Here is a proof of:

$$\lim_{n\rightarrow \infty}\frac{1}{n}\neq0, n \in \mathbb{N}$$

As per definition of a limit of a sequence:

$$ \lim_{n\to\infty}S_n=A\iff\forall \epsilon\in\Bbb R_{>0}: \exists N\in\Bbb N : \forall n\in \Bbb N_{>N} : |(S_n-A)|<\epsilon,$$

And the proof is simple, since $\frac{1}{n}$ is a rational number and $\epsilon$ is a real number, there always exists a real number between 0 and any rational number. Because of that there exists an $\epsilon$ for which the condition can not hold.

Why is this proof wrong?

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    $\begingroup$ You have to choose some $\epsilon >0$ and THEN find an $N$ $\endgroup$ – Peter Apr 22 '17 at 13:14
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    $\begingroup$ ... and what you are describing is precisely the opposite: you chose $n$ and THEN looking for $\epsilon$. $\endgroup$ – A.Γ. Apr 22 '17 at 13:15
  • $\begingroup$ @Peter I don't see how that solves the issue... Can I not simply choose $\epsilon$ that is a real number, and that is in-between 0 and the smallest rational number that the sequence can generate? Why would this not be equivalent? $\endgroup$ – Dole Apr 22 '17 at 13:18
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    $\begingroup$ @Dole there is no smallest rational number that the sequence can generate. $\endgroup$ – Antonio Vargas Apr 22 '17 at 13:18
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    $\begingroup$ You formula is badly written, it should be something like $$ \lim_{n\to\infty}S_n=A\iff\forall \epsilon\in\Bbb R_{>0}: \exists N\in\Bbb N : \forall n\in \Bbb N_{>N} : |S_n-A|<\epsilon,$$ (where subscripting a set by a condition designates its subset defined by that condition). Never write a trailing condition like $\epsilon\in\Bbb R$ to a logical formula, put it where it belongs in the formula itself (here that is at the quantifier introducing $\epsilon$).. $\endgroup$ – Marc van Leeuwen Apr 22 '17 at 13:53

The way it works : Choose $\epsilon>0$. It can be arbitary small, but it must be positive. THEN, you can find an $N$ (depending on $\epsilon$) with the given property.

But you cannot find a FIXED $N$ such that EVERY $\epsilon>0$ works. This must fail because the differences get arbitarily close to $0$.

The catch is that we need larger and larger $N$ if $\epsilon$ gets smaller and smaller.

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  • $\begingroup$ I am still not sure I understand. If I choose an epsilon such that it is between the smallest number that the sequence can generate and between 0 (there should be such epsilon as there are real numbers between any rationals), would that not imply that the definition doesn't hold? $\endgroup$ – Dole Apr 22 '17 at 13:44
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    $\begingroup$ @Dole: An infinite collection of real numbers does not need to have a smallest element, and indeed $\{\,\frac1n\mid n\in\Bbb N_{>0}\,\}$ does not have a smallest element. Also, the intersection of a nested sequence of sets can be empty although each set in the sequence is nonempty; for instance the sequence of "upper parts $\Bbb N_{\geq n}$ of the natural numbers", one for each $n\in\Bbb N$, is nested and each part in the sequence is nonempty (even infinite), yet the intersection of all of them is empty. $\endgroup$ – Marc van Leeuwen Apr 22 '17 at 14:00
  • $\begingroup$ @MarcvanLeeuwen So is: $\{\,\frac1n\mid n\in\Bbb N_{>0}\,\} $not a subset of rational numbers then? $\endgroup$ – Dole Apr 22 '17 at 14:10
  • $\begingroup$ Yes, it is a subset of the rational numbers too (and all rational numbers are real numbers). It is a subset of the rational numbers without any smallest element; such subsets exists for the rational numbers just as well as for the real numbers. $\endgroup$ – Marc van Leeuwen Apr 22 '17 at 15:40

You are confused by the order of the quantifiers. If you take a look at the formula you listed, it says that given any positive number there is some positive integer such that every term of the sequence with index no less than the positive integer is close to the "limit" by at most the positive number. So the upper bound $\varepsilon$ is given first, not the other way around.

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