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Here is a part of the proof:

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why the author put the equality on the inequality after taking $n \rightarrow \infty$, could anyone clarify this for me please?

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    $\begingroup$ E.g, $1-{1\over n}<1$ for every positive integer $n$, but $\lim_{n\rightarrow\infty} (1-{1\over n})=1$. $\endgroup$ – David Mitra Oct 18 '17 at 11:52
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Recall that the process of taking limits does not in general preserve strict inequalities. For example consider $\frac{1}{n} > 0$ for any $n \in \mathbb{N}$ but we clearly have equality in the limit $n \to \infty$.

Taking limits does however preserve weak inequalities which is enough to give the authors result.

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  • $\begingroup$ "but we clearly have equality in the limit $n\rightarrow \infty $" , "Taking limits does however preserve weak inequalities which is enough to give the authors result " could you clarify these 2 statements please? what do you mean by "weak inequalities", why u said "weak"? $\endgroup$ – Intuition Oct 18 '17 at 12:25
  • $\begingroup$ First $\lim_{n \to \infty} \frac{1}{n} = 0$ (i.e. we have equality in the limit). Weak is a standard term to describe inequalities with "$\leq$" (rather than strict inequalities with "$<$"). It is an easy result from basic real analysis that if $a_n \leq a$ for all $n$ then if $\lim_{n \to \infty} a_n$ exists $\lim_{n \to \infty} a_n \leq a$. That is applying limits to the weak inequality $a_n \leq a$ preserves the weak inequality. $\endgroup$ – Rhys Steele Oct 18 '17 at 12:32
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This is standard when working with inequalities and limits. If $a_{n}<b_{n}$ and $lim a_{n}=a$ and $lim b_{n}=b$ then the most we can say is $a \leq b$. For example take $a_{n}=0$ and $b_{n}:=1/n$. We know that $0<1/n$, but the limits of both sides are equal.

This answer is copied from this link:

Problems with the proof that $\ell^p$ is complete

Which contains a very smart explanation for completeness of $l_{p}$.

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