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Suppose that $a_n \leq b_n$ for all $n \in \mathbb{N}$, where $a_n$ and $b_n$ are two convergent sequences. Is there a constructive proof that $\lim_{n \to \infty} a_n \leq \lim_{n \to \infty} b_n$? The proofs that I have seen (e.g. this one) are nonconstructive.

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  • $\begingroup$ The proof you cited is constructive. It can be shown constructively that $L \leq M \iff \neg (M < L)$, though the precise proof depends on the construction of the reals and the definitions of $<$ and $\leq$. $\endgroup$ Commented Sep 16, 2022 at 18:55
  • $\begingroup$ @MarkSaving Do you happen to have a source discussing this? $\endgroup$
    – ಠ_ಠ
    Commented Sep 19, 2022 at 2:26

3 Answers 3

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Let $a = \lim a_n, b= \lim b_n$. Choose $\varepsilon >0$ Choose $N$ such that $$n\ge N \Rightarrow |a_n -a| <\frac{\varepsilon}{2}, |b_n -b| <\frac{\varepsilon}{2} $$ Then $$ a \le a_n + \frac{\varepsilon}{2}\le b_n + \frac{\varepsilon}{2} \le b + \varepsilon$$ Since $\varepsilon$ has been chosen arbitrarily, it follows that $a\le b$

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  • $\begingroup$ Beautiful! Thank you! $\endgroup$
    – ಠ_ಠ
    Commented Sep 12, 2022 at 19:58
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I'm assuming by "constructive" you mean not using proof by contradiction.

Let $a = \lim_{n \to \infty} a_n$ and $b = \lim_{n \to \infty} b_n$. Let $\varepsilon > 0$. Then there exists an $N \in \mathbb{N}$ such that $|a_n - a| < \varepsilon/2$ and $|b_n - b| < \varepsilon/2$. Thus,

$$ a- b = a - a_n - (b - b_n) + a_n - b_n < |a - a_n| + |b - b_n| + a_n - b_n < \varepsilon $$ Thus, $a \leq b$. Now, I'm assuming you've proven $a < b + \varepsilon$ for all $\varepsilon > 0$ implies $a \leq b$. But most people prove this fact by contradiction as well. So let's prove it without it.

Let $\eta = a - b$. Then $b + \eta = b + a - b = a$. So $a$ cannot be strictly less than $b + \eta$. This implies $\eta$ cannot be strictly positive. Thus, $\eta \leq 0$, which completes the proof. In the last statement I am assuming the law of the excluded middle, which, depending on your definition, makes the proof nonconstructive.

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  • $\begingroup$ The last paragraph I cannot follow. As far as I see, from $\eta=a-b$ and $b+\eta=a$ it does not follow that $\eta\le 0$. Take $a=2,b=1$ and $\eta=1$. $\endgroup$
    – Kurt G.
    Commented Sep 12, 2022 at 13:10
  • $\begingroup$ @KurtG. Note that I am assuming $a < b + \varepsilon$ for all $\varepsilon > 0$, which does not hold in your example. $\endgroup$ Commented Sep 12, 2022 at 13:17
  • $\begingroup$ Excluded middle (LEM) is precisely what constructive mathematics avoids (and the axiom of choice since that implies LEM). There is no issue with proof by contradiction in constructive mathematics—it is just the standard way that you prove a negation of a proposition. The issue is rather using double negation elimination, which is a consequence of LEM. $\endgroup$
    – ಠ_ಠ
    Commented Sep 12, 2022 at 19:46
  • $\begingroup$ @ಠ_ಠ I do not use LEM in the proof of the statement you wanted proven constructively, do I? $\endgroup$ Commented Sep 12, 2022 at 21:50
  • $\begingroup$ @Nirai Sorry, you are correct---I skimmed your answer and saw the use of LEM, but it was not for the statement I asked for. I've upvoted your answer now. $\endgroup$
    – ಠ_ಠ
    Commented Sep 12, 2022 at 23:10
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Given that $a_n\leq b_n$ for each $n$

Suppose $z_n=b_n-a_n\geq 0$

Let $\lim z_n<0$ if it diverges to $-\infty$ there are infintly points when $z_n<0$ which is not the case

In other case we will say if it converges to $l<0$ so we can pick some $\epsilon>0$ such that $l+\epsilon<0$ then $-\epsilon+l<z_n<\epsilon+l<0$ for $n\geq m$ again it is contradiction that $z_n<0$ so Hence

$\lim z_n\geq 0$ $\Rightarrow \lim a_n\leq \lim b_n$

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    $\begingroup$ Do you say that this proof is constructive? Welcome to Math Stack. $\endgroup$
    – 311411
    Commented Sep 12, 2022 at 17:38
  • $\begingroup$ Sorry, but this is not constructive. Constructive mathematics does not use the Law of Excluded Middle or the Axiom of Choice. $\endgroup$
    – ಠ_ಠ
    Commented Sep 12, 2022 at 19:46

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