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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.

Arctic Char
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idm
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  • $\varepsilon=\frac{\ell-\ell'}{2}$ – R.N Sep 17 '15 at 15:05
  • @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 $l$. – layman Sep 17 '15 at 15:09
  • @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 $(l' - (\frac{l - l'}{2}), l' + (\frac{l - l'}{2}))$. – layman Sep 17 '15 at 15:10
  • @idm $\varepsilon>0$. how you show that $\frac{\ell+\ell'}{2}>0$ – R.N Sep 17 '15 at 15:10
  • @idm Similarly, $l'$ is not in $(l - (\frac{l - l'}{2}), l + (\frac{l - l'}{2}))$. That's why $y_{N} < x_{N}$. – layman Sep 17 '15 at 15:10
  • Yes, thanks :-) I was thinking in fact at $$[\ell',\frac{\ell+\ell'}{2}]$$ which is $[\ell',\ell+\frac{\ell-\ell'}{2}]$ as you both said :-) – idm Sep 17 '15 at 15:23
  • Introducing $z_n=y_n-x_n\ge 0$ it is equivalent to prove that $\lim z_n\ge 0$, but the latter is the same as to say that $[0,+\infty)$ is a closed set. – A.Γ. Sep 17 '15 at 16:42

1 Answers1

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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.

Ben Grossmann
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  • In this proof, are $x$ and $y$ the limits of $x_n$ and $y_n$ respectively? – user21 Feb 25 '17 at 21:00
  • @user21 yes!! Left that out apparently – Ben Grossmann Feb 25 '17 at 21:16
  • 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. – user21 Feb 25 '17 at 21:31
  • @user21 it's easier than that: $a\geq -|a|$ for any $a\in \Bbb R$. – Ben Grossmann Feb 25 '17 at 21:49
  • 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$\epsilon$? – user21 Feb 25 '17 at 22:02
  • @user21 note that $y_n\geq x_n$, according to the question statement, and by our supposition – Ben Grossmann Feb 25 '17 at 22:07
  • Actually, I understand that. I'm confused on the next step though. If we know that $\epsilon>0$, then shouldn't $-2\epsilon<0$? Then could we say that $y-x\geq0$? – user21 Feb 25 '17 at 22:30
  • @user21 "for every $\epsilon>0$." Think about it. – Ben Grossmann Feb 25 '17 at 22:46
  • Hmm....Okay then. – user21 Feb 25 '17 at 23:38
  • To prove the last line explicitly I think you need a contradiction Mr. Grossmann! – Derek Luna Oct 31 '20 at 20:03
  • @Derek I don’t see why you need a contradiction to prove that the supremum of the set of negative numbers is $0$ – Ben Grossmann Nov 01 '20 at 15:46
  • Well when you put it that way...But it all comes down to how one translates statements to other equivalent statements. I feel like in these sorts of proofs dealing with epsilon, many direct proofs are written in a way such that if questioned, the response would be "Because otherwise...", which makes me think contradiction is inherent in the argument. The same can be said for proving the statement in your last comment. Why is $0$ the LEAST upper bound? Because if one supposed that there was a smaller upper bound, you could find a negative number between that smaller upper bound and $0$. – Derek Luna Nov 01 '20 at 16:07
  • If you had no intuition for what supremum meant, then you would have to prove it this way. Otherwise, your proof is relying implicitly on your intuition on what the word supremum represents to you. – Derek Luna Nov 01 '20 at 16:14
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    @Derek That's a fair point. At the very least, I have (without contradiction) reduced the problem to the simpler statement that "if $p \in \Bbb R$ is such that $p > q$ for every negative $q$, then $p$ is non-negative." – Ben Grossmann Nov 02 '20 at 02:45