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Theorem: If $ ( a_n ) $ is a cauchy sequence then it converges.

Note: I know that the common proof of this theorem uses subsequences and the fact that if we have a cauchy sequence then it is bounded, but I first tried to prove the above theorem on my own and stumbled upon a difficulty I haven't found elsewhere.

My attempt of proof: Suppose $ ( a_n ) $ is a cauchy sequence. Let $ \epsilon > 0 $ be arbitrary, hence there exists $ N \in \mathbb{N} $ s.t. $ \forall n,m>N. | a_n- a_m | < \epsilon$. Choose $ m=N+1 $, hence we have $ \forall n>N. |a_n - a_{N+1} | < \epsilon $. Since $ \epsilon >0 $ was arbitrary, we have $ \forall \epsilon>0 \exists N \in \mathbb{N} \forall n>N. |a_n - a_{N+1} | < \epsilon $, Therefore $ \lim a_n = a_{N+1} $, hence the sequence converges.

Question: Obviously there is a fallacy with the proof since following the reasoning in my proof, $ ( a_n) $ could converge also to $ a_{N+2} , a_{N+3}, ... $ and so forth which is obviously false since only one limit can exist for a sequence that converges. However I don't fully understand what went wrong, maybe something is wrong with the logic? can you please help me as to why my proof is wrong?

hazelnut_116
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    the issue with your proof is that you have overlooked the fact that the $N$ in your proof depends on the $\epsilon$ choosen, so you cannot suddenly assume $\epsilon$ is arbitrary. – skeptic Mar 22 '21 at 10:52

2 Answers2

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In the definition of a limit, the limit is a constant not depending on $N$. Formally, a sequence $(a_n)$ converges to a number $a$ iff

$\forall \epsilon >0 \exists N\in \mathbb{N}: \forall n\ge N, |a-a_n|<\epsilon$

So you see, in the defintion of a limit, we start out by choosing the limit $a$. You choose this after choosing $\epsilon$, so $a_N$ is not necessarily a limit of the sequence.

Richard Jensen
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  • In $ \forall \epsilon >0 \exists N{(\epsilon)}\in \mathbb{N}: \forall n\ge N, |a-a_n|<\epsilon $. Does the index $ n $ also depend on $ \epsilon $? ( meaning that the definition can be written as $ \forall \epsilon >0 \exists N{(\epsilon)}\in \mathbb{N}: \forall n(\epsilon) \ge N(\epsilon), |a-a_{n(\epsilon)}|<\epsilon $ , the reasoning is because if $ N(\epsilon) $ depends on $ \epsilon $ then $ n $ also should. ) – hazelnut_116 Mar 22 '21 at 11:19
  • Well, yes and no :). We are already given $N(\epsilon)$, and we only choose $n$ bigger than that. So it wouldn't really make sense to write $n(\epsilon)$, but writing $N(\epsilon)$ makes perfect sense. – Richard Jensen Mar 22 '21 at 11:38
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Please note that if $(a_n)$ is a Cauchy sequence in $\mathbb R$ (or in general in a complete metric space) then it converges. For example $(a_n):a_n=\frac 1n$ is a Cauchy sequence in $(0,\infty)$ but does not converge in $(0,\infty)$.

Coming to your proof, It is true that for every $\epsilon\gt 0,\exists N_\epsilon\in \mathbb N$ such that $m,n \gt N\implies |a_m-a_n|\lt \epsilon$. Note the subscript of $N_\epsilon$ here. If your change $\epsilon\gt 0$, then $N_\epsilon$ will also change and so will $a_{N_\epsilon}$. Limit of a sequence is unique so $a_{N_\epsilon}$ is not limit of the sequence unless the sequence is eventually constant sequence (i.e., there exists $M\in \mathbb N:m\gt M\implies a_m=a_{m+i}$ for all $i\in \mathbb N$.

Koro
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