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Let $(x_n)$ and $(y_n)$ convergent sequence. Is there an easier way to prove that if $x_n\leq y_n$ for all $n$, then $\lim_{n\to\infty }x_n\leq \lim_{n\to\infty }y_n$ ?

This is how I do:

Let $\ell=\lim_{n\to\infty }x_n$ and $\ell'=\lim_{n\to\infty }y_n$. Suppose $\ell'<\ell$. Let $\varepsilon=\frac{\ell-\ell'}{2}$. In particular, there is a $N$ such that $y_N\in]\ell'-\varepsilon,\ell'+\varepsilon[$ and $x_N\in]\ell-\varepsilon,\ell+\varepsilon[$ and thus $y_N<x_N$ which is a contradiction with the fact that $x_n\leq y_n$ for all $n$.

I would like to know if there is another proof that doesn't use contradiction.

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marked as duplicate by Martin Sleziak, JonMark Perry, M. Vinay, Davide Giraudo real-analysis Jul 7 '16 at 9:43

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  • $\begingroup$ $\varepsilon=\frac{\ell-\ell'}{2}$ $\endgroup$ – R.N Sep 17 '15 at 15:05
  • $\begingroup$ @idm If you add the two and then divide by two, you get the midpoint between the two. That's a pretty big number, so we will get $l \in (l' - \epsilon, l' + \epsilon)$. We want this interval to exclude $l$. $\endgroup$ – layman Sep 17 '15 at 15:09
  • $\begingroup$ @idm So instead we take $\frac{l - l'}{2}$. Since $l > l'$, $l - l'$ is the distance between $l$ and $l'$. Taking this distance and dividing it by $2$, it follows that $l$ is not in $(l' - (\frac{l - l'}{2}), l' + (\frac{l - l'}{2}))$. $\endgroup$ – layman Sep 17 '15 at 15:10
  • $\begingroup$ @idm $\varepsilon>0$. how you show that $\frac{\ell+\ell'}{2}>0$ $\endgroup$ – R.N Sep 17 '15 at 15:10
  • $\begingroup$ @idm Similarly, $l'$ is not in $(l - (\frac{l - l'}{2}), l + (\frac{l - l'}{2}))$. That's why $y_{N} < x_{N}$. $\endgroup$ – layman Sep 17 '15 at 15:10
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A proof without contradiction:

Suppose that $x_n \leq y_n$ for all $n$. For any $\epsilon > 0$, there exists an $N$ such that $n>N$ implies that $|x_n - x| \leq \epsilon$ and $|y_n - y| \leq \epsilon$. We then note that $$ y - x = (y - y_n) + (y_n - x_n) + (x_n - x) \geq\\ -|y-y_n| + (y_n - x_n) -|x_n - x| >\\ (y_n - x_n) - 2 \epsilon \geq \\ -2 \epsilon $$ So, we have $y - x > -2 \epsilon$ for every $\epsilon > 0$. It follows that $y - x \geq 0$, as desired.

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  • $\begingroup$ In this proof, are $x$ and $y$ the limits of $x_n$ and $y_n$ respectively? $\endgroup$ – user21 Feb 25 '17 at 21:00
  • $\begingroup$ @user21 yes!! Left that out apparently $\endgroup$ – Omnomnomnom Feb 25 '17 at 21:16
  • $\begingroup$ That's okay. How do you go from your first line to your third line in the equation? $(y-y_n)+(y_n-x_n)+(x_n-x)\geq-|y-y_n|+(y_n-x_n)-|x_n-x|$? I'm guessing triangle inequality. $\endgroup$ – user21 Feb 25 '17 at 21:31
  • $\begingroup$ @user21 it's easier than that: $a\geq -|a|$ for any $a\in \Bbb R$. $\endgroup$ – Omnomnomnom Feb 25 '17 at 21:49
  • $\begingroup$ Okay, that makes sense. I understand how you get to $(y_n-x_n)-2\epsilon$...how do you say that that is greater than -2$\epsilon$? $\endgroup$ – user21 Feb 25 '17 at 22:02

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