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What step in this "proof" that $\lim_{n \to\infty} a_{2n} = L$ implies $\lim_{n \to\infty} a_n = L$ is wrong:

1) $\lim_{n \to\infty} a_{2n} = L$

2) $\Rightarrow \forall \epsilon > 0$ $\exists N \in\mathbb{N} \ni$ if $n \ge N$ then $|a_{2n} - L| < \epsilon$

3) let $k = 2n$ $\wedge$ $n \ge N$ $\iff$ $n = k/2 \ge N$ $\iff$ $k \ge 2N$

4) $\Rightarrow$ $\forall \epsilon > 0$ $\exists N' \in\mathbb{N} \ni$ if $k \ge N'$ then $|a_k -L| < \epsilon$ where $N' = 2N$

5) $\Rightarrow$ $\lim_{k \to\infty} a_k = L$

The theorem is obviously wrong, just take $a_n = (-1)^n$ as a counterexample.

[Update]

To be precise, I'm only interested in what step in the above "proof" is wrong, not why the above "theorem" is wrong in general.

[Update]

Some people have had issues with my definition of the limit, specifically the "if $n \ge N$" clause. I got this straight from "Foundations of Mathematical Analysis" by Johnsonbaugh and Pfaffenberger (a Stanford University textbook):

$\lim_{n \to\infty} a_n = L$ if $\forall \epsilon > 0$ $\exists N \in \mathbb{N} \ni$ if $n \ge N$ then $|a_n - L| < \epsilon$

Note, some have taken issue with the fact that I've substituted quantifiers for some wording from text in the above definition, so here is the exact definition from the text:

$\{a_n\}_{n=1}^{\infty}$ has limit $L \in \mathbb{R}$ if for every $\epsilon > 0$, there exists a positive integer $N$, such that if $n \ge N$, then $|a_n -L| < \epsilon$

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  • $\begingroup$ The sequence $a_k$ is not the original sequence $a_n$ if $k = 2n$. $\endgroup$ – dannum May 4 '18 at 16:26
  • $\begingroup$ $k = 2n$ - you proved what you started from. $\endgroup$ – NickD May 4 '18 at 16:27
  • $\begingroup$ All the following terms after $a_{k}$ needs also to be within this interval. Or take $a_{k+1}$, you can't divide it by two since your $k$ is even, so you don't have a proof, since a following term in your sequence doesn't follow the rule of the limit. Your limit definition is wrong, it is \forall{n} \get $\endgroup$ – Jean Rostan May 4 '18 at 16:27
  • $\begingroup$ the fact is that $\lim_{n \to\infty} a_{2n} = L$ can't give any information on the behavour of $a_k$ when $k$ is odd $\endgroup$ – user May 4 '18 at 16:30
  • $\begingroup$ You did not justify your writing of "$\Rightarrow$" $\endgroup$ – Hagen von Eitzen May 4 '18 at 16:31
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You can note that $n$ is a positive integer and hence $k=2n$ implies that $k$ is an even positive integer. Your final statement $(4)$ thus deals with all even $k>N'$ and not all $k$ as required by definition of limit.

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  • $\begingroup$ Thanks, that's the crux of the problem. To symbolizes it, in step 4) I implicitly say $...\forall k[k \in \mathbb{2N} \wedge k \ge N' => |a_k -L| < \epsilon]...$ but the definition of the limit requires $...\forall k[k \in \mathbb{N} \wedge k \ge N' => |a_k -L| < \epsilon]...$ and you can't get the latter from the former. $\endgroup$ – Arnaut B May 6 '18 at 7:12
  • $\begingroup$ @ArnautB: forget my last deleted comment. I am glad you understood the issue here. $\endgroup$ – Paramanand Singh May 6 '18 at 7:39
  • $\begingroup$ @ArnautB: just to confirm your symbolic interpretation is exactly same as in my answer and what you have wrote in your comment is the correct argument. $\endgroup$ – Paramanand Singh May 6 '18 at 8:06
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The problem lies in your definition of a limit:

$$ \forall \epsilon >0, \exists N \in \mathbb{N}, \forall n\geq N, |a_n-L|<\epsilon $$.

The second $\forall$ is key.

See your mistake?

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  • $\begingroup$ Bill.Exactly, +1. $\endgroup$ – Peter Szilas May 4 '18 at 16:51
  • $\begingroup$ Are you saying my definition of the limit is incorrect? $\endgroup$ – Arnaut B May 4 '18 at 16:59
  • $\begingroup$ I think your definition is quite ambiguous and it is that ambiguity that lead you to this mistake: "if $n \geq N$" does not denote a set but rather a property where $n$ would have been introduced beforehand. $\endgroup$ – Bill O'Haran May 4 '18 at 17:05
  • $\begingroup$ As I noted elsewhere this definition is right out of Foundations of Mathematical Analysis by Johnsonbaugh and Pfaffenberger, a textbook used in Stanford. Sorry, I fail to see the ambiguity, can you elaborate? $n$ is implictly a natural number since $a_n$ is a sequence. $\endgroup$ – Arnaut B May 4 '18 at 17:15
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    $\begingroup$ Again, a book being used somewhere is no argument. I never said the definition given in the book was incorrect, having looked at it, it is actually perfectly equivalent to the one I gave. I'll be honest with you: I do not have any reference to provide. However, I would like you to understand that the whole problem you have comes down to translating that if correctly. Since you seem a little protective of what you think, remember that you (rightfully) accepted that your proof is wrong, so accept that there is something wrong with your proof ;) $\endgroup$ – Bill O'Haran May 4 '18 at 18:52
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Your highly abbreviated language hides the misstatement.

It is incorrect to say that "$\lim\limits_{n\to\infty} a_{2n}=L$" implies that "if $n \geq N$ then $|a_{2n}-L|<\epsilon$".

Rather, the correct conclusion is that "if $n$ is an integer with $n \geq N$ then $|a_{2n}-L|<\epsilon$".

This is implicit when using sequences, but it must be considered nonetheless.

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  • $\begingroup$ The textbook Foundations of Mathematical Analysis by Johnsonbaugh and Pfaffenberger states it like that. This is a textbook used in Stanford. As you say it is implicit $n$ is an integer since we're using sequences, therefore I don't think the wording is incorrect. $\endgroup$ – Arnaut B May 4 '18 at 16:50
  • $\begingroup$ @ArnautB : I don’t mean to say your wording is incorrect—it is perfectly acceptable. I just mean you have to be careful to only use expressions which represent integers for subscripts. I think that’s what got glossed over in the proof. $\endgroup$ – MPW May 4 '18 at 22:55

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