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I was wondering why the monotone convergence theorem (in the context of sequences) is not simply proven by contradiction? Why does the "official" proof include the argument with $\sup$ or $\inf$ respectively?

If you instead assume that a monotone sequence is not convergent then due to its monotonicity it must be unbounded which would be a contradiction. Hence it must be convergent.

Somehow I am sure that my reasoning must be flawed but I don't see where?

EDIT

If a sequence $\left(a_n\right)_{n\in\mathbb{N}}$ in the real numbers is not convergent it is divergent, hence for all $a\in\mathbb{R}~\exists \epsilon >0$ such that for all $n_0 \in\mathbb{N}~\exists N> n_0$ such that $|a_{N}-a|\geq\epsilon$. This means that either $\left(a_n\right)_{n\in\mathbb{N}}$ is unbounded or if it is bounded then the sequence can't be monotone and divergent at the same time. To see this, let's assume it were monotonously increasing, bounded and divergent. Then the absolute value $|a_n-a|$ for any $a$ is also bounded. This means that at some index we must find members $a_m>a_n$ where $m<n$ which contradicts increasing monotonicity or we reach an index $m$ where $a_m=a_{m+1}=a_{m+2}...$ for all subsequent members which contradicts divergence. Hence, if $\left(a_n\right)_{n\in\mathbb{N}}$ is not convergent it must be unbounded. (The case where $\left(a_n\right)_{n\in\mathbb{N}}$ is monotounously decreasing is handled in the same way)

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    $\begingroup$ “... then due to its monotonicity it must be unbounded ...” – why? Can you formulate this as a strict proof? That would make it easier to compare your approach with the “official” one. $\endgroup$ – Martin R Oct 12 '20 at 11:37
  • $\begingroup$ "If you instead assume that a monotone sequence is not convergent then due to its monotonicity it must be unbounded which would be a contradiction." this really amounts to proving the theorem by simply saying it's true. $\endgroup$ – David C. Ullrich Oct 12 '20 at 11:52
  • $\begingroup$ @DavidC.Ullrich, why do I assume the theorem to be true in this case? $\endgroup$ – Philipp Oct 12 '20 at 11:55
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    $\begingroup$ "If it's not convergent then it must be unbounded" is exactly what needs to be proved; you need to prove it, instead of just stating it. $\endgroup$ – David C. Ullrich Oct 12 '20 at 11:58
  • $\begingroup$ I would add that you should always prefer a direct proof than a proof by contradiction, if both are possible. $\endgroup$ – TheSilverDoe Oct 12 '20 at 14:16
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The proof of the monotone convergence theorem uses the least-upper-bound property of the real numbers:

Any non-empty set of real numbers that has an upper bound must have a least upper bound in real numbers.

This least upper bound is then called “supremum” of the set.

One cannot prove the monotone convergence theorem without using the existence of a supremum because it does not necessarily hold in ordered sets without the least-upper-bound property. As an example, consider the sequence $(x_n)$ in $\Bbb Q$ defined recursively as $$ x_0 = 0 \, , \, x_{n+1} = \frac{2x_n+2}{x_n+2} \, . $$ One can show that $(x_n)$ is increasing and bounded above. But the sequence is not convergent in $\Bbb Q$ because the existence of $L = \lim_{n \to \infty} x_n$ would imply that $L^2 = 2$, and there is no rational number $L$ with that property.

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  • $\begingroup$ I understand your point. I think that my sketch of a proof via contradiction indirectly makes use of the supremum-principle of the real numbers. However, I was not aware of this fact. If I elaborate my reasoning of that proof I always reach a point where I need to assume a supremum. $\endgroup$ – Philipp Oct 16 '20 at 10:29

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