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If you have a sequence $ a_n $ which is bounded and monontonic increasing the theorem tells us it converges to a limit $ L $

But having looked at the proof is it implicit that this limit $ L= Sup \left \{a_n : n \in \mathbb{N} \right \} $ ?

and does the proof also implicitly tell you if $ a_n $ is bounded and monontonic decreasing then $ L= Inf \left \{a_n : n \in \mathbb{N} \right\} $ ?

I.e asking to show that $ L= Sup/Inf \left \{a_n : n \in \mathbb{N} \right \} $ is the same as asking for proof of bounded and monontonic sequence theorem? (edit: given the sequence is bounded and monotone)

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  • $\begingroup$ Yes, you can prove it (in fact, I proved that theorem by showing that $\sup$/$\inf$ was the limit of the sequence) $\endgroup$ – CIJ Aug 7 '15 at 18:12
  • $\begingroup$ That answers my question then :) This proof of yours does not involve the approximation property does it? $\endgroup$ – Arcane1729 Aug 7 '15 at 18:20
  • $\begingroup$ You don't "show" that $L$ equals that sup. You define $L$ to be that sup, and then show that the limit equals $L$. $\endgroup$ – David C. Ullrich Aug 7 '15 at 18:21
  • $\begingroup$ Technically you need to be careful in your last statement in your question. It's possible that the limit is equal to either supremum or infimum of the sequence, even though the sequence is not monotonic (i.e. even never monotonic even a little bit, i.e. the sequence could keep alternating with each term between being increasing and decreasing). But the other direction (e.g. if sequence is increasing and bounded, then limit equals supremum) is true. $\endgroup$ – user2566092 Aug 7 '15 at 18:37
  • $\begingroup$ There was ambiguity in that statement- I couldnt be bothered to change it and was hoping everyone would assume the sequence was bounded and monotone given thats whats being discussed. $\endgroup$ – Arcane1729 Aug 8 '15 at 11:30

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