence $a_n = a, forall n in mathbb{N}$. Clearly $(a_n)_n$ converges to $a$, but the set ${a_n : n in mathbb{N}} = {a}$ has no accumulation points by your definition.



          The proper definition is:




          $x in mathbb{R}$ is an accumulation point of a sequence $(a_n)_n$ if for every $varepsilon > 0$ the interval $langle x-varepsilon, x+varepsilonrangle$ contains infinitely many terms of the sequence $(a_n)_n$, i.e. for every $n in mathbb{N}$ there exists $m in mathbb{N}, m > n$ such that $|x-a_m| < varepsilon$.




          Try to show your lemma now.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Ah, very useful answer indeed. That's a very good point you make. I din't know that actually, thank you.
            $endgroup$
            – Wesley Strik
            Jan 8 at 19:59


















          5












          $begingroup$

          An accumulation point of a sequence $(a_n)_n$ is not the same as the accumulation point of its set of values ${a_n : n in mathbb{N}}$.



          Indeed, consider the constant sequence $a_n = a, forall n in mathbb{N}$. Clearly $(a_n)_n$ converges to $a$, but the set ${a_n : n in mathbb{N}} = {a}$ has no accumulation points by your definition.



          The proper definition is:




          $x in mathbb{R}$ is an accumulation point of a sequence $(a_n)_n$ if for every $varepsilon > 0$ the interval $langle x-varepsilon, x+varepsilonrangle$ contains infinitely many terms of the sequence $(a_n)_n$, i.e. for every $n in mathbb{N}$ there exists $m in mathbb{N}, m > n$ such that $|x-a_m| < varepsilon$.




          Try to show your lemma now.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Ah, very useful answer indeed. That's a very good point you make. I din't know that actually, thank you.
            $endgroup$
            – Wesley Strik
            Jan 8 at 19:59
















          5












          5








          5





          $begingroup$

          An accumulation point of a sequence $(a_n)_n$ is not the same as the accumulation point of its set of values ${a_n : n in mathbb{N}}$.



          Indeed, consider the constant sequence $a_n = a, forall n in mathbb{N}$. Clearly $(a_n)_n$ converges to $a$, but the set ${a_n : n in mathbb{N}} = {a}$ has no accumulation points by your definition.



          The proper definition is:




          $x in mathbb{R}$ is an accumulation point of a sequence $(a_n)_n$ if for every $varepsilon > 0$ the interval $langle x-varepsilon, x+varepsilonrangle$ contains infinitely many terms of the sequence $(a_n)_n$, i.e. for every $n in mathbb{N}$ there exists $m in mathbb{N}, m > n$ such that $|x-a_m| < varepsilon$.




          Try to show your lemma now.






          share|cite|improve this answer









          $endgroup$



          An accumulation point of a sequence $(a_n)_n$ is not the same as the accumulation point of its set of values ${a_n : n in mathbb{N}}$.



          Indeed, consider the constant sequence $a_n = a, forall n in mathbb{N}$. Clearly $(a_n)_n$ converges to $a$, but the set ${a_n : n in mathbb{N}} = {a}$ has no accumulation points by your definition.



          The proper definition is:




          $x in mathbb{R}$ is an accumulation point of a sequence $(a_n)_n$ if for every $varepsilon > 0$ the interval $langle x-varepsilon, x+varepsilonrangle$ contains infinitely many terms of the sequence $(a_n)_n$, i.e. for every $n in mathbb{N}$ there exists $m in mathbb{N}, m > n$ such that $|x-a_m| < varepsilon$.




          Try to show your lemma now.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 8 at 14:13









          mechanodroidmechanodroid

          29k62648




          29k62648












          • $begingroup$
            Ah, very useful answer indeed. That's a very good point you make. I din't know that actually, thank you.
            $endgroup$
            – Wesley Strik
            Jan 8 at 19:59




















          • $begingroup$
            Ah, very useful answer indeed. That's a very good point you make. I din't know that actually, thank you.
            $endgroup$
            – Wesley Strik
            Jan 8 at 19:59


















          $begingroup$
          Ah, very useful answer indeed. That's a very good point you make. I din't know that actually, thank you.
          $endgroup$
          – Wesley Strik
          Jan 8 at 19:59






          $begingroup$
          Ah, very useful answer indeed. That's a very good point you make. I din't know that actually, thank you.
          $endgroup$
          – Wesley Strik
          Jan 8 at 19:59




















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