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If a monotone increasing sequence is not convergent, then is the sequence unbounded?

I know that a monotone increasing sequence that is convergent must be bounded.

But $P \to Q$ being true doesn't necessarily mean $-P \to -Q$ must be true.

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  • $\begingroup$ This is the "completeness axiom" for the real numbers $\endgroup$ – Tim kinsella Feb 19 '17 at 22:37
  • $\begingroup$ A monotone increasing sequence is convergent if and only if it is bounded above $\endgroup$ – joeb Feb 19 '17 at 22:40
  • $\begingroup$ @joeb And below too $\ddot \smile$ $\endgroup$ – Git Gud Feb 19 '17 at 22:40
  • $\begingroup$ @user3000482 What is $P$ and what is $Q$ to you? $\endgroup$ – Git Gud Feb 19 '17 at 22:41
  • $\begingroup$ Consider rational numbers $1, \frac75, \frac{41}{29}, \ldots$ where if a term is $q$ the next is $\frac{4+3q}{3+2q}$. This is monotone increasing and is clearly bounded above by $2$, but does not converge to a rational number $\endgroup$ – Henry Feb 19 '17 at 22:52
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Let $(a_n)$ an increasing and divergent sequence of real numbers.

If we suppose that this sequence has an upper bound, being increasing and having an upper bound, it would converge to some real limit : a contradiction. So this sequence doesn't have any upper bound (and in particular, it's unbounded).

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